\(\int (c+d x)^3 \tanh ^3(e+f x) \, dx\) [11]

Optimal result
Mathematica [B] (verified)
Rubi [C] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 16, antiderivative size = 237 \[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=-\frac {3 d (c+d x)^2}{2 f^2}+\frac {(c+d x)^3}{2 f}-\frac {(c+d x)^4}{4 d}+\frac {3 d^2 (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^3}+\frac {(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 d^3 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^4}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}-\frac {3 d (c+d x)^2 \tanh (e+f x)}{2 f^2}-\frac {(c+d x)^3 \tanh ^2(e+f x)}{2 f} \] Output:

-3/2*d*(d*x+c)^2/f^2+1/2*(d*x+c)^3/f-1/4*(d*x+c)^4/d+3*d^2*(d*x+c)*ln(1+ex 
p(2*f*x+2*e))/f^3+(d*x+c)^3*ln(1+exp(2*f*x+2*e))/f+3/2*d^3*polylog(2,-exp( 
2*f*x+2*e))/f^4+3/2*d*(d*x+c)^2*polylog(2,-exp(2*f*x+2*e))/f^2-3/2*d^2*(d* 
x+c)*polylog(3,-exp(2*f*x+2*e))/f^3+3/4*d^3*polylog(4,-exp(2*f*x+2*e))/f^4 
-3/2*d*(d*x+c)^2*tanh(f*x+e)/f^2-1/2*(d*x+c)^3*tanh(f*x+e)^2/f
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(496\) vs. \(2(237)=474\).

Time = 6.81 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.09 \[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=\frac {-8 c \left (1+e^{2 e}\right ) \left (3 d^2+c^2 f^2\right ) x+\frac {2 \left (c^2 f^2+2 c d f^2 x+d^2 \left (3+f^2 x^2\right )\right )^2}{d f^2}+\frac {12 d \left (1+e^{2 e}\right ) \left (d^2+c^2 f^2\right ) x \log \left (1+e^{-2 (e+f x)}\right )}{f}+12 c d^2 \left (1+e^{2 e}\right ) f x^2 \log \left (1+e^{-2 (e+f x)}\right )+4 d^3 \left (1+e^{2 e}\right ) f x^3 \log \left (1+e^{-2 (e+f x)}\right )+\frac {4 c \left (1+e^{2 e}\right ) \left (3 d^2+c^2 f^2\right ) \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac {6 d \left (1+e^{2 e}\right ) \left (d^2+c^2 f^2\right ) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f^2}-12 c d^2 \left (1+e^{2 e}\right ) x \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )-6 d^3 \left (1+e^{2 e}\right ) x^2 \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )-\frac {6 c d^2 \left (1+e^{2 e}\right ) \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )}{f}-\frac {6 d^3 \left (1+e^{2 e}\right ) x \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )}{f}-\frac {3 d^3 \left (1+e^{2 e}\right ) \operatorname {PolyLog}\left (4,-e^{-2 (e+f x)}\right )}{f^2}}{4 \left (1+e^{2 e}\right ) f^2}+\frac {(c+d x)^3 \text {sech}^2(e+f x)}{2 f}-\frac {3 \text {sech}(e) \text {sech}(e+f x) \left (c^2 d \sinh (f x)+2 c d^2 x \sinh (f x)+d^3 x^2 \sinh (f x)\right )}{2 f^2}+\frac {1}{4} x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \tanh (e) \] Input:

Integrate[(c + d*x)^3*Tanh[e + f*x]^3,x]
 

Output:

