Integrand size = 24, antiderivative size = 307 \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\frac {(e x)^{2 n}}{2 a e n}-\frac {b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}+\frac {b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n}+\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n} \]
1/2*(e*x)^(2*n)/a/e/n-b*(e*x)^(2*n)*ln(1+a*exp(c+d*x^n)/(b-(-a^2+b^2)^(1/2 )))/a/d/e/n/(x^n)/(-a^2+b^2)^(1/2)+b*(e*x)^(2*n)*ln(1+a*exp(c+d*x^n)/(b+(- a^2+b^2)^(1/2)))/a/d/e/n/(x^n)/(-a^2+b^2)^(1/2)-b*(e*x)^(2*n)*polylog(2,-a *exp(c+d*x^n)/(b-(-a^2+b^2)^(1/2)))/a/d^2/e/n/(x^(2*n))/(-a^2+b^2)^(1/2)+b *(e*x)^(2*n)*polylog(2,-a*exp(c+d*x^n)/(b+(-a^2+b^2)^(1/2)))/a/d^2/e/n/(x^ (2*n))/(-a^2+b^2)^(1/2)
Result contains complex when optimal does not.
Time = 3.70 (sec) , antiderivative size = 859, normalized size of antiderivative = 2.80 \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\frac {(e x)^{2 n} \left (b+a \cosh \left (c+d x^n\right )\right ) \left (1+\frac {2 b x^{-2 n} \left (2 \left (c+d x^n\right ) \arctan \left (\frac {(a+b) \coth \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )+2 \left (c-i \arccos \left (-\frac {b}{a}\right )\right ) \arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )+\left (\arccos \left (-\frac {b}{a}\right )+2 \left (\arctan \left (\frac {(a+b) \coth \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )+\arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{-\frac {c}{2}-\frac {d x^n}{2}}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cosh \left (c+d x^n\right )}}\right )+\left (\arccos \left (-\frac {b}{a}\right )-2 \left (\arctan \left (\frac {(a+b) \coth \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )+\arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{\frac {1}{2} \left (c+d x^n\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cosh \left (c+d x^n\right )}}\right )-\left (\arccos \left (-\frac {b}{a}\right )+2 \arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (-a+b+i \sqrt {a^2-b^2}\right ) \left (-1+\tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+i \sqrt {a^2-b^2} \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )-\left (\arccos \left (-\frac {b}{a}\right )-2 \arctan \left (\frac {(a-b) \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (a-b+i \sqrt {a^2-b^2}\right ) \left (1+\tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+i \sqrt {a^2-b^2} \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )+i \left (\operatorname {PolyLog}\left (2,\frac {\left (b-i \sqrt {a^2-b^2}\right ) \left (a+b-i \sqrt {a^2-b^2} \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+i \sqrt {a^2-b^2} \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )-\operatorname {PolyLog}\left (2,\frac {\left (b+i \sqrt {a^2-b^2}\right ) \left (a+b-i \sqrt {a^2-b^2} \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+i \sqrt {a^2-b^2} \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )\right )\right )}{\sqrt {a^2-b^2} d^2}\right ) \text {sech}\left (c+d x^n\right )}{2 a e n \left (a+b \text {sech}\left (c+d x^n\right )\right )} \]
((e*x)^(2*n)*(b + a*Cosh[c + d*x^n])*(1 + (2*b*(2*(c + d*x^n)*ArcTan[((a + b)*Coth[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + 2*(c - I*ArcCos[-(b/a)])*ArcTa n[((a - b)*Tanh[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + (ArcCos[-(b/a)] + 2*(Ar cTan[((a + b)*Coth[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + ArcTan[((a - b)*Tanh [(c + d*x^n)/2])/Sqrt[a^2 - b^2]]))*Log[(Sqrt[a^2 - b^2]*E^(-1/2*c - (d*x^ n)/2))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cosh[c + d*x^n]])] + (ArcCos[-(b/a)] - 2*(ArcTan[((a + b)*Coth[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + ArcTan[((a - b) *Tanh[(c + d*x^n)/2])/Sqrt[a^2 - b^2]]))*Log[(Sqrt[a^2 - b^2]*E^((c + d*x^ n)/2))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cosh[c + d*x^n]])] - (ArcCos[-(b/a)] + 2*ArcTan[((a - b)*Tanh[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(-a + b + I*Sqrt[a^2 - b^2])*(-1 + Tanh[(c + d*x^n)/2]))/(a*(a + b + I*Sqrt[a^ 2 - b^2]*Tanh[(c + d*x^n)/2]))] - (ArcCos[-(b/a)] - 2*ArcTan[((a - b)*Tanh [(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*(a - b + I*Sqrt[a^2 - b^2] )*(1 + Tanh[(c + d*x^n)/2]))/(a*(a + b + I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^n )/2]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a + b - I*Sqrt[a^2 - b^2 ]*Tanh[(c + d*x^n)/2]))/(a*(a + b + I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^n)/2]) )] - PolyLog[2, ((b + I*Sqrt[a^2 - b^2])*(a + b - I*Sqrt[a^2 - b^2]*Tanh[( c + d*x^n)/2]))/(a*(a + b + I*Sqrt[a^2 - b^2]*Tanh[(c + d*x^n)/2]))])))/(S qrt[a^2 - b^2]*d^2*x^(2*n)))*Sech[c + d*x^n])/(2*a*e*n*(a + b*Sech[c + d*x ^n]))
Time = 0.85 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5963, 5959, 3042, 4679, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(e x)^{2 n-1}}{a+b \text {sech}\left (c+d x^n\right )} \, dx\) |
