Optimal. Leaf size=101 \[ -\frac{4 d^2 \text{PolyLog}\left (2,-i e^{e+f x}\right )}{a f^3}-\frac{4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a f}+\frac{(c+d x)^2}{a f} \]
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Rubi [A] time = 0.218708, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac{4 d^2 \text{PolyLog}\left (2,-i e^{e+f x}\right )}{a f^3}-\frac{4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a f}+\frac{(c+d x)^2}{a f} \]
Antiderivative was successfully verified.
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Rule 3318
Rule 4184
Rule 3716
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{(c+d x)^2}{a+i a \sinh (e+f x)} \, dx &=\frac{\int (c+d x)^2 \csc ^2\left (\frac{1}{2} \left (i e+\frac{\pi }{2}\right )+\frac{i f x}{2}\right ) \, dx}{2 a}\\ &=\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}-\frac{(2 d) \int (c+d x) \coth \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=\frac{(c+d x)^2}{a f}+\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}-\frac{(4 i d) \int \frac{e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)}{1+i e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{a f}\\ &=\frac{(c+d x)^2}{a f}-\frac{4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{\left (4 d^2\right ) \int \log \left (1+i e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=\frac{(c+d x)^2}{a f}-\frac{4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{\left (4 d^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}\right )}{a f^3}\\ &=\frac{(c+d x)^2}{a f}-\frac{4 d (c+d x) \log \left (1+i e^{e+f x}\right )}{a f^2}-\frac{4 d^2 \text{Li}_2\left (-i e^{e+f x}\right )}{a f^3}+\frac{(c+d x)^2 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}\\ \end{align*}
Mathematica [A] time = 2.21886, size = 150, normalized size = 1.49 \[ \frac{2 \left (2 d^2 \text{PolyLog}\left (2,i e^{-e-f x}\right )+\frac{f^2 (c+d x)^2 \sinh \left (\frac{f x}{2}\right )}{\left (\cosh \left (\frac{e}{2}\right )+i \sinh \left (\frac{e}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )}+\frac{i f (c+d x) \left (f (c+d x)+2 d \left (1+i e^e\right ) \log \left (1-i e^{-e-f x}\right )\right )}{e^e-i}\right )}{a f^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.056, size = 227, normalized size = 2.3 \begin{align*}{\frac{2\,i \left ({d}^{2}{x}^{2}+2\,cdx+{c}^{2} \right ) }{fa \left ({{\rm e}^{fx+e}}-i \right ) }}-4\,{\frac{d\ln \left ({{\rm e}^{fx+e}}-i \right ) c}{a{f}^{2}}}+4\,{\frac{d\ln \left ({{\rm e}^{fx+e}} \right ) c}{a{f}^{2}}}+2\,{\frac{{d}^{2}{x}^{2}}{fa}}+4\,{\frac{{d}^{2}ex}{a{f}^{2}}}+2\,{\frac{{d}^{2}{e}^{2}}{{f}^{3}a}}-4\,{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) x}{a{f}^{2}}}-4\,{\frac{{d}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) e}{{f}^{3}a}}-4\,{\frac{{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{fx+e}} \right ) }{{f}^{3}a}}+4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{fx+e}}-i \right ) }{{f}^{3}a}}-4\,{\frac{{d}^{2}e\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}a}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} d^{2}{\left (\frac{2 i \, x^{2}}{a f e^{\left (f x + e\right )} - i \, a f} - 4 i \, \int \frac{x}{a f e^{\left (f x + e\right )} - i \, a f}\,{d x}\right )} + 4 \, c d{\left (\frac{x e^{\left (f x + e\right )}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac{\log \left ({\left (e^{\left (f x + e\right )} - i\right )} e^{\left (-e\right )}\right )}{a f^{2}}\right )} - \frac{2 \, c^{2}}{{\left (i \, a e^{\left (-f x - e\right )} - a\right )} f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.69656, size = 483, normalized size = 4.78 \begin{align*} \frac{2 i \, d^{2} e^{2} - 4 i \, c d e f + 2 i \, c^{2} f^{2} -{\left (4 \, d^{2} e^{\left (f x + e\right )} - 4 i \, d^{2}\right )}{\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) + 2 \,{\left (d^{2} f^{2} x^{2} + 2 \, c d f^{2} x - d^{2} e^{2} + 2 \, c d e f\right )} e^{\left (f x + e\right )} +{\left (-4 i \, d^{2} e + 4 i \, c d f + 4 \,{\left (d^{2} e - c d f\right )} e^{\left (f x + e\right )}\right )} \log \left (e^{\left (f x + e\right )} - i\right ) +{\left (4 i \, d^{2} f x + 4 i \, d^{2} e - 4 \,{\left (d^{2} f x + d^{2} e\right )} e^{\left (f x + e\right )}\right )} \log \left (i \, e^{\left (f x + e\right )} + 1\right )}{a f^{3} e^{\left (f x + e\right )} - i \, a f^{3}} \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 (d x + c\right )}^{2}}{i \, a \sinh \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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