(-8*c*(1 + E^(2*e))*(3*d^2 + c^2*f^2)*x + (2*(c^2*f^2 + 2*c*d*f^2*x + d^2* 
(3 + f^2*x^2))^2)/(d*f^2) + (12*d*(1 + E^(2*e))*(d^2 + c^2*f^2)*x*Log[1 + 
E^(-2*(e + f*x))])/f + 12*c*d^2*(1 + E^(2*e))*f*x^2*Log[1 + E^(-2*(e + f*x 
))] + 4*d^3*(1 + E^(2*e))*f*x^3*Log[1 + E^(-2*(e + f*x))] + (4*c*(1 + E^(2 
*e))*(3*d^2 + c^2*f^2)*Log[1 + E^(2*(e + f*x))])/f - (6*d*(1 + E^(2*e))*(d 
^2 + c^2*f^2)*PolyLog[2, -E^(-2*(e + f*x))])/f^2 - 12*c*d^2*(1 + E^(2*e))* 
x*PolyLog[2, -E^(-2*(e + f*x))] - 6*d^3*(1 + E^(2*e))*x^2*PolyLog[2, -E^(- 
2*(e + f*x))] - (6*c*d^2*(1 + E^(2*e))*PolyLog[3, -E^(-2*(e + f*x))])/f - 
(6*d^3*(1 + E^(2*e))*x*PolyLog[3, -E^(-2*(e + f*x))])/f - (3*d^3*(1 + E^(2 
*e))*PolyLog[4, -E^(-2*(e + f*x))])/f^2)/(4*(1 + E^(2*e))*f^2) + ((c + d*x 
)^3*Sech[e + f*x]^2)/(2*f) - (3*Sech[e]*Sech[e + f*x]*(c^2*d*Sinh[f*x] + 2 
*c*d^2*x*Sinh[f*x] + d^3*x^2*Sinh[f*x]))/(2*f^2) + (x*(4*c^3 + 6*c^2*d*x + 
 4*c*d^2*x^2 + d^3*x^3)*Tanh[e])/4
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 1.59 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.19, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.438, Rules used = {3042, 26, 4203, 25, 26, 3042, 25, 26, 4201, 2620, 3011, 4203, 17, 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.

\(\displaystyle \int (c+d x)^3 \tanh ^3(e+f x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int i (c+d x)^3 \tan (i e+i f x)^3dx\)

\(\Big \downarrow \) 26

\(\displaystyle i \int (c+d x)^3 \tan (i e+i f x)^3dx\)

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

\(\displaystyle i \left (-i \int -i (c+d x)^3 \tan (i e+i f x)dx-\frac {3 i d \int -(c+d x)^2 \tan (i e+i f x)^2dx}{2 f}+\frac {i (c+d x)^3 \tanh ^2(e+f x)}{2 f}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle i \left (-i \int -i (c+d x)^3 \tan (i e+i f x)dx+\frac {3 i d \int (c+d x)^2 \tan (i e+i f x)^2dx}{2 f}+\frac {i (c+d x)^3 \tanh ^2(e+f x)}{2 f}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle i \left (-\int (c+d x)^3 \tan (i e+i f x)dx+\frac {3 i d \int (c+d x)^2 \tan (i e+i f x)^2dx}{2 f}+\frac {i (c+d x)^3 \tanh ^2(e+f x)}{2 f}\right )\)

\(\Big \downarrow \) 4201

\(\displaystyle i \left (-2 i \int \frac {e^{2 (e+f x)} (c+d x)^3}{1+e^{2 (e+f x)}}dx+\frac {3 i d \int (c+d x)^2 \tan (i e+i f x)^2dx}{2 f}+\frac {i (c+d x)^3 \tanh ^2(e+f x)}{2 f}+\frac {i (c+d x)^4}{4 d}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle i \left (-2 i \left (\frac {(c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {3 d \int (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )dx}{2 f}\right )+\frac {3 i d \int (c+d x)^2 \tan (i e+i f x)^2dx}{2 f}+\frac {i (c+d x)^3 \tanh ^2(e+f x)}{2 f}+\frac {i (c+d x)^4}{4 d}\right )\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4201

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

Int[(c + d*x)^3*Tanh[e + f*x]^3,x]
 

Output:

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

Defintions of rubi rules used

rule 17
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4203
Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 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)

Leaf count of result is larger than twice the leaf count of optimal. \(684\) vs. \(2(219)=438\).

