\(\int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [272]

Optimal result

Integrand size = 29, antiderivative size = 268 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {(e+f x)^2 \arctan \left (e^{c+d x}\right )}{a d}-\frac {f^2 \arctan (\sinh (c+d x))}{a d^3}+\frac {i f^2 \log (\cosh (c+d x))}{a d^3}-\frac {i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{c+d x}\right )}{a d^2}+\frac {i f (e+f x) \operatorname {PolyLog}\left (2,i e^{c+d x}\right )}{a d^2}+\frac {i f^2 \operatorname {PolyLog}\left (3,-i e^{c+d x}\right )}{a d^3}-\frac {i f^2 \operatorname {PolyLog}\left (3,i e^{c+d x}\right )}{a d^3}+\frac {f (e+f x) \text {sech}(c+d x)}{a d^2}+\frac {i (e+f x)^2 \text {sech}^2(c+d x)}{2 a d}-\frac {i f (e+f x) \tanh (c+d x)}{a d^2}+\frac {(e+f x)^2 \text {sech}(c+d x) \tanh (c+d x)}{2 a d} \] Output:

(f*x+e)^2*arctan(exp(d*x+c))/a/d-f^2*arctan(sinh(d*x+c))/a/d^3+I*f^2*ln(co 
sh(d*x+c))/a/d^3-I*f*(f*x+e)*polylog(2,-I*exp(d*x+c))/a/d^2+I*f*(f*x+e)*po 
lylog(2,I*exp(d*x+c))/a/d^2+I*f^2*polylog(3,-I*exp(d*x+c))/a/d^3-I*f^2*pol 
ylog(3,I*exp(d*x+c))/a/d^3+f*(f*x+e)*sech(d*x+c)/a/d^2+1/2*I*(f*x+e)^2*sec 
h(d*x+c)^2/a/d-I*f*(f*x+e)*tanh(d*x+c)/a/d^2+1/2*(f*x+e)^2*sech(d*x+c)*tan 
h(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 5.28 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.98 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\frac {\frac {(e+f x)^3}{f}+\frac {3 \left (1-i e^c\right ) (e+f x)^2 \log \left (1+i e^{-c-d x}\right )}{d}+\frac {6 i \left (i+e^c\right ) f \left (d (e+f x) \operatorname {PolyLog}\left (2,-i e^{-c-d x}\right )+f \operatorname {PolyLog}\left (3,-i e^{-c-d x}\right )\right )}{d^3}}{i+e^c}+\frac {3 d^2 e^2 x-12 f^2 x-3 \left (1+i e^c\right ) \left (d^2 e^2-4 f^2\right ) x+3 d^2 e f x^2+d^2 f^2 x^3+6 d e \left (1+i e^c\right ) f x \log \left (1-i e^{-c-d x}\right )+3 d \left (1+i e^c\right ) f^2 x^2 \log \left (1-i e^{-c-d x}\right )+\frac {3 \left (1+i e^c\right ) \left (d^2 e^2-4 f^2\right ) \log \left (i-e^{c+d x}\right )}{d}-6 e \left (1+i e^c\right ) f \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )-6 \left (1+i e^c\right ) f^2 x \operatorname {PolyLog}\left (2,i e^{-c-d x}\right )-\frac {6 \left (1+i e^c\right ) f^2 \operatorname {PolyLog}\left (3,i e^{-c-d x}\right )}{d}}{d^2 \left (-i+e^c\right )}-x \left (3 e^2+3 e f x+f^2 x^2\right ) \text {sech}(c)-\frac {3 i (e+f x)^2}{d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 i f (e+f x) \sinh \left (\frac {d x}{2}\right )}{d^2 \left (\cosh \left (\frac {c}{2}\right )+i \sinh \left (\frac {c}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )}}{6 a} \] Input:

Integrate[((e + f*x)^2*Sech[c + d*x])/(a + I*a*Sinh[c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.74 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.94, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {6105, 3042, 4674, 3042, 4257, 4668, 3011, 2720, 5974, 3042, 4672, 26, 3042, 26, 3956, 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.

