\(\int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [271]

Optimal result

Integrand size = 29, antiderivative size = 463 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 f^2 (e+f x) \arctan \left (e^{c+d x}\right )}{a d^3}+\frac {(e+f x)^3 \arctan \left (e^{c+d x}\right )}{a d}+\frac {3 i f^2 (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{2 a d^2}-\frac {3 i f^3 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{2 a d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 a d^4}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}-\frac {3 i f^3 \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{a d^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (4,i e^{c+d x}\right )}{a d^4}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 a d^2}+\frac {i (e+f x)^3 \text {sech}^2(c+d x)}{2 a d}-\frac {3 i f (e+f x)^2 \tanh (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \text {sech}(c+d x) \tanh (c+d x)}{2 a d} \] Output:

3*I*f^3*polylog(4,I*exp(d*x+c))/a/d^4-6*f^2*(f*x+e)*arctan(exp(d*x+c))/a/d 
^3+(f*x+e)^3*arctan(exp(d*x+c))/a/d-3*I*f^3*polylog(2,I*exp(d*x+c))/a/d^4- 
3/2*I*f*(f*x+e)^2*tanh(d*x+c)/a/d^2+3*I*f^2*(f*x+e)*polylog(3,-I*exp(d*x+c 
))/a/d^3+3/2*I*f^3*polylog(2,-exp(2*d*x+2*c))/a/d^4-3*I*f^2*(f*x+e)*polylo 
g(3,I*exp(d*x+c))/a/d^3+3/2*I*f*(f*x+e)^2*polylog(2,I*exp(d*x+c))/a/d^2-3* 
I*f^3*polylog(4,-I*exp(d*x+c))/a/d^4+3*I*f^3*polylog(2,-I*exp(d*x+c))/a/d^ 
4+1/2*I*(f*x+e)^3*sech(d*x+c)^2/a/d-3/2*I*f*(f*x+e)^2*polylog(2,-I*exp(d*x 
+c))/a/d^2+3/2*f*(f*x+e)^2*sech(d*x+c)/a/d^2-3/2*I*f*(f*x+e)^2/a/d^2+3*I*f 
^2*(f*x+e)*ln(1+exp(2*d*x+2*c))/a/d^3+1/2*(f*x+e)^3*sech(d*x+c)*tanh(d*x+c 
)/a/d
 

Mathematica [A] (verified)

Time = 7.82 (sec) , antiderivative size = 828, normalized size of antiderivative = 1.79 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[((e + f*x)^3*Sech[c + d*x])/(a + I*a*Sinh[c + d*x]),x]
 

Output:

-1/8*((e + f*x)^4/f + (4*(1 - I*E^c)*(e + f*x)^3*Log[1 + I*E^(-c - d*x)])/ 
d + ((12*I)*(I + E^c)*f*(d^2*(e + f*x)^2*PolyLog[2, (-I)*E^(-c - d*x)] + 2 
*f*(d*(e + f*x)*PolyLog[3, (-I)*E^(-c - d*x)] + f*PolyLog[4, (-I)*E^(-c - 
d*x)])))/d^4)/(a*(I + E^c)) - (-4*d^2*e*(1 + I*E^c)*f*(d^2*e^2 - 12*f^2)*x 
 + (-12*f^2 + d^2*(e + f*x)^2)^2 + 12*d*(1 + I*E^c)*f^2*(d^2*e^2 - 4*f^2)* 
x*Log[1 - I*E^(-c - d*x)] + 12*d^3*e*(1 + I*E^c)*f^3*x^2*Log[1 - I*E^(-c - 
 d*x)] + 4*d^3*(1 + I*E^c)*f^4*x^3*Log[1 - I*E^(-c - d*x)] + 4*d*e*(1 + I* 
E^c)*f*(d^2*e^2 - 12*f^2)*Log[I - E^(c + d*x)] + 12*(1 + I*E^c)*f^2*(-(d^2 
*e^2) + 4*f^2)*PolyLog[2, I*E^(-c - d*x)] - 24*d^2*e*(1 + I*E^c)*f^3*x*Pol 
yLog[2, I*E^(-c - d*x)] - 12*d^2*(1 + I*E^c)*f^4*x^2*PolyLog[2, I*E^(-c - 
d*x)] - 24*d*e*(1 + I*E^c)*f^3*PolyLog[3, I*E^(-c - d*x)] - 24*d*(1 + I*E^ 
c)*f^4*x*PolyLog[3, I*E^(-c - d*x)] - 24*(1 + I*E^c)*f^4*PolyLog[4, I*E^(- 
c - d*x)])/(8*a*d^4*(-I + E^c)*f) + (x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + 
f^3*x^3))/(8*a*(Cosh[c/2] - I*Sinh[c/2])*(Cosh[c/2] + I*Sinh[c/2])) + ((I/ 
2)*(e + f*x)^3)/(a*d*(Cosh[c/2 + (d*x)/2] + I*Sinh[c/2 + (d*x)/2])^2) - (( 
3*I)*(e^2*f*Sinh[(d*x)/2] + 2*e*f^2*x*Sinh[(d*x)/2] + f^3*x^2*Sinh[(d*x)/2 
]))/(a*d^2*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[c/2 + (d*x)/2] + I*Sinh[c/2 + ( 
d*x)/2]))
 