Time = 0.62 (sec) , antiderivative size = 685, normalized size of antiderivative = 2.89

method result size
risch \(\frac {c^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {6 c^{2} d e x}{f}-\frac {3 d^{3} e^{4}}{2 f^{4}}-\frac {2 c^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {3 d^{2} c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {6 e \,d^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {3 d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f^{3}}+\frac {d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{3}}{f}-d^{2} c \,x^{3}-\frac {3 d \,c^{2} x^{2}}{2}+c^{3} x -\frac {6 d^{3} e x}{f^{3}}-\frac {3 c^{2} d \,e^{2}}{f^{2}}+\frac {4 c \,d^{2} e^{3}}{f^{3}}+\frac {3 d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{2 f^{2}}-\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right ) x}{2 f^{3}}+\frac {2 d^{3} e^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {2 d^{3} f \,x^{3} {\mathrm e}^{2 f x +2 e}+6 c \,d^{2} f \,x^{2} {\mathrm e}^{2 f x +2 e}+6 c^{2} d f x \,{\mathrm e}^{2 f x +2 e}+3 d^{3} x^{2} {\mathrm e}^{2 f x +2 e}+2 c^{3} f \,{\mathrm e}^{2 f x +2 e}+6 c \,d^{2} x \,{\mathrm e}^{2 f x +2 e}+3 c^{2} d \,{\mathrm e}^{2 f x +2 e}+3 d^{3} x^{2}+6 c \,d^{2} x +3 d \,c^{2}}{f^{2} \left (1+{\mathrm e}^{2 f x +2 e}\right )^{2}}+\frac {6 c \,d^{2} e^{2} x}{f^{2}}+\frac {3 d^{2} c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}-\frac {2 d^{3} e^{3} x}{f^{3}}+\frac {3 c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}-\frac {3 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {3 c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {6 c \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {6 c^{2} d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {3 c^{2} d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}-\frac {d^{3} x^{4}}{4}+\frac {c^{4}}{4 d}+\frac {3 d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{4}}+\frac {3 d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 f x +2 e}\right )}{4 f^{4}}-\frac {3 d^{3} x^{2}}{f^{2}}-\frac {3 e^{2} d^{3}}{f^{4}}\) \(685\)

Input:

int((d*x+c)^3*tanh(f*x+e)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*c^3*ln(1+exp(2*f*x+2*e))-6/f*c^2*d*e*x-3/2/f^4*d^3*e^4-2/f*c^3*ln(exp( 
f*x+e))-6/f^3*d^2*c*ln(exp(f*x+e))+3/f^3*d^2*c*ln(1+exp(2*f*x+2*e))+6/f^4* 
e*d^3*ln(exp(f*x+e))+3/f^3*d^3*ln(1+exp(2*f*x+2*e))*x+1/f*d^3*ln(1+exp(2*f 
*x+2*e))*x^3-d^2*c*x^3-3/2*d*c^2*x^2+c^3*x-6/f^3*d^3*e*x-3/f^2*c^2*d*e^2+4 
/f^3*c*d^2*e^3+3/2/f^2*d^3*polylog(2,-exp(2*f*x+2*e))*x^2-3/2/f^3*d^3*poly 
log(3,-exp(2*f*x+2*e))*x+2/f^4*d^3*e^3*ln(exp(f*x+e))+(2*d^3*f*x^3*exp(2*f 
*x+2*e)+6*c*d^2*f*x^2*exp(2*f*x+2*e)+6*c^2*d*f*x*exp(2*f*x+2*e)+3*d^3*x^2* 
exp(2*f*x+2*e)+2*c^3*f*exp(2*f*x+2*e)+6*c*d^2*x*exp(2*f*x+2*e)+3*c^2*d*exp 
(2*f*x+2*e)+3*d^3*x^2+6*c*d^2*x+3*d*c^2)/f^2/(1+exp(2*f*x+2*e))^2+6/f^2*c* 
d^2*e^2*x+3/f*d^2*c*ln(1+exp(2*f*x+2*e))*x^2-2/f^3*d^3*e^3*x+3/2/f^2*c^2*d 
*polylog(2,-exp(2*f*x+2*e))-3/2/f^3*c*d^2*polylog(3,-exp(2*f*x+2*e))+3/f^2 
*c*d^2*polylog(2,-exp(2*f*x+2*e))*x-6/f^3*c*d^2*e^2*ln(exp(f*x+e))+6/f^2*c 
^2*d*e*ln(exp(f*x+e))+3/f*c^2*d*ln(1+exp(2*f*x+2*e))*x-1/4*d^3*x^4+1/4/d*c 
^4+3/2*d^3*polylog(2,-exp(2*f*x+2*e))/f^4+3/4*d^3*polylog(4,-exp(2*f*x+2*e 
))/f^4-3/f^2*d^3*x^2-3/f^4*e^2*d^3
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 5569, normalized size of antiderivative = 23.50 \[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^3*tanh(f*x+e)^3,x, algorithm="fricas")
 

