In order to solve Eq. (1), we use the following wave transformation [11–16]
$$\begin{aligned} q(x,t)=U(\xi )\hbox {e}^{i\Phi \left( x,t\right) } \end{aligned}$$
(2)
where \(U(\xi )\) represents the shape of the pulse and
$$\begin{aligned}&\xi =x-vt,\end{aligned}$$
(3)
$$\begin{aligned}&\Phi \left( x,t\right) =-\kappa x+\omega t+\theta . \end{aligned}$$
(4)
In Eq. (2), the function \(\Phi \left( x,t\right) \) is the phase component of the soliton. Then, in Eq. (4), \(\kappa \) is the soliton frequency, while \(\omega \) is the wave number of the soliton and \(\theta \) is the phase constant. Finally in Eq. (3), v is the velocity of the soliton. After substituting (2) in (1) and decomposing into real and imaginary parts lead to
$$\begin{aligned}&(a-bv+3\kappa \gamma )U''\nonumber \\&\quad -\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa +\gamma \kappa ^{3}\right) U +cF(U^2)U\nonumber \\&\quad +\left( s\kappa +\theta \kappa -\theta _{1}\kappa ^{2}-\theta _{2}\kappa ^{2}-\theta _{3}\kappa ^{2}\right) U^{3}\nonumber \\&\quad +\,6\theta _{1}U(U')^2+(3\theta _{1}+\theta _{2}+\theta _{3})U^{2}U''=0, \end{aligned}$$
(5)
and
$$\begin{aligned}&(-v-2a\kappa +b\omega +b\kappa v+\lambda -3\gamma \kappa ^{2})U'\nonumber \\&\quad +\,(3s+2\mu +\theta -2\kappa (3\theta _{1}+\theta _{2}-\theta _{3}))U^2U'\nonumber \\&\quad +\,\gamma U'''=0. \end{aligned}$$
(6)
The imaginary part Eq. (6) implies the relations
$$\begin{aligned}&v=-\frac{2a\kappa -b\omega -\lambda }{1-b\kappa },\end{aligned}$$
(7)
$$\begin{aligned}&\gamma =0,\end{aligned}$$
(8)
$$\begin{aligned}&3s+2\mu +\theta -2\kappa (3\theta _{1}+\theta _{2}-\theta _{3})=0, \end{aligned}$$
(9)
and
$$\begin{aligned}&(a-bv)U''-\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) U\nonumber \\&\quad +\,cF(U^2)U+\left( s\kappa +\theta \kappa -\kappa ^2\theta _{1}-\kappa ^2\theta _{2}-\kappa ^2\theta _{3}\right) U^{3}\nonumber \\&\quad +\left( 3\theta _{1}+\theta _{2}+\theta _{3}\right) U^{2}U'' +6\theta _{1}U\left( U'\right) ^{2}=0. \end{aligned}$$
(10)
To obtain the analytic solution, the transformations \(\theta _{1} =0, \theta _{2}=-\theta _{3}\) and \(s=-\theta \) are applied in Eq. (10) and give
$$\begin{aligned}&(a-bv)U''-\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) U\nonumber \\&\quad +\,cF(U^2)U=0, \end{aligned}$$
(11)
where
$$\begin{aligned} \kappa = \frac{\theta -\mu }{2\theta _{3}}. \end{aligned}$$
(12)
2.1 Kerr law
For Kerr law nonlinearity
$$\begin{aligned} F(u)=u, \end{aligned}$$
(13)
so that (1) reduces to
$$\begin{aligned}&iq_{t}+aq_{xx}+bq_{xt}+c|q|^{2}q=\nonumber \\&\quad -\,i\lambda q_x -is\left( |q|^{2}q\right) _{x}-i\mu \left( |q|^{2}\right) _{x}q-i\theta |q|^{2}q_{x}\nonumber \\&\quad -\,i\gamma q_{xxx}-\theta _{1}\left( |q|^{2}q\right) _{xx}-\theta _{2}|q|^{2}q_{xx}\qquad \nonumber \\&-\,\theta _{3}q^{2}q^{*}_{xx}, \end{aligned}$$
(14)
and Eq. (11) simplifies to
$$\begin{aligned}&(a-bv)U''-\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) \nonumber \\&\quad U+cU^{3}=0. \end{aligned}$$
(15)
In this subsection, we would like to extend the extended trial equation method [6–10] to solve the NLSE with Kerr law nonlinearity. Suppose that the solution of Eq. (15) can be given by
$$\begin{aligned} U=\sum _{i=0}^{\varsigma }\tau _i\Psi ^{i}, \end{aligned}$$
(16)
where
$$\begin{aligned} (\Psi ')^2= & {} \Lambda (\Psi )=\frac{\Phi (\Psi )}{\Upsilon (\Psi )}\nonumber \\= & {} \frac{\mu _\sigma \Psi ^\sigma +\cdot \cdot \cdot +\mu _1\Psi +\mu _0}{\chi _\rho \Psi ^\rho +\cdot \cdot \cdot +\chi _1\Psi +\chi _0}. \end{aligned}$$
(17)
Using the relations (16) and (17), we can derive the terms \((U')^2\) and \(U''\) as
$$\begin{aligned} (U')^2=\frac{\Phi (\Psi )}{\Upsilon (\Psi )}\left( \sum _{i=0}^{\varsigma }i\tau _i\Psi ^{i-1}\right) ^2, \end{aligned}$$
(18)
and
$$\begin{aligned} U''= & {} \frac{\Phi '(\Psi )\Upsilon (\Psi )-\Phi (\Psi )\Upsilon '(\Psi )}{2\Upsilon ^2(\Psi )} \left( \sum _{i=0}^{\varsigma }i\tau _i\Psi ^{i-1}\right) \nonumber \\&+\frac{\Phi (\Psi )}{\Upsilon (\Psi )}\left( \sum _{i=0}^{\varsigma }i(i-1)\tau _i\Psi ^{i-2}\right) , \end{aligned}$$
(19)
where \(\Phi (\Psi )\) and \(\Upsilon (\Psi )\) are polynomials of \(\Psi \). We can reduce Eq. (17) to the elementary integral form as follows:
$$\begin{aligned} \pm (\xi -\xi _0)=\int \frac{\hbox {d}\Psi }{\sqrt{\Lambda (\Psi )}}=\int \sqrt{\frac{\Upsilon (\Psi )}{\Phi (\Psi )}}\hbox {d}\Psi . \end{aligned}$$
(20)
Substituting Eqs. (16) and (19) in Eq. (15), and using the balance principle, we determine a relation of \(\sigma , \rho \) and \(\varsigma \) as
$$\begin{aligned} \sigma =\rho +2\varsigma +2. \end{aligned}$$
(21)
If we take \(\sigma =4, \rho =0\) and \(\varsigma =1\) in Eq. (21), then
$$\begin{aligned}&U=\tau _0+\tau _1\Psi ,\quad \end{aligned}$$
(22)
$$\begin{aligned}&U''=\frac{\tau _1(4\mu _4\Psi ^3+3\mu _3\Psi ^2+2\mu _2\Psi +\mu _1)}{2\chi _0}, \end{aligned}$$
(23)
where \(\mu _4\ne 0, \chi _0\ne 0.\) Substituting Eqs. (22) and (23) in Eq. (15), collecting the coefficients of \(\Psi \) and solving the resulting system, we have
$$\begin{aligned} \mu _0= & {} \mu _0,\quad \mu _1=\mu _1, \quad \mu _2= \displaystyle \frac{\mu _1 \tau _1}{2 \tau _0}-\frac{2 c \tau _0^2 \chi _0}{a-bv} , \nonumber \\ \mu _3= & {} \displaystyle -\frac{2 c \tau _0 \tau _1\chi _0}{a-b v} , \quad \mu _4= \displaystyle -\frac{c \tau _1^2 \chi _0}{2(a-b v)} ,\nonumber \\ \chi _0= & {} \chi _0,\quad \tau _0=\tau _0,\quad \tau _1=\tau _1, \nonumber \\ \omega= & {} \displaystyle \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} .\nonumber \\ \end{aligned}$$
(24)
Substituting the solution set (24) in Eqs. (17) and (20), we find that
$$\begin{aligned} \pm (\xi -\xi _0)= W \int {\frac{\hbox {d}\Psi }{\sqrt{\Lambda (\Psi )}}}, \end{aligned}$$