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

\(\Big \downarrow \) 6105

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4674

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4668

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 5974

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3956

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

\(\Big \downarrow \) 7143

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

Input:

Int[((e + f*x)^2*Sech[c + d*x])/(a + I*a*Sinh[c + d*x]),x]
 

Output:

(-((f^2*ArcTan[Sinh[c + d*x]])/d^3) + ((2*(e + f*x)^2*ArcTan[E^(c + d*x)]) 
/d + ((2*I)*f*(-(((e + f*x)*PolyLog[2, (-I)*E^(c + d*x)])/d) + (f*PolyLog[ 
3, (-I)*E^(c + d*x)])/d^2))/d - ((2*I)*f*(-(((e + f*x)*PolyLog[2, I*E^(c + 
 d*x)])/d) + (f*PolyLog[3, I*E^(c + d*x)])/d^2))/d)/2 + (f*(e + f*x)*Sech[ 
c + d*x])/d^2 + ((e + f*x)^2*Sech[c + d*x]*Tanh[c + d*x])/(2*d))/a - (I*(- 
1/2*((e + f*x)^2*Sech[c + d*x]^2)/d + (f*(-((f*Log[Cosh[c + d*x]])/d^2) + 
((e + f*x)*Tanh[c + d*x])/d))/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 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 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 3956

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 549 vs. \(2 (248 ) = 496\).

Time = 13.68 (sec) , antiderivative size = 550, normalized size of antiderivative = 2.05

method	result	size
risch	\(\frac {d \,x^{2} f^{2} {\mathrm e}^{d x +c}+2 d e f x \,{\mathrm e}^{d x +c}+d \,e^{2} {\mathrm e}^{d x +c}-2 i f^{2} x +2 f^{2} x \,{\mathrm e}^{d x +c}-2 i e f +2 e f \,{\mathrm e}^{d x +c}}{\left ({\mathrm e}^{d x +c}-i\right )^{2} d^{2} a}+\frac {i f^{2} \operatorname {polylog}\left (3, -i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {i \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right ) f^{2} x}{a \,d^{2}}+\frac {i e f \operatorname {polylog}\left (2, i {\mathrm e}^{d x +c}\right )}{a \,d^{2}}+\frac {i f^{2} \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{a \,d^{3}}+\frac {e^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{a d}+\frac {c^{2} f^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {i \ln \left (1+i {\mathrm e}^{d x +c}\right ) c^{2} f^{2}}{2 a \,d^{3}}-\frac {2 i f^{2} \ln \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}+\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) f^{2} x^{2}}{2 a d}+\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) e f x}{a d}-\frac {2 c e f \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {i e f \ln \left (1+i {\mathrm e}^{d x +c}\right ) c}{d^{2} a}-\frac {i e f \operatorname {polylog}\left (2, -i {\mathrm e}^{d x +c}\right )}{a \,d^{2}}-\frac {i f^{2} \operatorname {polylog}\left (3, i {\mathrm e}^{d x +c}\right )}{a \,d^{3}}-\frac {i \ln \left (1+i {\mathrm e}^{d x +c}\right ) f^{2} x^{2}}{2 a d}+\frac {i \operatorname {polylog}\left (2, i {\mathrm e}^{d x +c}\right ) f^{2} x}{a \,d^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) c^{2} f^{2}}{2 a \,d^{3}}-\frac {i e f \ln \left (1+i {\mathrm e}^{d x +c}\right ) x}{d a}+\frac {i \ln \left (1-i {\mathrm e}^{d x +c}\right ) c e f}{a \,d^{2}}-\frac {2 f^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{a \,d^{3}}\)	\(550\)