Rubi [A] (verified)

Time = 3.07 (sec) , antiderivative size = 442, normalized size of antiderivative = 0.95, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.724, Rules used = {6105, 3042, 4674, 3042, 4668, 2715, 2838, 3011, 5974, 3042, 4672, 26, 3042, 26, 4201, 2620, 2715, 2838, 7163, 2720, 7143}

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.

\(\displaystyle \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 6105

\(\displaystyle \frac {\int (e+f x)^3 \text {sech}^3(c+d x)dx}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e+f x)^3 \csc \left (i c+i d x+\frac {\pi }{2}\right )^3dx}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {-\frac {3 f^2 \int (e+f x) \text {sech}(c+d x)dx}{d^2}+\frac {1}{2} \int (e+f x)^3 \text {sech}(c+d x)dx+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {3 f^2 \int (e+f x) \csc \left (i c+i d x+\frac {\pi }{2}\right )dx}{d^2}+\frac {1}{2} \int (e+f x)^3 \csc \left (i c+i d x+\frac {\pi }{2}\right )dx+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {-\frac {3 f^2 \left (-\frac {i f \int \log \left (1-i e^{c+d x}\right )dx}{d}+\frac {i f \int \log \left (1+i e^{c+d x}\right )dx}{d}+\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}\right )}{d^2}+\frac {1}{2} \left (-\frac {3 i f \int (e+f x)^2 \log \left (1-i e^{c+d x}\right )dx}{d}+\frac {3 i f \int (e+f x)^2 \log \left (1+i e^{c+d x}\right )dx}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {-\frac {3 f^2 \left (-\frac {i f \int e^{-c-d x} \log \left (1-i e^{c+d x}\right )de^{c+d x}}{d^2}+\frac {i f \int e^{-c-d x} \log \left (1+i e^{c+d x}\right )de^{c+d x}}{d^2}+\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}\right )}{d^2}+\frac {1}{2} \left (-\frac {3 i f \int (e+f x)^2 \log \left (1-i e^{c+d x}\right )dx}{d}+\frac {3 i f \int (e+f x)^2 \log \left (1+i e^{c+d x}\right )dx}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {1}{2} \left (-\frac {3 i f \int (e+f x)^2 \log \left (1-i e^{c+d x}\right )dx}{d}+\frac {3 i f \int (e+f x)^2 \log \left (1+i e^{c+d x}\right )dx}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \int (e+f x)^3 \text {sech}^2(c+d x) \tanh (c+d x)dx}{a}\)

\(\Big \downarrow \) 5974

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (\frac {3 f \int (e+f x)^2 \text {sech}^2(c+d x)dx}{2 d}-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \int (e+f x)^2 \csc \left (i c+i d x+\frac {\pi }{2}\right )^2dx}{2 d}\right )}{a}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}-\frac {2 i f \int -i (e+f x) \tanh (c+d x)dx}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}-\frac {2 f \int (e+f x) \tanh (c+d x)dx}{d}\right )}{2 d}-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}\right )}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}-\frac {2 f \int -i (e+f x) \tan (i c+i d x)dx}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \int (e+f x) \tan (i c+i d x)dx}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 4201