(25)
where
$$\begin{aligned}&\Lambda (\Psi )=\Psi ^4+\frac{\mu _3}{\mu _4}\Psi ^3+\frac{\mu _2}{\mu _4}\Psi ^2+\frac{\mu _1}{\mu _4}\Psi +\frac{\mu _0}{\mu _4},\nonumber \\&\quad W=\sqrt{\frac{\chi _0}{\mu _4}}. \end{aligned}$$
(26)
Integrating Eq. (25), and inserting the result in Eq. (22), we obtain the exact solutions to Eq. (15). Consequently, we have the traveling wave solutions to the NLSE with Kerr law nonlinearity (14) as the following:
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^4\), we obtain
$$\begin{aligned} q(x,t)= & {} \left\{ \displaystyle \tau _0+\tau _1\lambda _1\pm \frac{\tau _1W}{x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t -\xi _{0}}\right\} \nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } .\nonumber \\ \end{aligned}$$
(27)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^3(\Psi -\lambda _2)\) and \(\lambda _2>\lambda _1\), we get
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _1+\frac{4W^2(\lambda _2-\lambda _1)\tau _1}{4W^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t -\xi _{0}\right) \right] ^2}\right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } . \end{aligned}$$
(28)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)^2\), we have
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _2+\frac{(\lambda _2-\lambda _1)\tau _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t -\xi _{0}\right) \right] -1}\right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(29)
and
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _1+\frac{(\lambda _1-\lambda _2)\tau _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right] -1}\right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } . \end{aligned}$$
(30)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)(\Psi -\lambda _3)\) and \(\lambda _1>\lambda _2>\lambda _3\), we attain
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _1-\frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)\tau _1}{2\lambda _1-\lambda _2-\lambda _3+(\lambda _3-\lambda _2) \cosh \left( \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W} \left[ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right] \right) }\right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } . \end{aligned}$$
(31)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)(\Psi -\lambda _2)(\Psi -\lambda _3)(\Psi -\lambda _4)\) and \(\lambda _1>\lambda _2>\lambda _3>\lambda _4\), we find
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _2+\frac{\tau _1(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _4-\lambda _2+(\lambda _1-\lambda _4) {{\mathrm{sn}}}^2\left[ \pm \frac{\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W} \left[ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t -\xi _{0} \right] , l \right] }\right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(32)
where
$$\begin{aligned} l^2=\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}. \end{aligned}$$
(33)
Also, \(\lambda _i (i=1,\ldots ,4)\) are the roots of the polynomial equation
$$\begin{aligned} \Lambda (\Psi )=0. \end{aligned}$$
(34)
When \(\tau _0=-\tau _1\lambda _1\) and \(\xi _0=0\), we can reduce the solutions (27)–(31) to plane wave solutions
$$\begin{aligned}&q(x,t) = \left\{ \pm \displaystyle \frac{\tau _1W}{ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t }\right\} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } ,\end{aligned}$$
(35)
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left\{ \frac{4W^2(\lambda _2-\lambda _1)\tau _1}{4W^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] ^2}\right\} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(36)
singular soliton solutions
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \frac{(\lambda _2-\lambda _1)\tau _1}{2}\left\{ 1\mp \coth \left[ \frac{\lambda _1-\lambda _2}{2W}\right. \right. \nonumber \\&\left. \left. \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right\} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(37)
and bright soliton solution
$$\begin{aligned}&q(x,t) = \left\{ \frac{A}{C+\cosh \left[ B\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] } \right\} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(38)
where
$$\begin{aligned} A= & {} \frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)\tau _1}{\lambda _3-\lambda _2},\nonumber \\ B= & {} \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W},\nonumber \\ C= & {} \frac{2\lambda _1-\lambda _2-\lambda _3}{\lambda _3-\lambda _2}. \end{aligned}$$
(39)
Here, A is the amplitude of the soliton, while B is the inverse width of the soliton. These solitons exist for \(\tau _1<0\). Furthermore, when \(\tau _0=-\tau _1\lambda _2\) and \(\xi _0=0\), we can write the Jacobi elliptic function solution (32) as
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left\{ \frac{A_1}{C_1+ {{\mathrm{sn}}}^2\left[ B_j \left[ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right] ,\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)} \right] } \right\} \nonumber \\&\qquad \times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(40)
where
$$\begin{aligned} A_1= & {} \frac{\tau _1(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _1-\lambda _4}, \quad C_1= \frac{\lambda _4-\lambda _2}{\lambda _1-\lambda _4}, \nonumber \\ B_j= & {} \frac{(-1)^j\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W},\ \ (j=1,2). \end{aligned}$$
(41)
Remark 1
When the modulus \(l\rightarrow 1\), a second form of singular optical soliton solutions fall out:
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_1}{C_1+ \tanh ^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] }\right\} \nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } ,\nonumber \\ \end{aligned}$$
(42)
where \(\lambda _3=\lambda _4\).