| \(\Big \downarrow \) 5963 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^{2 n-1}}{a+b \text {sech}\left (d x^n+c\right )}dx}{e}\) |
| \(\Big \downarrow \) 5959 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{a+b \text {sech}\left (d x^n+c\right )}dx^n}{e n}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{a+b \csc \left (i d x^n+i c+\frac {\pi }{2}\right )}dx^n}{e n}\) |
| \(\Big \downarrow \) 4679 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \left (\frac {x^n}{a}-\frac {b x^n}{a \left (b+a \cosh \left (d x^n+c\right )\right )}\right )dx^n}{e n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \left (-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{d x^n+c}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{d x^n+c}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b x^n \log \left (\frac {a e^{c+d x^n}}{b-\sqrt {b^2-a^2}}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {b x^n \log \left (\frac {a e^{c+d x^n}}{\sqrt {b^2-a^2}+b}+1\right )}{a d \sqrt {b^2-a^2}}+\frac {x^{2 n}}{2 a}\right )}{e n}\) |
((e*x)^(2*n)*(x^(2*n)/(2*a) - (b*x^n*Log[1 + (a*E^(c + d*x^n))/(b - Sqrt[- a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (b*x^n*Log[1 + (a*E^(c + d*x^n))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (b*PolyLog[2, -((a*E^(c + d*x^n))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) + (b*PolyLog[2, -((a*E^(c + d*x^n))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2)))/ (e*n*x^(2*n))
3.1.80.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt Q[m, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] )^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m]) Int[x^m* (a + b*Sech[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.78 (sec) , antiderivative size = 585, normalized size of antiderivative = 1.91
| method | result | size |
| risch | \(\frac {x \,{\mathrm e}^{\frac {\left (2 n -1\right ) \left (-i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right ) \pi +i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi +i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi -i \operatorname {csgn}\left (i e x \right )^{3} \pi +2 \ln \left (e \right )+2 \ln \left (x \right )\right )}{2}}}{2 a n}-\frac {2 b \,{\mathrm e}^{-i \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )} {\mathrm e}^{i \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}} {\mathrm e}^{i \pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}} {\mathrm e}^{-i \pi n \operatorname {csgn}\left (i e x \right )^{3}} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right ) \pi }{2}} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi }{2}} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi }{2}} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i e x \right )^{3} \pi }{2}} e^{2 n} {\mathrm e}^{c} \left (\frac {x^{n} d \left (\ln \left (\frac {-a \,{\mathrm e}^{2 c +d \,x^{n}}-{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}{-{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}\right )-\ln \left (\frac {a \,{\mathrm e}^{2 c +d \,x^{n}}+{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}{{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}\right )\right )}{2 \sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}+\frac {\operatorname {dilog}\left (\frac {-a \,{\mathrm e}^{2 c +d \,x^{n}}-{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}{-{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}\right )-\operatorname {dilog}\left (\frac {a \,{\mathrm e}^{2 c +d \,x^{n}}+{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}{{\mathrm e}^{c} b +\sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}\right )}{2 \sqrt {{\mathrm e}^{2 c} b^{2}-a^{2} {\mathrm e}^{2 c}}}\right )}{a e n \,d^{2}}\) | \(585\) |
1/2/a/n*x*exp(1/2*(2*n-1)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I*csgn(I* e)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+2*ln(e )+2*ln(x)))-2*b/a*exp(-I*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(I*Pi*n* csgn(I*e)*csgn(I*e*x)^2)*exp(I*Pi*n*csgn(I*x)*csgn(I*e*x)^2)*exp(-I*Pi*n*c sgn(I*e*x)^3)*exp(1/2*I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(-1/2*I*Pi* csgn(I*e)*csgn(I*e*x)^2)*exp(-1/2*I*Pi*csgn(I*x)*csgn(I*e*x)^2)*exp(1/2*I* Pi*csgn(I*e*x)^3)*(e^n)^2/e*exp(c)/n/d^2*(1/2*x^n*d*(ln((-a*exp(2*c+d*x^n) -exp(c)*b+(exp(2*c)*b^2-a^2*exp(2*c))^(1/2))/(-exp(c)*b+(exp(2*c)*b^2-a^2* exp(2*c))^(1/2)))-ln((a*exp(2*c+d*x^n)+exp(c)*b+(exp(2*c)*b^2-a^2*exp(2*c) )^(1/2))/(exp(c)*b+(exp(2*c)*b^2-a^2*exp(2*c))^(1/2))))/(exp(2*c)*b^2-a^2* exp(2*c))^(1/2)+1/2*(dilog((-a*exp(2*c+d*x^n)-exp(c)*b+(exp(2*c)*b^2-a^2*e xp(2*c))^(1/2))/(-exp(c)*b+(exp(2*c)*b^2-a^2*exp(2*c))^(1/2)))-dilog((a*ex p(2*c+d*x^n)+exp(c)*b+(exp(2*c)*b^2-a^2*exp(2*c))^(1/2))/(exp(c)*b+(exp(2* c)*b^2-a^2*exp(2*c))^(1/2))))/(exp(2*c)*b^2-a^2*exp(2*c))^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 1286 vs. \(2 (287) = 574\).