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \int \frac {e^{2 (c+d x)} (e+f x)}{1+e^{2 (c+d x)}}dx-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}-\frac {f \int \log \left (1+e^{2 (c+d x)}\right )dx}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}-\frac {f \int e^{-2 (c+d x)} \log \left (1+e^{2 (c+d x)}\right )de^{2 (c+d x)}}{4 d^2}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \int (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )dx}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{d}-\frac {f \int \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )dx}{d}\right )}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{d}-\frac {f \int \operatorname {PolyLog}\left (3,i e^{c+d x}\right )dx}{d}\right )}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 i f \left (\frac {2 f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{d}-\frac {f \int e^{-c-d x} \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )de^{c+d x}}{d^2}\right )}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{d}-\frac {f \int e^{-c-d x} \operatorname {PolyLog}\left (3,i e^{c+d x}\right )de^{c+d x}}{d^2}\right )}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}+\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}\right )-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {-\frac {3 f^2 \left (\frac {2 (e+f x) \arctan \left (e^{c+d x}\right )}{d}-\frac {i f \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2}+\frac {i f \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d^2}\right )}{d^2}+\frac {1}{2} \left (\frac {2 (e+f x)^3 \arctan \left (e^{c+d x}\right )}{d}+\frac {3 i f \left (\frac {2 f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{d}-\frac {f \operatorname {PolyLog}\left (4,-i e^{c+d x}\right )}{d^2}\right )}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{d}\right )}{d}-\frac {3 i f \left (\frac {2 f \left (\frac {(e+f x) \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{d}-\frac {f \operatorname {PolyLog}\left (4,i e^{c+d x}\right )}{d^2}\right )}{d}-\frac {(e+f x)^2 \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{d}\right )}{d}\right )+\frac {3 f (e+f x)^2 \text {sech}(c+d x)}{2 d^2}+\frac {(e+f x)^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d}}{a}-\frac {i \left (-\frac {(e+f x)^3 \text {sech}^2(c+d x)}{2 d}+\frac {3 f \left (\frac {(e+f x)^2 \tanh (c+d x)}{d}+\frac {2 i f \left (2 i \left (\frac {f \operatorname {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{4 d^2}+\frac {(e+f x) \log \left (e^{2 (c+d x)}+1\right )}{2 d}\right )-\frac {i (e+f x)^2}{2 f}\right )}{d}\right )}{2 d}\right )}{a}\)

Input:

Int[((e + f*x)^3*Sech[c + d*x])/(a + I*a*Sinh[c + d*x]),x]
 

Output:

((-3*f^2*((2*(e + f*x)*ArcTan[E^(c + d*x)])/d - (I*f*PolyLog[2, (-I)*E^(c 
+ d*x)])/d^2 + (I*f*PolyLog[2, I*E^(c + d*x)])/d^2))/d^2 + ((2*(e + f*x)^3 
*ArcTan[E^(c + d*x)])/d + ((3*I)*f*(-(((e + f*x)^2*PolyLog[2, (-I)*E^(c + 
d*x)])/d) + (2*f*(((e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/d - (f*PolyLog[ 
4, (-I)*E^(c + d*x)])/d^2))/d))/d - ((3*I)*f*(-(((e + f*x)^2*PolyLog[2, I* 
E^(c + d*x)])/d) + (2*f*(((e + f*x)*PolyLog[3, I*E^(c + d*x)])/d - (f*Poly 
Log[4, I*E^(c + d*x)])/d^2))/d))/d)/2 + (3*f*(e + f*x)^2*Sech[c + d*x])/(2 
*d^2) + ((e + f*x)^3*Sech[c + d*x]*Tanh[c + d*x])/(2*d))/a - (I*(-1/2*((e 
+ f*x)^3*Sech[c + d*x]^2)/d + (3*f*(((2*I)*f*(((-1/2*I)*(e + f*x)^2)/f + ( 
2*I)*(((e + f*x)*Log[1 + E^(2*(c + d*x))])/(2*d) + (f*PolyLog[2, -E^(2*(c 
+ d*x))])/(4*d^2))))/d + ((e + f*x)^2*Tanh[c + d*x])/d))/(2*d)))/a
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 4668

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 4674

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo 
l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n 
- 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 
2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) 
Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ 
(n - 1))   Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c 
, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
 

rule 5974

Int[((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.)*Tanh[(a_.) + 
(b_.)*(x_)]^(p_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Sech[a + b*x]^n/(b*n)) 
, x] + Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Sech[a + b*x]^n, x], x] /; 
FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
 

rule 6105

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_ 
.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :> Simp[1/a   Int[(e + f*x)^m*Sech[ 
c + d*x]^(n + 2), x], x] + Simp[1/b   Int[(e + f*x)^m*Sech[c + d*x]^(n + 1) 
*Tanh[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && 
EqQ[a^2 + b^2, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1079 vs. \(2 (416 ) = 832\).