Remark 2
However, if \(l\rightarrow 0\), periodic singular solutions are listed as follows:
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_1}{C_1+ \sin ^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] } \right\} \nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _1 \tau _1 (bv-a)-2 \tau _0 \chi _0 \left[ \kappa (\lambda -a \kappa )+c \tau _0^2\right] }{2 \tau _0 \chi _0 (b\kappa -1)} \right) t + \theta \right\} } ,\nonumber \\ \end{aligned}$$
(43)
where \(\lambda _2=\lambda _3\).
2.2 Power law
In this case,
$$\begin{aligned} F(u)=u^{n}, \end{aligned}$$
(44)
for power law nonlinear medium. Therefore, (1) takes the form
$$\begin{aligned}&iq_{t}+aq_{xx}+bq_{xt}+c|q|^{2n}q=\nonumber \\&\quad -i\lambda q_x -is\left( |q|^{2}q\right) _{x}-i\mu \left( |q|^{2}\right) _{x}q\nonumber \\&\quad -i\theta |q|^{2}q_{x}-i\gamma q_{xxx}-\theta _{1}\left( |q|^{2}q\right) _{xx}\nonumber \\&\quad -\theta _{2}|q|^{2}q_{xx}-\theta _{3}q^{2}q^{*}_{xx} . \end{aligned}$$
(45)
In this case, Eq. (11) simplifies to
$$\begin{aligned}&(a-bv)U''-\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) U\nonumber \\&\quad +\,cU^{2n+1}=0. \end{aligned}$$
(46)
Balancing \(U''\) with \(U^{2n+1}\) in Eq. (46) gives \(N=\frac{1}{n}.\) In order to obtain closed-form solutions, we use the transformation
$$\begin{aligned} U=V^{\frac{1}{2n}}, \end{aligned}$$
(47)
that will reduce Eq. (46) to the ODE
$$\begin{aligned}&(a-bv)\left( (1-2n)(V')^{2}+2nVV''\right) \nonumber \\&\quad -4n^{2}\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) V^2\nonumber \\&\quad +\,4cn^2V^3=0. \end{aligned}$$
(48)
In this subsection, we will utilize the extended trial equation method to construct the exact solutions to the NLSE with power law nonlinearity. Substituting Eqs. (16), (18) and (19) in Eq. (48), and using the balance principle, we determine a relation of \(\sigma , \rho \) and \(\varsigma \) as
$$\begin{aligned} \sigma =\rho +\varsigma +2. \end{aligned}$$
(49)
Case 1: If we take \(\sigma =3, \rho =0\) and \(\varsigma =1\) in Eq. (49), then
$$\begin{aligned}&V=\tau _0+\tau _1\Psi ,\end{aligned}$$
(50)
$$\begin{aligned}&(V')^2=\frac{\tau _1^2(\mu _3\Psi ^3+\mu _2\Psi ^2+\mu _1\Psi +\mu _0)}{\chi _0},\end{aligned}$$
(51)
$$\begin{aligned}&V''=\frac{\tau _1(3\mu _3\Psi ^2+2\mu _2\Psi +\mu _1)}{2\chi _0}, \end{aligned}$$
(52)
where \(\mu _3\ne 0, \chi _0\ne 0.\) Substituting Eqs. (50)–(52) in Eq. (48), and solving the resulting system of algebraic equations we have
$$\begin{aligned}&\mu _1= \displaystyle \frac{2 \mu _0 \tau _1}{\tau _0}-\frac{4 c n^2 \tau _0^2 \chi _0}{\tau _1 (1+n) (a-b v)} , \nonumber \\&\mu _2= \displaystyle \frac{\mu _0 \tau _1^2}{\tau _0^2}-\frac{8 c n^2 \tau _0 \chi _0}{(1+n) (a-b v)} ,\nonumber \\&\mu _3= \displaystyle -\frac{4 c n^2 \tau _1 \chi _0}{(1+n) (a-b v)} , \nonumber \\&\omega = \displaystyle \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} ,\nonumber \\&\mu _0 = \mu _0,\quad \chi _0=\chi _0,\quad \tau _0=\tau _0,\quad \tau _1=\tau _1. \end{aligned}$$
(53)
Substituting the solution set (53) in Eqs. (17) and (20), we find that
$$\begin{aligned} \pm (\xi -\xi _0)=\sqrt{W_1} \int {\frac{\hbox {d}\Psi }{\sqrt{ \Lambda (\Psi )}}}, \end{aligned}$$
(54)
where
$$\begin{aligned} \Lambda (\Psi )=\Psi ^3+\frac{\mu _2}{\mu _3}\Psi ^2+\frac{\mu _1}{\mu _3}\Psi +\frac{\mu _0}{\mu _3},\quad W_1= \displaystyle \frac{\chi _0}{\mu _3} . \end{aligned}$$
(55)
Consequently, we have the traveling wave solutions to the NLSE with power law nonlinearity (45) as the following:
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^3\), we find rational function solution as
$$\begin{aligned}&q(x,t)=\displaystyle \left\{ \tau _0+\tau _1\lambda _1+\displaystyle \frac{4\tau _1W_1}{\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) ^2}\right\} ^{\frac{1}{2n}} \nonumber \\&\quad \ \ \times \, \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} }.\nonumber \\ \end{aligned}$$
(56)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)\) and \(\lambda _2>\lambda _1\), we obtain solitary wave solution as
$$\begin{aligned}&q(x,t)=\left\{ \tau _0+\tau _1\lambda _2+ \tau _1(\lambda _1-\lambda _2)\right. \nonumber \\&\quad \left. \times \,\tanh ^2\left( \frac{1}{2}\sqrt{\frac{\lambda _1-\lambda _2}{W_1}} \displaystyle \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right) \right\} ^{\frac{1}{2n}} \nonumber \\&\quad \times \, \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} }.\nonumber \\ \end{aligned}$$
(57)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)(\Psi -\lambda _2)^2\) and \(\lambda _1>\lambda _2\), we have hyperbolic function solution as
$$\begin{aligned}&q(x,t)=\left\{ \tau _0+\tau _1\lambda _1+ \tau _1(\lambda _1-\lambda _2)\right. \nonumber \\&\quad \left. \times \,\text {cosech}^2\left[ \frac{1}{2}\sqrt{\frac{\lambda _1-\lambda _2}{W_1}} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right\} ^{\frac{1}{2n}} \nonumber \\&\quad \times \, \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(58)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)(\Psi -\lambda _2)(\Psi -\lambda _3)\) and \(\lambda _1>\lambda _2>\lambda _3\), we get Jacobi elliptic function solutions as
$$\begin{aligned}&q(x,t)=\left\{ \tau _0+\tau _1\lambda _3+ \tau _1(\lambda _2-\lambda _3)\right. \nonumber \\&\left. \quad \times \,{{\mathrm{sn}}}^2\left( \mp \frac{1}{2}\sqrt{\frac{\lambda _1-\lambda _3}{W_1}} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) ,l\right) \right\} ^{\frac{1}{2n}} \nonumber \\&\quad \times \, \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(59)
where
$$\begin{aligned} l^2=\frac{\lambda _2-\lambda _3}{\lambda _1-\lambda _3}. \end{aligned}$$
(60)
Also, \(\lambda _i\,(i=1,2,3)\) are the roots of the polynomial equation
$$\begin{aligned} \Lambda (\Psi )=0. \end{aligned}$$
(61)
