Optimal. Leaf size=287 \[ \frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}+\frac{2 f^2 \cosh (c+d x)}{a d^3}-\frac{i f^2 \sinh (c+d x) \cosh (c+d x)}{4 a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^2 \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f} \]
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Rubi [A] time = 0.546854, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.419, Rules used = {5557, 3311, 32, 2635, 8, 3296, 2638, 3318, 4184, 3716, 2190, 2279, 2391} \[ \frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}+\frac{2 f^2 \cosh (c+d x)}{a d^3}-\frac{i f^2 \sinh (c+d x) \cosh (c+d x)}{4 a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x)^2 \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f} \]
Antiderivative was successfully verified.
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Rule 5557
Rule 3311
Rule 32
Rule 2635
Rule 8
Rule 3296
Rule 2638
Rule 3318
Rule 4184
Rule 3716
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac{(e+f x)^2 \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac{i \int (e+f x)^2 \sinh ^2(c+d x) \, dx}{a}\\ &=-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}+\frac{i \int (e+f x)^2 \, dx}{2 a}+\frac{\int (e+f x)^2 \sinh (c+d x) \, dx}{a}-\frac{\left (i f^2\right ) \int \sinh ^2(c+d x) \, dx}{2 a d^2}-\int \frac{(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\\ &=\frac{i (e+f x)^3}{6 a f}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-i \int \frac{(e+f x)^2}{a+i a \sinh (c+d x)} \, dx+\frac{i \int (e+f x)^2 \, dx}{a}-\frac{(2 f) \int (e+f x) \cosh (c+d x) \, dx}{a d}+\frac{\left (i f^2\right ) \int 1 \, dx}{4 a d^2}\\ &=\frac{i f^2 x}{4 a d^2}+\frac{i (e+f x)^3}{2 a f}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{i \int (e+f x)^2 \csc ^2\left (\frac{1}{2} \left (i c+\frac{\pi }{2}\right )+\frac{i d x}{2}\right ) \, dx}{2 a}+\frac{\left (2 f^2\right ) \int \sinh (c+d x) \, dx}{a d^2}\\ &=\frac{i f^2 x}{4 a d^2}+\frac{i (e+f x)^3}{2 a f}+\frac{2 f^2 \cosh (c+d x)}{a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(2 i f) \int (e+f x) \coth \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}\\ &=\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f}+\frac{2 f^2 \cosh (c+d x)}{a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(4 f) \int \frac{e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )} (e+f x)}{1+i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{a d}\\ &=\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f}+\frac{2 f^2 \cosh (c+d x)}{a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (4 i f^2\right ) \int \log \left (1+i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f}+\frac{2 f^2 \cosh (c+d x)}{a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (4 i f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right )}{a d^3}\\ &=\frac{i f^2 x}{4 a d^2}-\frac{i (e+f x)^2}{a d}+\frac{i (e+f x)^3}{2 a f}+\frac{2 f^2 \cosh (c+d x)}{a d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{a d}+\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{4 i f^2 \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}-\frac{2 f (e+f x) \sinh (c+d x)}{a d^2}-\frac{i f^2 \cosh (c+d x) \sinh (c+d x)}{4 a d^3}-\frac{i (e+f x)^2 \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f (e+f x) \sinh ^2(c+d x)}{2 a d^2}-\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}\\ \end{align*}