Output:

Too large to include
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=\int \left (c + d x\right )^{3} \tanh ^{3}{\left (e + f x \right )}\, dx \] Input:

integrate((d*x+c)**3*tanh(f*x+e)**3,x)
 

Output:

Integral((c + d*x)**3*tanh(e + f*x)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 595 vs. \(2 (217) = 434\).

Time = 0.21 (sec) , antiderivative size = 595, normalized size of antiderivative = 2.51 \[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^3*tanh(f*x+e)^3,x, algorithm="maxima")
 

Output:

c^3*(x + e/f + log(e^(-2*f*x - 2*e) + 1)/f + 2*e^(-2*f*x - 2*e)/(f*(2*e^(- 
2*f*x - 2*e) + e^(-4*f*x - 4*e) + 1))) - 6*c*d^2*x/f^2 + 3/2*(2*f^2*x^2*lo 
g(e^(2*f*x + 2*e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - polylog(3, -e^(2* 
f*x + 2*e)))*c*d^2/f^3 + 3*c*d^2*log(e^(2*f*x + 2*e) + 1)/f^3 + 1/4*(d^3*f 
^2*x^4 + 4*c*d^2*f^2*x^3 + 24*c*d^2*x + 12*c^2*d + 6*(c^2*d*f^2 + 2*d^3)*x 
^2 + (d^3*f^2*x^4*e^(4*e) + 4*c*d^2*f^2*x^3*e^(4*e) + 6*c^2*d*f^2*x^2*e^(4 
*e))*e^(4*f*x) + 2*(d^3*f^2*x^4*e^(2*e) + 4*(c*d^2*f^2*e^(2*e) + d^3*f*e^( 
2*e))*x^3 + 6*c^2*d*e^(2*e) + 6*(c^2*d*f^2*e^(2*e) + 2*c*d^2*f*e^(2*e) + d 
^3*e^(2*e))*x^2 + 12*(c^2*d*f*e^(2*e) + c*d^2*e^(2*e))*x)*e^(2*f*x))/(f^2* 
e^(4*f*x + 4*e) + 2*f^2*e^(2*f*x + 2*e) + f^2) + 1/3*(4*f^3*x^3*log(e^(2*f 
*x + 2*e) + 1) + 6*f^2*x^2*dilog(-e^(2*f*x + 2*e)) - 6*f*x*polylog(3, -e^( 
2*f*x + 2*e)) + 3*polylog(4, -e^(2*f*x + 2*e)))*d^3/f^4 + 3/2*(c^2*d*f^2 + 
 d^3)*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))/f^4 - 1/2 
*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*(c^2*d*f^2 + d^3)*f^2*x^2)/f^4
 