When \(\tau _0=-\tau _1\lambda _1\) and \(\xi _0=0\), we can reduce the solutions (56)–(58) to plane wave solution
$$\begin{aligned}&q(x,t)=\left\{ \displaystyle \frac{\widetilde{A}}{x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t}\right\} ^{\frac{1}{n}}\nonumber \\&\quad \times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(62)
1-soliton solution
$$\begin{aligned}&q(x,t) =\left\{ \frac{A_2}{\cosh ^{\frac{1}{n}}\left[ B_3\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] }\right\} \nonumber \\&\quad \times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(63)
and singular soliton solution
$$\begin{aligned}&q(x,t) =\left\{ \frac{{A_3}}{\sinh ^{\frac{1}{n}}\left[ B_3\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] }\right\} \nonumber \\&\quad \times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(64)
where
$$\begin{aligned}&\displaystyle {\widetilde{A}}=2\sqrt{\tau _1W_1},\quad A_2=[\tau _1(\lambda _2-\lambda _1)]^{\frac{1}{2n}},\nonumber \\&A_3= [\tau _1(\lambda _1-\lambda _2)]^{\frac{1}{2n}}, \quad B_3=\frac{1}{2}\sqrt{\frac{\lambda _1-\lambda _2}{W_1}}. \end{aligned}$$
(65)
Here, \(A_2\) and \(A_3\) are respectively the amplitudes of 1-soliton and singular soliton, while \(B_3\) is the inverse width of the solitons. These solitons exist for \(\tau _1>0\). Furthermore, when \(\tau _0=-\tau _1\lambda _3\) and \(\xi _0=0\), we can simplify the solution (59) as follows:
$$\begin{aligned}&\displaystyle q(x,t)=A_4{{\mathrm{sn}}}^{\frac{1}{n}}\left[ B_j\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) ,\frac{\lambda _2-\lambda _3}{\lambda _1-\lambda _3}\right] \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(66)
where
$$\begin{aligned} A_4= & {} [\tau _1(\lambda _2-\lambda _3)]^{\frac{1}{2n}}, \nonumber \\ B_j= & {} \frac{(-1)^j}{2}\sqrt{\frac{\lambda _1-\lambda _3}{W_1}},\ \ (j=4,5). \end{aligned}$$
(67)
Remark 3
When the modulus \(l\rightarrow 1\), dark soliton solutions fall out:
$$\begin{aligned}&\displaystyle q(x,t)=A_4\tanh ^{\frac{1}{n}}\left[ B_j\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (a-b v)+4 n^2 \tau _0^2 \chi _0\left[ \kappa (1+n) (a \kappa -\lambda ) -c \tau _0\right] }{4 n^2 \tau _0^2\chi _0 (1+n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(68)
where \(\lambda _1=\lambda _2\).
Case 2: If we take \(\sigma =4, \rho =0\) and \(\varsigma =2\) in Eq. (49), then
$$\begin{aligned}&V=\tau _0+\tau _1\Psi +\tau _2\Psi ^2,\end{aligned}$$
(69)
$$\begin{aligned}&(V')^2=\frac{(\tau _1+2\tau _2\Psi )^2(\mu _4\Psi ^4+\mu _3\Psi ^3+\mu _2\Psi ^2+\mu _1\Psi +\mu _0)}{\chi _0},\nonumber \\\end{aligned}$$
(70)
$$\begin{aligned}&V''=\frac{(\tau _1+2\tau _2\Psi )(4\mu _4\Psi ^3+3\mu _3\Psi ^2+2\mu _2\Psi +\mu _1)}{2\chi _0}\nonumber \\&\qquad \qquad + \frac{2\tau _2(\mu _4\Psi ^4+\mu _3\Psi ^3+\mu _2\Psi ^2+\mu _1\Psi +\mu _0)}{\chi _0}, \end{aligned}$$
(71)
where \(\mu _4\ne 0, \chi _0\ne 0.\) Substituting Eqs. (69)–(71) in Eq. (48), and solving the resulting system of algebraic equations, we have
$$\begin{aligned} \mu _0= & {} \displaystyle -\frac{c n^2 \tau _0^2 \chi _0}{\tau _2 (1+n) (a-b v)} , \nonumber \\ \mu _1= & {} \displaystyle -\frac{2 c n^2 \tau _0 \tau _1 \chi _0}{\tau _2 (1+n) (a-b v)} , \nonumber \\ \mu _2= & {} \displaystyle -\frac{c n^2 \chi _0 \left( \tau _1^2+2 \tau _0\tau _2\right) }{\tau _2 (1+n) (a-b v)} ,\nonumber \\ \mu _3= & {} \displaystyle -\frac{2 c n^2 \tau _1 \chi _0}{(1+n)(a-b v)}, \quad \mu _4= \displaystyle -\frac{c n^2 \tau _2 \chi _0}{(1+n)(a-b v)}, \nonumber \\ \omega= & {} \displaystyle \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} ,\nonumber \\ \chi _0= & {} \chi _0,\quad \tau _0=\tau _0,\quad \tau _1=\tau _1,\quad \tau _2=\tau _2. \end{aligned}$$
(72)
Substituting the solution set (72) in Eqs. (17) and (20), we find that
$$\begin{aligned} \pm (\xi -\xi _0)= W_2 \int {\frac{\hbox {d}\Psi }{\sqrt{\Lambda (\Psi )}}}, \end{aligned}$$
(73)
where
$$\begin{aligned} \Lambda (\Psi )= & {} \Psi ^4+\frac{\mu _3}{\mu _4}\Psi ^3+\frac{\mu _2}{\mu _4}\Psi ^2+\frac{\mu _1}{\mu _4}\Psi +\frac{\mu _0}{\mu _4},\nonumber \\ W_2= & {} \sqrt{\frac{\chi _0}{\mu _4}}. \end{aligned}$$
(74)
Consequently, taking \(\xi _0=0\), we have the traveling wave solutions to the NLSE with power law nonlinearity (45) as the following:
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^4\), we obtain
$$\begin{aligned} q(x,t)= & {} \left[ \sum _{i=0}^{2}\tau _i\left( \lambda _1\pm \displaystyle \frac{W_2}{ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t }\right) ^i \right] ^{\frac{1}{2n}}\nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(75)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^3(\Psi -\lambda _2)\) and \(\lambda _2>\lambda _1\), we get
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left[ \sum _{i=0}^{2}\tau _i\left( \lambda _1+\frac{4W_2^2(\lambda _2-\lambda _1)}{4W_2^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] ^2} \right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(76)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)^2\), we have
$$\begin{aligned}&q(x,t)\nonumber \\&\quad =\left[ \sum _{i=0}^{2}\tau _i\left( \lambda _2+\frac{\lambda _2-\lambda _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W_2} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) \right] -1} \right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(77)
and
$$\begin{aligned}&q(x,t)\nonumber \\&\quad =\left[ \sum _{i=0}^{2}\tau _i\left( \lambda _1+\frac{\lambda _1-\lambda _2}{\exp \left[ \frac{\lambda _1-\lambda _2}{W_2}\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) \right] -1}\right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\qquad \times \, \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(78)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)(\Psi -\lambda _3)\) and \(\lambda _1>\lambda _2>\lambda _3\), we attain