Mathematica [B] time = 5.13595, size = 1661, normalized size = 5.79 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.128, size = 508, normalized size = 1.8 \begin{align*}{\frac{-2\,i{f}^{2}{c}^{2}}{a{d}^{3}}}+{\frac{{\frac{i}{16}} \left ( 2\,{f}^{2}{x}^{2}{d}^{2}+4\,{d}^{2}efx+2\,{d}^{2}{e}^{2}+2\,d{f}^{2}x+2\,efd+{f}^{2} \right ){{\rm e}^{-2\,dx-2\,c}}}{a{d}^{3}}}-{\frac{2\,i{f}^{2}{x}^{2}}{da}}-{\frac{4\,i\ln \left ({{\rm e}^{dx+c}} \right ) ef}{a{d}^{2}}}+{\frac{ \left ({f}^{2}{x}^{2}{d}^{2}+2\,{d}^{2}efx+{d}^{2}{e}^{2}-2\,d{f}^{2}x-2\,efd+2\,{f}^{2} \right ){{\rm e}^{dx+c}}}{2\,a{d}^{3}}}+{\frac{ \left ({f}^{2}{x}^{2}{d}^{2}+2\,{d}^{2}efx+{d}^{2}{e}^{2}+2\,d{f}^{2}x+2\,efd+2\,{f}^{2} \right ){{\rm e}^{-dx-c}}}{2\,a{d}^{3}}}+{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}+2\,{\frac{{x}^{2}{f}^{2}+2\,efx+{e}^{2}}{da \left ({{\rm e}^{dx+c}}-i \right ) }}-{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-{\frac{{\frac{i}{16}} \left ( 2\,{f}^{2}{x}^{2}{d}^{2}+4\,{d}^{2}efx+2\,{d}^{2}{e}^{2}-2\,d{f}^{2}x-2\,efd+{f}^{2} \right ){{\rm e}^{2\,dx+2\,c}}}{a{d}^{3}}}+{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{\frac{3\,i}{2}}{e}^{2}x}{a}}-{\frac{4\,i{f}^{2}cx}{a{d}^{2}}}+{\frac{{\frac{i}{2}}{x}^{3}{f}^{2}}{a}}+{\frac{4\,i{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{{\frac{3\,i}{2}}ef{x}^{2}}{a}}+{\frac{4\,i\ln \left ({{\rm e}^{dx+c}}-i \right ) ef}{a{d}^{2}}}+{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.67821, size = 1419, normalized size = 4.94 \begin{align*} \frac{2 \, d^{2} f^{2} x^{2} + 2 \, d^{2} e^{2} + 2 \, d e f + f^{2} + 2 \,{\left (2 \, d^{2} e f + d f^{2}\right )} x +{\left (64 i \, f^{2} e^{\left (3 \, d x + 3 \, c\right )} + 64 \, f^{2} e^{\left (2 \, d x + 2 \, c\right )}\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) +{\left (-2 i \, d^{2} f^{2} x^{2} - 2 i \, d^{2} e^{2} + 2 i \, d e f - i \, f^{2} +{\left (-4 i \, d^{2} e f + 2 i \, d f^{2}\right )} x\right )} e^{\left (5 \, d x + 5 \, c\right )} +{\left (6 \, d^{2} f^{2} x^{2} + 6 \, d^{2} e^{2} - 14 \, d e f + 15 \, f^{2} + 2 \,{\left (6 \, d^{2} e f - 7 \, d f^{2}\right )} x\right )} e^{\left (4 \, d x + 4 \, c\right )} +{\left (8 i \, d^{3} f^{2} x^{3} - 8 i \, d^{2} e^{2} +{\left (-64 i \, c + 16 i\right )} d e f +{\left (32 i \, c^{2} - 16 i\right )} f^{2} +{\left (24 i \, d^{3} e f - 40 i \, d^{2} f^{2}\right )} x^{2} +{\left (24 i \, d^{3} e^{2} - 80 i \, d^{2} e f + 16 i \, d f^{2}\right )} x\right )} e^{\left (3 \, d x + 3 \, c\right )} + 8 \,{\left (d^{3} f^{2} x^{3} + 5 \, d^{2} e^{2} - 2 \,{\left (4 \, c - 1\right )} d e f + 2 \,{\left (2 \, c^{2} + 1\right )} f^{2} +{\left (3 \, d^{3} e f + d^{2} f^{2}\right )} x^{2} +{\left (3 \, d^{3} e^{2} + 2 \, d^{2} e f + 2 \, d f^{2}\right )} x\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (-6 i \, d^{2} f^{2} x^{2} - 6 i \, d^{2} e^{2} - 14 i \, d e f - 15 i \, f^{2} +{\left (-12 i \, d^{2} e f - 14 i \, d f^{2}\right )} x\right )} e^{\left (d x + c\right )} +{\left ({\left (64 i \, d e f - 64 i \, c f^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )} + 64 \,{\left (d e f - c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left ({\left (64 i \, d f^{2} x + 64 i \, c f^{2}\right )} e^{\left (3 \, d x + 3 \, c\right )} + 64 \,{\left (d f^{2} x + c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{16 \, a d^{3} e^{\left (3 \, d x + 3 \, c\right )} - 16 i \, a d^{3} e^{\left (2 \, d x + 2 \, c\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2} \sinh \left (d x + c\right )^{3}}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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