Time = 33.68 (sec) , antiderivative size = 1080, normalized size of antiderivative = 2.33

method	result	size
risch	\(\text {Expression too large to display}\)	\(1080\)

Input:

int((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-3*I*f^3*polylog(4,-I*exp(d*x+c))/a/d^4+6/a/d^4*f^3*c*arctan(exp(d*x+c))-1 
/a/d^4*f^3*c^3*arctan(exp(d*x+c))-6/a/d^3*e*f^2*arctan(exp(d*x+c))+6*I/a/d 
^4*f^3*polylog(2,-I*exp(d*x+c))-3*I/a/d^2*f^3*x^2-3*I/a/d^4*f^3*c^2+1/a/d* 
e^3*arctan(exp(d*x+c))+3*I*f^3*polylog(4,I*exp(d*x+c))/a/d^4+6*I/a/d^4*f^3 
*ln(1+I*exp(d*x+c))*c+3/2*I/a/d^2*e^2*f*polylog(2,I*exp(d*x+c))-3/2*I/a/d^ 
2*e^2*f*polylog(2,-I*exp(d*x+c))-3*I/a/d^4*f^3*c*ln(1+exp(2*d*x+2*c))+6*I/ 
a/d^4*f^3*c*ln(exp(d*x+c))+1/2*I/a/d*f^3*ln(1-I*exp(d*x+c))*x^3+3/2*I/a/d^ 
2*f^3*polylog(2,I*exp(d*x+c))*x^2-3*I/a/d^3*f^3*polylog(3,I*exp(d*x+c))*x- 
1/2*I/a/d*f^3*ln(1+I*exp(d*x+c))*x^3-3/2*I/a/d^2*f^3*polylog(2,-I*exp(d*x+ 
c))*x^2+3*I/a/d^3*f^3*polylog(3,-I*exp(d*x+c))*x+1/2*I/a/d^4*f^3*ln(1-I*ex 
p(d*x+c))*c^3-1/2*I/a/d^4*f^3*ln(1+I*exp(d*x+c))*c^3-3*I/a/d^3*e*f^2*polyl 
og(3,I*exp(d*x+c))+3*I/a/d^3*e*f^2*polylog(3,-I*exp(d*x+c))+3*I/a/d^3*e*f^ 
2*ln(1+exp(2*d*x+2*c))-6*I/a/d^3*e*f^2*ln(exp(d*x+c))+3/a/d^3*f^2*c^2*e*ar 
ctan(exp(d*x+c))-3/a/d^2*e^2*f*c*arctan(exp(d*x+c))-6*I/a/d^3*f^3*c*x+6*I/ 
a/d^3*f^3*ln(1+I*exp(d*x+c))*x+(d*f^3*x^3*exp(d*x+c)+3*d*e*f^2*x^2*exp(d*x 
+c)+3*d*e^2*f*x*exp(d*x+c)+d*e^3*exp(d*x+c)+3*f^3*x^2*exp(d*x+c)-3*I*f^3*x 
^2+6*e*f^2*x*exp(d*x+c)-6*I*e*f^2*x+3*e^2*f*exp(d*x+c)-3*I*e^2*f)/(exp(d*x 
+c)-I)^2/d^2/a+3/2*I/a/d*e^2*f*ln(1-I*exp(d*x+c))*x+3/2*I/a/d^2*e^2*f*ln(1 
-I*exp(d*x+c))*c-3/2*I/a/d*e^2*f*ln(1+I*exp(d*x+c))*x-3/2*I/a/d^2*e^2*f*ln 
(1+I*exp(d*x+c))*c+3/2*I/a/d*e*f^2*ln(1-I*exp(d*x+c))*x^2+3*I/a/d^2*e*f...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1461 vs. \(2 (390) = 780\).