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left[ \sum _{i=0}^{2}\tau _i\left( \lambda _1-\frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}{2\lambda _1-\lambda _2-\lambda _3+(\lambda _3-\lambda _2)\cosh \left( \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W_2} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) \right) } \right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\qquad \quad \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(79)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)(\Psi -\lambda _2)(\Psi -\lambda _3)(\Psi -\lambda _4)\) and \(\lambda _1>\lambda _2>\lambda _3>\lambda _4\), we achieve
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left[ \sum _{i=0}^{2}\tau _i\left( \lambda _2+\frac{(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _4-\lambda _2+(\lambda _1-\lambda _4){{\mathrm{sn}}}^2\left[ \pm \frac{\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_2} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) , l \right] } \right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\qquad \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(80)
where
$$\begin{aligned} l^2=\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}. \end{aligned}$$
(81)
Also, \(\lambda _i\, (i=1,\ldots ,4)\) are the roots of the polynomial equation
$$\begin{aligned} \Lambda (\Psi )=0. \end{aligned}$$
(82)
Remark 4
When the modulus \(l\rightarrow 1\), we write the Jacobi elliptic function solutions (80) as
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left[ \sum _{i=0}^{2}\tau _i\left( \lambda _2+\frac{(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _4-\lambda _2+(\lambda _1-\lambda _4)\tanh ^2\left[ \pm \frac{\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_2} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) \right] } \right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\qquad \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(83)
where \(\lambda _3=\lambda _4\).
Remark 5
When the modulus \(l\rightarrow 0\), we write the Jacobi elliptic function solutions (80) as
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left[ \sum _{i=0}^{2}\tau _i\left( \lambda _2+\frac{(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _4-\lambda _2+(\lambda _1-\lambda _4)\sin ^2\left[ \pm \frac{\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_2} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) \right] } \right) ^i\right] ^{\frac{1}{2n}} \nonumber \\&\qquad \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{c \tau _1^2+4 \tau _2 \left[ \kappa (1+n) (a\kappa -\lambda )-c \tau _0\right] }{4\tau _2 (1+n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(84)
where \(\lambda _2=\lambda _3\).
2.3 Parabolic law
In this case,
$$\begin{aligned} F(u)=u+\eta u^{2}, \end{aligned}$$
(85)
where \(\eta \) is a real-valued constant. Therefore, Eq. (1) takes the form
$$\begin{aligned}&iq_{t}+aq_{xx}+bq_{xt}+c(|q|^{2}+\eta |q|^{4})q=\nonumber \\&\quad -\,i\lambda q_x -is\left( |q|^{2}q\right) _{x}-i\mu \left( |q|^{2}\right) _{x}q-i\theta |q|^{2}q_{x}\nonumber \\&\quad -\,i\gamma q_{xxx}-\theta _{1}\left( |q|^{2}q\right) _{xx}-\theta _{2}|q|^{2}q_{xx}-\theta _{3}q^{2}q^{*}_{xx} .\nonumber \\ \end{aligned}$$
(86)
In this case, Eq. (11) simplifies to
$$\begin{aligned}&(a-bv)U''-\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) U\nonumber \\&\quad +\,cU^{3}+c\eta U^{5}=0. \end{aligned}$$
(87)
Balancing \(U''\) with \(U^{5}\) gives \(N=\frac{1}{2}.\) In order to obtain closed-form solution, we use the transformation
$$\begin{aligned} U=V^{\frac{1}{2}}, \end{aligned}$$
(88)
that will reduce Eq. (87) to the ODE
$$\begin{aligned}&(a-bv)\left( 2VV''-(V')^{2}\right) \nonumber \\&\quad -\,4\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) V^{2} +4cV^{3}\nonumber \\&\quad +\,4c\eta V^{4}=0. \end{aligned}$$
(89)
In this subsection, we will implement the extended trial equation method to obtain the exact solutions of the NLSE with parabolic law nonlinearity. Substituting Eqs. (16), (18) and (19) in Eq. (89), and using the balance principle, we determine a relation of \(\sigma , \rho \) and \(\varsigma \) as
$$\begin{aligned} \sigma =\rho +2\varsigma +2. \end{aligned}$$
(90)
If we take \(\sigma =4, \rho =0\) and \(\varsigma =1\) in Eq. (90), then
$$\begin{aligned}&V=\tau _0+\tau _1\Psi ,\end{aligned}$$
(91)
$$\begin{aligned}&(V')^2=\frac{\tau _1^2(\mu _4\Psi ^4+\mu _3\Psi ^3+\mu _2\Psi ^2+\mu _1\Psi +\mu _0)}{\chi _0},\nonumber \\ \end{aligned}$$
(92)
$$\begin{aligned}&V''=\frac{\tau _1(4\mu _4\Psi ^3+3\mu _3\Psi ^2+2\mu _2\Psi +\mu _1)}{2\chi _0}, \end{aligned}$$
(93)
where \(\mu _4\ne 0, \chi _0\ne 0.\) Substituting Eqs. (91)–(93) in Eq. (89), collecting the coefficients of \(\Psi \), and solving the resulting system, we have
$$\begin{aligned} \mu _0= & {} \displaystyle \frac{\tau _0}{\tau _1^2}\left( -\mu _2 \tau _0+\mu _1 \tau _1-\frac{2c \tau _0^2 \chi _0 \left( 1+2 \eta \tau _0\right) }{a-b v}\right) , \nonumber \\ \mu _3= & {} \displaystyle -\frac{2 c \tau _1 \chi _0\left( 3+8 \eta \tau _0\right) }{3(a-b v)} ,\nonumber \\ \mu _4= & {} \displaystyle -\frac{4 c \eta \tau _1^2\chi _0}{3 (a-b v)} , \nonumber \\ \omega= & {} \displaystyle \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} ,\nonumber \\ \mu _1= & {} \mu _1,\quad \mu _2=\mu _2,\quad \chi _0=\chi _0,\quad \tau _0=\tau _0,\nonumber \\ \tau _1= & {} \tau _1. \end{aligned}$$
(94)
Substituting the solution set (94) in Eqs. (17) and (20), we find that
$$\begin{aligned} \pm (\xi -\xi _0)= W_3 \int {\frac{\hbox {d}\Psi }{\sqrt{\Lambda (\Psi )}}}, \end{aligned}$$
(95)
where
$$\begin{aligned} \Lambda (\Psi )= & {} \Psi ^4+\frac{\mu _3}{\mu _4}\Psi ^3+\frac{\mu _2}{\mu _4}\Psi ^2+\frac{\mu _1}{\mu _4}\Psi +\frac{\mu _0}{\mu _4},\nonumber \\ W_3= & {} \sqrt{\frac{\chi _0}{\mu _4}}. \end{aligned}$$
(96)