Giac [F]

\[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=\int { {\left (d x + c\right )}^{3} \tanh \left (f x + e\right )^{3} \,d x } \] Input:

integrate((d*x+c)^3*tanh(f*x+e)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*tanh(f*x + e)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=\int {\mathrm {tanh}\left (e+f\,x\right )}^3\,{\left (c+d\,x\right )}^3 \,d x \] Input:

int(tanh(e + f*x)^3*(c + d*x)^3,x)
 

Output:

int(tanh(e + f*x)^3*(c + d*x)^3, x)
 

Reduce [F]

\[ \int (c+d x)^3 \tanh ^3(e+f x) \, dx=\text {too large to display} \] Input:

int((d*x+c)^3*tanh(f*x+e)^3,x)
 

Output:

( - 128*e**(4*e + 4*f*x)*int(x**3/(e**(6*e + 6*f*x) + 3*e**(4*e + 4*f*x) + 
 3*e**(2*e + 2*f*x) + 1),x)*d**3*f**4 - 384*e**(4*e + 4*f*x)*int(x**2/(e** 
(6*e + 6*f*x) + 3*e**(4*e + 4*f*x) + 3*e**(2*e + 2*f*x) + 1),x)*c*d**2*f** 
4 - 288*e**(4*e + 4*f*x)*int(x**2/(e**(6*e + 6*f*x) + 3*e**(4*e + 4*f*x) + 
 3*e**(2*e + 2*f*x) + 1),x)*d**3*f**3 - 384*e**(4*e + 4*f*x)*int(x/(e**(6* 
e + 6*f*x) + 3*e**(4*e + 4*f*x) + 3*e**(2*e + 2*f*x) + 1),x)*c**2*d*f**4 - 
 576*e**(4*e + 4*f*x)*int(x/(e**(6*e + 6*f*x) + 3*e**(4*e + 4*f*x) + 3*e** 
(2*e + 2*f*x) + 1),x)*c*d**2*f**3 - 720*e**(4*e + 4*f*x)*int(x/(e**(6*e + 
6*f*x) + 3*e**(4*e + 4*f*x) + 3*e**(2*e + 2*f*x) + 1),x)*d**3*f**2 + 64*e* 
*(4*e + 4*f*x)*log(e**(2*e + 2*f*x) + 1)*c**3*f**3 + 144*e**(4*e + 4*f*x)* 
log(e**(2*e + 2*f*x) + 1)*c**2*d*f**2 + 360*e**(4*e + 4*f*x)*log(e**(2*e + 
 2*f*x) + 1)*c*d**2*f + 234*e**(4*e + 4*f*x)*log(e**(2*e + 2*f*x) + 1)*d** 
3 - 64*e**(4*e + 4*f*x)*c**3*f**4*x - 64*e**(4*e + 4*f*x)*c**3*f**3 + 96*e 
**(4*e + 4*f*x)*c**2*d*f**4*x**2 - 288*e**(4*e + 4*f*x)*c**2*d*f**3*x - 72 
*e**(4*e + 4*f*x)*c**2*d*f**2 + 64*e**(4*e + 4*f*x)*c*d**2*f**4*x**3 - 720 
*e**(4*e + 4*f*x)*c*d**2*f**2*x + 36*e**(4*e + 4*f*x)*c*d**2*f + 16*e**(4* 
e + 4*f*x)*d**3*f**4*x**4 - 468*e**(4*e + 4*f*x)*d**3*f*x + 45*e**(4*e + 4 
*f*x)*d**3 - 256*e**(2*e + 2*f*x)*int(x**3/(e**(6*e + 6*f*x) + 3*e**(4*e + 
 4*f*x) + 3*e**(2*e + 2*f*x) + 1),x)*d**3*f**4 - 768*e**(2*e + 2*f*x)*int( 
x**2/(e**(6*e + 6*f*x) + 3*e**(4*e + 4*f*x) + 3*e**(2*e + 2*f*x) + 1),x...