Input:

int((f*x+e)^2*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

(d*x^2*f^2*exp(d*x+c)+2*d*e*f*x*exp(d*x+c)+d*e^2*exp(d*x+c)-2*I*f^2*x+2*f^ 
2*x*exp(d*x+c)-2*I*e*f+2*e*f*exp(d*x+c))/(exp(d*x+c)-I)^2/d^2/a+I*f^2*poly 
log(3,-I*exp(d*x+c))/a/d^3-I/a/d^2*polylog(2,-I*exp(d*x+c))*f^2*x+I/a/d^2* 
e*f*polylog(2,I*exp(d*x+c))+I/a/d^3*f^2*ln(1+exp(2*d*x+2*c))+1/a/d*e^2*arc 
tan(exp(d*x+c))+1/a/d^3*c^2*f^2*arctan(exp(d*x+c))+1/2*I/a/d^3*ln(1+I*exp( 
d*x+c))*c^2*f^2-2*I/a/d^3*f^2*ln(exp(d*x+c))+1/2*I/a/d*ln(1-I*exp(d*x+c))* 
f^2*x^2+I/a/d*ln(1-I*exp(d*x+c))*e*f*x-2/a/d^2*c*e*f*arctan(exp(d*x+c))-I/ 
a/d^2*ln(1+I*exp(d*x+c))*c*e*f-I/a/d^2*e*f*polylog(2,-I*exp(d*x+c))-I*f^2* 
polylog(3,I*exp(d*x+c))/a/d^3-1/2*I/a/d*ln(1+I*exp(d*x+c))*f^2*x^2+I/a/d^2 
*polylog(2,I*exp(d*x+c))*f^2*x-1/2*I/a/d^3*ln(1-I*exp(d*x+c))*c^2*f^2-I/a/ 
d*ln(1+I*exp(d*x+c))*e*f*x+I/a/d^2*ln(1-I*exp(d*x+c))*c*e*f-2/a/d^3*f^2*ar 
ctan(exp(d*x+c))
 

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. 805 vs. \(2 (235) = 470\).

Time = 0.13 (sec) , antiderivative size = 805, normalized size of antiderivative = 3.00 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((f*x+e)^2*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.44 \[ \int \frac {(e+f x)^2 \text {sech}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {1}{2} \, e^{2} {\left (\frac {4 \, e^{\left (-d x - c\right )}}{-2 \, {\left (-2 i \, a e^{\left (-d x - c\right )} - a e^{\left (-2 \, d x - 2 \, c\right )} + a\right )} d} + \frac {i \, \log \left (e^{\left (-d x - c\right )} + i\right )}{a d} - \frac {i \, \log \left (i \, e^{\left (-d x - c\right )} + 1\right )}{a d}\right )} + \frac {-2 i \, f^{2} x - 2 i \, e f + {\left (d f^{2} x^{2} e^{c} + 2 \, e f e^{c} + 2 \, {\left (d e f + f^{2}\right )} x e^{c}\right )} e^{\left (d x\right )}}{a d^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 i \, a d^{2} e^{\left (d x + c\right )} - a d^{2}} - \frac {i \, {\left (d x \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac {i \, {\left (d x \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} - \frac {2 i \, f^{2} x}{a d^{2}} - \frac {i \, {\left (d^{2} x^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (d x + c\right )})\right )} f^{2}}{2 \, a d^{3}} + \frac {i \, {\left (d^{2} x^{2} \log \left (-i \, e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (i \, e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(i \, e^{\left (d x + c\right )})\right )} f^{2}}{2 \, a d^{3}} + \frac {2 i \, f^{2} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{a d^{3}} \] Input:

integrate((f*x+e)^2*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
 

Output:

-1/2*e^2*(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)) + 
(-2*I*f^2*x - 2*I*e*f + (d*f^2*x^2*e^c + 2*e*f*e^c + 2*(d*e*f + 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) - I*(d* 
x*log(I*e^(d*x + c) + 1) + dilog(-I*e^(d*x + c)))*e*f/(a*d^2) + I*(d*x*log 
(-I*e^(d*x + c) + 1) + dilog(I*e^(d*x + c)))*e*f/(a*d^2) - 2*I*f^2*x/(a*d^ 
2) - 1/2*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)))*f^2/(a*d^3) + 1/2*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)))*f^2/ 
(a*d^3) + 2*I*f^2*log(I*e^(d*x + c) + 1)/(a*d^3)
 

Giac [F]

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

integrate((f*x+e)^2*sech(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
 

Output:

integrate((f*x + e)^2*sech(d*x + c)/(I*a*sinh(d*x + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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