Consequently, we obtain the traveling wave solutions to the NLSE with parabolic law nonlinearity (86) as the following:
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^4\), we have
$$\begin{aligned} q(x,t)= & {} \left\{ \displaystyle \tau _0+\tau _1\lambda _1\pm \frac{\tau _1W_3}{ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}}\right\} ^{\frac{1}{2}}\nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }.\nonumber \\ \end{aligned}$$
(97)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^3(\Psi -\lambda _2)\) and \(\lambda _2>\lambda _1\), we find
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left\{ \tau _0+\tau _1\lambda _1+\frac{4W_3^2(\lambda _2-\lambda _1)\tau _1}{4W_3^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t -\xi _{0}\right) \right] ^2}\right\} ^{\frac{1}{2}} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(98)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)^2\), we get
$$\begin{aligned}&q(x,t)\nonumber \\&\quad =\left\{ \tau _0+\tau _1\lambda _2+\frac{(\lambda _2-\lambda _1)\tau _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W_3} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right] -1}\right\} ^{\frac{1}{2}} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(99)
and
$$\begin{aligned}&q(x,t)\nonumber \\&\quad =\left\{ \tau _0+\tau _1\lambda _1+\frac{(\lambda _1-\lambda _2)\tau _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W_3} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right] -1}\right\} ^{\frac{1}{2}} \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(100)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)(\Psi -\lambda _3)\) and \(\lambda _1>\lambda _2>\lambda _3\), we attain
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left\{ \tau _0+\tau _1\lambda _1-\frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)\tau _1}{2\lambda _1-\lambda _2-\lambda _3+(\lambda _3-\lambda _2) \cosh \left( \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W_3} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t\right) \right) }\right\} ^{\frac{1}{2}} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(101)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)(\Psi -\lambda _2)(\Psi -\lambda _3)(\Psi -\lambda _4)\) and \(\lambda _1>\lambda _2>\lambda _3>\lambda _4\), we obtain
$$\begin{aligned}&q(x,t)\nonumber \\&\quad =\left\{ \tau _0+\tau _1\lambda _2+\frac{\tau _1(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _4-\lambda _2+(\lambda _1-\lambda _4){{\mathrm{sn}}}^2\left[ \pm \frac{\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_3} \left[ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t -\xi _{0} \right] , l \right] }\right\} ^{\frac{1}{2}} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(102)
where
$$\begin{aligned} l^2=\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}. \end{aligned}$$
(103)
Also, \(\lambda _i\,(i=1,\ldots ,4)\) are the roots of the polynomial equation
$$\begin{aligned} \Lambda (\Psi )=0. \end{aligned}$$
(104)
When \(\tau _0=-\tau _1\lambda _1\) and \(\xi _0=0\), we can reduce the solutions (97)–(101) to rational function solutions
$$\begin{aligned}&q(x,t) = \left\{ \pm \displaystyle \frac{\tau _1W_3}{ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t }\right\} ^{\frac{1}{2}}\nonumber \\&\quad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} },\end{aligned}$$
(105)
$$\begin{aligned}&q(x,t) = \left\{ \frac{4W_3^2(\lambda _2-\lambda _1)\tau _1}{4W_3^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] ^2}\right\} ^{\frac{1}{2}}\nonumber \\&\quad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(106)
traveling wave solutions
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \Bigg \{ \frac{(\lambda _2-\lambda _1)\tau _1}{2}\Bigg (1\mp \coth \Bigg [\frac{\lambda _1-\lambda _2}{2W_3}\nonumber \\&\qquad \times \Bigg ( x + \Bigg (\frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \Bigg ) t \Bigg )\Bigg ]\Bigg )\Bigg \}^{\frac{1}{2}} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(107)
and soliton solution
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_5}{\left( C_2+\cosh \left[ B_6\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) ^{\frac{1}{2}}} \right\} \nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(108)
where
$$\begin{aligned} A_5= & {} \left( \frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)\tau _1}{\lambda _3-\lambda _2}\right) ^{\frac{1}{2}},\nonumber \\ B_6= & {} \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W_3},\nonumber \\ C_2= & {} \frac{2\lambda _1-\lambda _2-\lambda _3}{\lambda _3-\lambda _2}. \end{aligned}$$
(109)
Here, \(A_5\) is the amplitude of the soliton, while \(B_6\) is the inverse width of the soliton. These solitons exist for \(\tau _1<0\). On the other hand, when \(\tau _0=-\tau _1\lambda _2\) and \(\xi _0=0\), we can write the Jacobi elliptic function solution (102) as
$$\begin{aligned}&q(x,t)\nonumber \\&\quad = \left\{ \frac{A_6}{ \left( C_3+ {{\mathrm{sn}}}^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) ,\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)} \right] \right) ^{\frac{1}{2}}} \right\} \nonumber \\&\qquad \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(110)
where
$$\begin{aligned} A_6= & {} \left( \frac{\tau _1(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _1-\lambda _4} \right) ^{\frac{1}{2}}, \nonumber \\ C_3= & {} \frac{\lambda _4-\lambda _2}{\lambda _1-\lambda _4}, \quad B_j= \frac{(-1)^j\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_3},\nonumber \\&(j=7,8). \end{aligned}$$
(111)
Remark 6
When the modulus \(l\rightarrow 1\), hyperbolic function solutions fall out:
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_6}{\left( C_3+ \tanh ^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) ^{\frac{1}{2}} }\right\} \nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(112)
where \(\lambda _3=\lambda _4\).