Time = 0.30 (sec) , antiderivative size = 1286, normalized size of antiderivative = 4.19 \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\text {Too large to display} \]
1/2*((a^2 - b^2)*d^2*cosh((2*n - 1)*log(e))*cosh(n*log(x))^2 + (a^2 - b^2) *d^2*cosh(n*log(x))^2*sinh((2*n - 1)*log(e)) + ((a^2 - b^2)*d^2*cosh((2*n - 1)*log(e)) + (a^2 - b^2)*d^2*sinh((2*n - 1)*log(e)))*sinh(n*log(x))^2 + 2*(a*b*sqrt(-(a^2 - b^2)/a^2)*cosh((2*n - 1)*log(e)) + a*b*sqrt(-(a^2 - b^ 2)/a^2)*sinh((2*n - 1)*log(e)))*dilog(-((a*sqrt(-(a^2 - b^2)/a^2) + b)*cos h(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + (a*sqrt(-(a^2 - b^2)/a^2) + b )*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + a)/a + 1) - 2*(a*b*sqrt( -(a^2 - b^2)/a^2)*cosh((2*n - 1)*log(e)) + a*b*sqrt(-(a^2 - b^2)/a^2)*sinh ((2*n - 1)*log(e)))*dilog(((a*sqrt(-(a^2 - b^2)/a^2) - b)*cosh(d*cosh(n*lo g(x)) + d*sinh(n*log(x)) + c) + (a*sqrt(-(a^2 - b^2)/a^2) - b)*sinh(d*cosh (n*log(x)) + d*sinh(n*log(x)) + c) - a)/a + 1) + 2*(a*b*c*sqrt(-(a^2 - b^2 )/a^2)*cosh((2*n - 1)*log(e)) + a*b*c*sqrt(-(a^2 - b^2)/a^2)*sinh((2*n - 1 )*log(e)))*log(2*a*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + 2*a*sin h(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - 2*(a*b*c*sqrt(-(a^2 - b^2)/a^2)*cosh((2*n - 1)*log(e)) + a*b*c*sqrt (-(a^2 - b^2)/a^2)*sinh((2*n - 1)*log(e)))*log(2*a*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + 2*a*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) + 2*(a*b*d*sqrt(-(a^2 - b^2)/a^2)*cos h((2*n - 1)*log(e))*cosh(n*log(x)) + a*b*c*sqrt(-(a^2 - b^2)/a^2)*cosh((2* n - 1)*log(e)) + (a*b*d*sqrt(-(a^2 - b^2)/a^2)*cosh(n*log(x)) + a*b*c*s...
\[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\int \frac {\left (e x\right )^{2 n - 1}}{a + b \operatorname {sech}{\left (c + d x^{n} \right )}}\, dx \]
\[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{b \operatorname {sech}\left (d x^{n} + c\right ) + a} \,d x } \]
-2*b*e^(2*n)*integrate(e^(d*x^n + 2*n*log(x) + c)/(a^2*e*x*e^(2*d*x^n + 2* c) + 2*a*b*e*x*e^(d*x^n + c) + a^2*e*x), x) + 1/2*e^(2*n - 1)*x^(2*n)/(a*n )
\[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{b \operatorname {sech}\left (d x^{n} + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {sech}\left (c+d x^n\right )} \, dx=\int \frac {{\left (e\,x\right )}^{2\,n-1}}{a+\frac {b}{\mathrm {cosh}\left (c+d\,x^n\right )}} \,d x \]