Time = 0.11 (sec) , antiderivative size = 1461, normalized size of antiderivative = 3.16 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
 

Output:

1/2*(-6*I*d^2*e^2*f + 12*I*c*d*e*f^2 - 6*I*c^2*f^3 - 3*(I*d^2*f^3*x^2 + 2* 
I*d^2*e*f^2*x + I*d^2*e^2*f + (-I*d^2*f^3*x^2 - 2*I*d^2*e*f^2*x - I*d^2*e^ 
2*f)*e^(2*d*x + 2*c) - 2*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f)*e^(d*x 
+ c))*dilog(I*e^(d*x + c)) - 3*(-I*d^2*f^3*x^2 - 2*I*d^2*e*f^2*x - I*d^2*e 
^2*f + 4*I*f^3 + (I*d^2*f^3*x^2 + 2*I*d^2*e*f^2*x + I*d^2*e^2*f - 4*I*f^3) 
*e^(2*d*x + 2*c) + 2*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + d^2*e^2*f - 4*f^3)*e^( 
d*x + c))*dilog(-I*e^(d*x + c)) - 6*(I*d^2*f^3*x^2 + 2*I*d^2*e*f^2*x + 2*I 
*c*d*e*f^2 - I*c^2*f^3)*e^(2*d*x + 2*c) + 2*(d^3*f^3*x^3 + d^3*e^3 + 3*d^2 
*e^2*f - 12*c*d*e*f^2 + 6*c^2*f^3 + 3*(d^3*e*f^2 - d^2*f^3)*x^2 + 3*(d^3*e 
^2*f - 2*d^2*e*f^2)*x)*e^(d*x + c) + (-I*d^3*e^3 + 3*I*c*d^2*e^2*f - 3*I*c 
^2*d*e*f^2 + I*c^3*f^3 + (I*d^3*e^3 - 3*I*c*d^2*e^2*f + 3*I*c^2*d*e*f^2 - 
I*c^3*f^3)*e^(2*d*x + 2*c) + 2*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - 
c^3*f^3)*e^(d*x + c))*log(e^(d*x + c) + I) + (I*d^3*e^3 - 3*I*c*d^2*e^2*f 
- 3*(-I*c^2 + 4*I)*d*e*f^2 + (-I*c^3 + 12*I*c)*f^3 + (-I*d^3*e^3 + 3*I*c*d 
^2*e^2*f - 3*(I*c^2 - 4*I)*d*e*f^2 + (I*c^3 - 12*I*c)*f^3)*e^(2*d*x + 2*c) 
 - 2*(d^3*e^3 - 3*c*d^2*e^2*f + 3*(c^2 - 4)*d*e*f^2 - (c^3 - 12*c)*f^3)*e^ 
(d*x + c))*log(e^(d*x + c) - I) + (I*d^3*f^3*x^3 + 3*I*d^3*e*f^2*x^2 + 3*I 
*c*d^2*e^2*f - 3*I*c^2*d*e*f^2 + (I*c^3 - 12*I*c)*f^3 - 3*(-I*d^3*e^2*f + 
4*I*d*f^3)*x + (-I*d^3*f^3*x^3 - 3*I*d^3*e*f^2*x^2 - 3*I*c*d^2*e^2*f + 3*I 
*c^2*d*e*f^2 + (-I*c^3 + 12*I*c)*f^3 - 3*(I*d^3*e^2*f - 4*I*d*f^3)*x)*e...
 

Sympy [F]

\[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \left (\int \frac {e^{3} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {f^{3} x^{3} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e f^{2} x^{2} \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx + \int \frac {3 e^{2} f x \operatorname {sech}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx\right )}{a} \] Input:

integrate((f*x+e)**3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x)
 

Output:

-I*(Integral(e**3*sech(c + d*x)/(sinh(c + d*x) - I), x) + Integral(f**3*x* 
*3*sech(c + d*x)/(sinh(c + d*x) - I), x) + Integral(3*e*f**2*x**2*sech(c + 
 d*x)/(sinh(c + d*x) - I), x) + Integral(3*e**2*f*x*sech(c + d*x)/(sinh(c 
+ d*x) - I), x))/a
 

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 685, normalized size of antiderivative = 1.48 \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
 

Output:

-1/2*e^3*(4*e^(-d*x - c)/((4*I*a*e^(-d*x - c) + 2*a*e^(-2*d*x - 2*c) - 2*a 
)*d) + I*log(e^(-d*x - c) + I)/(a*d) - I*log(I*e^(-d*x - c) + 1)/(a*d)) + 
3/2*I*(d*x*log(-I*e^(d*x + c) + 1) + dilog(I*e^(d*x + c)))*e^2*f/(a*d^2) - 
 6*I*e*f^2*x/(a*d^2) + (-3*I*f^3*x^2 - 6*I*e*f^2*x - 3*I*e^2*f + (d*f^3*x^ 
3*e^c + 3*e^2*f*e^c + 3*(d*e*f^2 + f^3)*x^2*e^c + 3*(d*e^2*f + 2*e*f^2)*x* 
e^c)*e^(d*x))/(a*d^2*e^(2*d*x + 2*c) - 2*I*a*d^2*e^(d*x + c) - a*d^2) - 3/ 
2*I*(d^2*x^2*log(I*e^(d*x + c) + 1) + 2*d*x*dilog(-I*e^(d*x + c)) - 2*poly 
log(3, -I*e^(d*x + c)))*e*f^2/(a*d^3) + 3/2*I*(d^2*x^2*log(-I*e^(d*x + c) 
+ 1) + 2*d*x*dilog(I*e^(d*x + c)) - 2*polylog(3, I*e^(d*x + c)))*e*f^2/(a* 
d^3) + 6*I*e*f^2*log(I*e^(d*x + c) + 1)/(a*d^3) - 1/2*I*(d^3*x^3*log(I*e^( 
d*x + c) + 1) + 3*d^2*x^2*dilog(-I*e^(d*x + c)) - 6*d*x*polylog(3, -I*e^(d 
*x + c)) + 6*polylog(4, -I*e^(d*x + c)))*f^3/(a*d^4) + 1/2*I*(d^3*x^3*log( 
-I*e^(d*x + c) + 1) + 3*d^2*x^2*dilog(I*e^(d*x + c)) - 6*d*x*polylog(3, I* 
e^(d*x + c)) + 6*polylog(4, I*e^(d*x + c)))*f^3/(a*d^4) - 3/2*I*(d^2*e^2*f 
 - 4*f^3)*(d*x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))/(a*d^4) - 1 
/8*(I*d^4*f^3*x^4 + 4*I*d^4*e*f^2*x^3 + 6*I*d^4*e^2*f*x^2)/(a*d^4) + 1/8*( 
I*d^4*f^3*x^4 + 4*I*d^4*e*f^2*x^3 - 6*(-I*d^2*e^2*f + 4*I*f^3)*d^2*x^2)/(a 
*d^4)
 

Giac [F]

\[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \operatorname {sech}\left (d x + c\right )}{i \, a \sinh \left (d x + c\right ) + a} \,d x } \] Input:

integrate((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)^3*sech(d*x + c)/(I*a*sinh(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\int \frac {{\left (e+f\,x\right )}^3}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:

int((e + f*x)^3/(cosh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)
 

Output:

int((e + f*x)^3/(cosh(c + d*x)*(a + a*sinh(c + d*x)*1i)), x)
 

Reduce [F]

\[ \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\left (\int \frac {\mathrm {sech}\left (d x +c \right )}{\sinh \left (d x +c \right ) i +1}d x \right ) e^{3}+\left (\int \frac {\mathrm {sech}\left (d x +c \right ) x^{3}}{\sinh \left (d x +c \right ) i +1}d x \right ) f^{3}+3 \left (\int \frac {\mathrm {sech}\left (d x +c \right ) x^{2}}{\sinh \left (d x +c \right ) i +1}d x \right ) e \,f^{2}+3 \left (\int \frac {\mathrm {sech}\left (d x +c \right ) x}{\sinh \left (d x +c \right ) i +1}d x \right ) e^{2} f}{a} \] Input:

int((f*x+e)^3*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x)
 

Output:

(int(sech(c + d*x)/(sinh(c + d*x)*i + 1),x)*e**3 + int((sech(c + d*x)*x**3 
)/(sinh(c + d*x)*i + 1),x)*f**3 + 3*int((sech(c + d*x)*x**2)/(sinh(c + d*x 
)*i + 1),x)*e*f**2 + 3*int((sech(c + d*x)*x)/(sinh(c + d*x)*i + 1),x)*e**2 
*f)/a