Remark 7
However, if \(l\rightarrow 0\), periodic wave solutions are as listed below:
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_6}{\left( C_3+ \sin ^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) ^{\frac{1}{2}} } \right\} \nonumber \\&\times \,\hbox {e}^{i\left\{ -\kappa x + \left( \frac{\mu _2 (b v-a)+2 \chi _0 \left[ 2 \kappa (a \kappa -\lambda )-c \tau _0 \left( 3+4 \eta \tau _0\right) \right] }{4 \chi _0(b \kappa -1)} \right) t + \theta \right\} },\nonumber \\ \end{aligned}$$
(113)
where \(\lambda _2=\lambda _3\).
2.4 Dual-power law
In this case,
$$\begin{aligned} F(u)=u^{n}+\eta u^{2n}, \end{aligned}$$
(114)
where \(\eta \) is a real-valued constant. Therefore, Eq. (1) takes the form
$$\begin{aligned}&iq_{t}+aq_{xx}+bq_{xt}+c(|q|^{2n}+\eta |q|^{4n})q=\nonumber \\&\quad -\,i\lambda q_x -is\left( |q|^{2}q\right) _{x}-i\mu \left( |q|^{2}\right) _{x}q\nonumber \\&\quad -\,i\theta |q|^{2}q_{x}-i\gamma q_{xxx}-\theta _{1}\left( |q|^{2}q\right) _{xx}\nonumber \\&\quad -\,\theta _{2}|q|^{2}q_{xx}-\theta _{3}q^{2}q^{*}_{xx} . \end{aligned}$$
(115)
In this case, Eq. (11) simplifies to
$$\begin{aligned}&(a-bv)U''-\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) U\nonumber \\&\quad +\,cU^{2n+1}+c\eta U^{4n+1} =0. \end{aligned}$$
(116)
Balancing \(U''\) with \(U^{4n+1}\) gives \(N=\frac{1}{2n}.\) In order to obtain closed-form solutions, we use the transformation
$$\begin{aligned} U=V^{\frac{1}{2n}}, \end{aligned}$$
(117)
that will reduce Eq. (116) to the ODE
$$\begin{aligned}&(a-bv)\left( (1-2n)(V')^{2}+2nVV''\right) \nonumber \\&\quad -\,4n^{2}\left( (1-b\kappa )\omega +a\kappa ^2-\lambda \kappa \right) V^{2}\nonumber \\&\quad +4cn^{2}V^{3}+4cn^{2}\eta V^{4}=0. \end{aligned}$$
(118)
In this subsection, we will apply the extended trial equation method to solve the NLSE with dual-power law nonlinearity. Substituting Eqs. (16), (18) and (19) in Eq. (118), and using the balance principle, we determine a relation of \(\sigma , \rho \) and \(\varsigma \) as
$$\begin{aligned} \sigma =\rho +2\varsigma +2. \end{aligned}$$
(119)
If we take \(\sigma =4, \rho =0\) and \(\varsigma =1\) in Eq. (119), then
$$\begin{aligned}&V=\tau _0+\tau _1\Psi ,\end{aligned}$$
(120)
$$\begin{aligned}&(V')^2\!=\!\frac{\tau _1^2(\mu _4\Psi ^4\!+\!\mu _3\Psi ^3\!+\! \mu _2\Psi ^2\!+\!\mu _1\Psi \!+\!\mu _0)}{\chi _0}, \end{aligned}$$
(121)
$$\begin{aligned}&V''=\frac{\tau _1(4\mu _4\Psi ^3+3\mu _3\Psi ^2+2\mu _2\Psi +\mu _1)}{2\chi _0}, \end{aligned}$$
(122)
where \(\mu _4\ne 0, \chi _0\ne 0.\) Substituting Eqs. (120)–(122) in Eq. (118), collecting the coefficients of \(\Psi \), and solving the resulting algebraic equations system, we have
$$\begin{aligned} \mu _1= & {} \displaystyle \frac{2 \mu _0 \tau _1}{\tau _0}-\frac{4 c n^2 \tau _0^2 \chi _0 \left[ 1+2n+2 \eta \tau _0 (1+n) \right] }{\tau _1 (1+n) (1+2 n) (a-b v)} , \\ \mu _2= & {} \displaystyle \frac{\mu _0 \tau _1^2}{\tau _0^2}-\frac{4 c n^2 \tau _0 \chi _0 \left[ 2+4n+5 \eta \tau _0 (1+n) \right] }{(1+n) (1+2 n) (a-b v)} ,\\ \mu _3= & {} \displaystyle -\frac{4 c n^2 \tau _1 \chi _0 \left[ 1+2n+4 \eta \tau _0 (1+n) \right] }{(1+n) (1+2 n) (a-b v)} , \\ \mu _4= & {} \displaystyle -\frac{4 c n^2 \eta \tau _1^2 \chi _0}{(1+2 n) (a-b v)}, \end{aligned}$$
$$\begin{aligned} \omega= & {} \displaystyle \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} , \end{aligned}$$
$$\begin{aligned} \mu _0=\mu _0,\quad \chi _0=\chi _0,\quad \tau _0=\tau _0,\quad \tau _1=\tau _1 . \end{aligned}$$
(123)
Substituting the solution set (123) in Eqs. (17) and (20), we find that
$$\begin{aligned} \pm (\xi -\xi _0)= W_4 \int {\frac{\hbox {d}\Psi }{\sqrt{\Lambda (\Psi )}}}, \end{aligned}$$
(124)
where
$$\begin{aligned} \Lambda (\Psi )= & {} \Psi ^4+\frac{\mu _3}{\mu _4}\Psi ^3+\frac{\mu _2}{\mu _4}\Psi ^2+\frac{\mu _1}{\mu _4}\Psi +\frac{\mu _0}{\mu _4}, \nonumber \\ W_4= & {} \sqrt{\frac{\chi _0}{\mu _4}}. \end{aligned}$$
(125)
Consequently, we have the traveling wave solutions to the NLSE with dual-power law nonlinearity (115) as the following:
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^4\), we get
$$\begin{aligned}&q(x,t)= \left\{ \displaystyle \tau _0+\tau _1\lambda _1\pm \frac{\tau _1W_4}{ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}}\right\} ^{\frac{1}{2n}} \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(126)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^3(\Psi -\lambda _2)\) and \(\lambda _2>\lambda _1\), we obtain
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _1+\frac{4W_4^2(\lambda _2-\lambda _1)\tau _1}{4W_4^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right] ^2}\right\} ^{\frac{1}{2n}} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(127)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)^2\), we find
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _2+\frac{(\lambda _2-\lambda _1)\tau _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W_4} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right] -1}\right\} ^{\frac{1}{2n}} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(128)
and
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _1+\frac{(\lambda _1-\lambda _2)\tau _1}{\exp \left[ \frac{\lambda _1-\lambda _2}{W_4} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0}\right) \right] -1}\right\} ^{\frac{1}{2n}} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(129)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)^2(\Psi -\lambda _2)(\Psi -\lambda _3)\) and \(\lambda _1>\lambda _2>\lambda _3\), we attain
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _1-\frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)\tau _1}{2\lambda _1-\lambda _2-\lambda _3+(\lambda _3-\lambda _2) \cosh \left( \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W_4} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right) }\right\} ^{\frac{1}{2n}} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }. \end{aligned}$$
(130)
When \(\Lambda (\Psi )=(\Psi -\lambda _1)(\Psi -\lambda _2)(\Psi -\lambda _3)(\Psi -\lambda _4)\) and \(\lambda _1>\lambda _2>\lambda _3>\lambda _4\), we have
$$\begin{aligned} q(x,t)= & {} \left\{ \tau _0+\tau _1\lambda _2+\frac{\tau _1(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _4-\lambda _2+(\lambda _1-\lambda _4){{\mathrm{sn}}}^2\left[ \pm \frac{\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_4} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t-\xi _{0} \right) , l \right] }\right\} ^{\frac{1}{2n}}\nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(131)
where
$$\begin{aligned} l^2=\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}. \end{aligned}$$
(132)
Also, \(\lambda _i\,(i=1,\ldots ,4)\) are the roots of the polynomial equation
$$\begin{aligned} \Lambda (\Psi )=0. \end{aligned}$$
(133)
When \(\tau _0=-\tau _1\lambda _1\) and \(\xi _0=0\), we can reduce the solutions (126)–(130) to rational function solutions
$$\begin{aligned}&q(x,t)= \left\{ \pm \displaystyle \frac{\tau _1W_4}{ x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t }\right\} ^{\frac{1}{2n}} \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} },\end{aligned}$$
(134)
$$\begin{aligned}&q(x,t)= \left\{ \frac{4W_4^2(\lambda _2-\lambda _1)\tau _1}{4W_4^2-\left[ (\lambda _1-\lambda _2) \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] ^2}\right\} ^{\frac{1}{2n}} \nonumber \\&\quad \ \ \times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(135)
traveling wave solutions
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{(\lambda _2-\lambda _1)\tau _1}{2}\left( 1\mp \coth \left[ \frac{\lambda _1-\lambda _2}{2W_4} \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) \right\} ^{\frac{1}{2n}} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(136)
and soliton solution
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_7}{\left( C_4+\cosh \left[ B_9\left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) ^{\frac{1}{2n}}} \right\} \nonumber \\ &\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(137)
where
$$\begin{aligned} A_7= & {} \left( \frac{2(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)\tau _1}{\lambda _3-\lambda _2}\right) ^{\frac{1}{2n}},\nonumber \\ B_9= & {} \frac{\sqrt{(\lambda _1-\lambda _2)(\lambda _1-\lambda _3)}}{W_4},\nonumber \\ C_4= & {} \frac{2\lambda _1-\lambda _2-\lambda _3}{\lambda _3-\lambda _2}. \end{aligned}$$
(138)
Here, \(A_7\) is the amplitude of the soliton, while \(B_9\) is the inverse width of the soliton. These solitons exist for \(\tau _1<0\). Also, when \(\tau _0=-\tau _1\lambda _2\) and \(\xi _0=0\), we can write the Jacobi elliptic function solution (131) as
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_8}{ \left( C_5+ {{\mathrm{sn}}}^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) ,\frac{(\lambda _2-\lambda _3)(\lambda _1-\lambda _4)}{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)} \right] \right) ^{\frac{1}{2n}}} \right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} }, \end{aligned}$$
(139)
where
$$\begin{aligned} A_8= & {} \left( \frac{\tau _1(\lambda _1-\lambda _2)(\lambda _4-\lambda _2)}{\lambda _1-\lambda _4} \right) ^{\frac{1}{2n}}, \nonumber \\ C_5= & {} \frac{\lambda _4-\lambda _2}{\lambda _1-\lambda _4}, \nonumber \\ B_j= & {} \frac{(-1)^j\sqrt{(\lambda _1-\lambda _3)(\lambda _2-\lambda _4)}}{2W_4},\ \ (j=10,11).\nonumber \\ \end{aligned}$$
(140)
Remark 8
When the modulus \(l\rightarrow 1\), hyperbolic function solutions fall out:
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_8}{\left( C_5+ \tanh ^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) ^{\frac{1}{2n}} }\right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(141)
where \(\lambda _3=\lambda _4\).
Remark 9
However, if \(l\rightarrow 0\), periodic wave solutions are as listed below:
$$\begin{aligned} q(x,t)= & {} \left\{ \frac{A_8}{\left( C_5+ \sin ^2\left[ B_j \left( x + \left( \frac{2a\kappa -b\omega -\lambda }{1-b\kappa } \right) t \right) \right] \right) ^{\frac{1}{2n}} } \right\} \nonumber \\&\times \hbox {e}^{i\left\{ -\kappa x + \left( \frac{-\mu _0 \tau _1^2 (1+n) (1+2 n) (a-b v)+4 n^2 \tau _0^2 \chi _0 \left[ \kappa (1+n) (1+2 n)(a \kappa -\lambda )-c \tau _0 \left( 1+2n+\eta \tau _0 (1+n)\right) \right] }{4 n^2 \tau _0^2 \chi _0 (1+n) (1+2n) (b \kappa -1)} \right) t + \theta \right\} } , \end{aligned}$$
(142)
where \(\lambda _2=\lambda _3\).
