Optimal. Leaf size=59 \[ -\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b} \]
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Rubi [A] time = 0.0605293, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5419, 4182, 2279, 2391} \[ -\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 5419
Rule 4182
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x^2 \coth (a+b x) \text{csch}(a+b x) \, dx &=-\frac{x^2 \text{csch}(a+b x)}{b}+\frac{2 \int x \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \int \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac{2 \int \log \left (1+e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac{2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}\\ \end{align*}
Mathematica [B] time = 0.830012, size = 133, normalized size = 2.25 \[ -\frac{-2 \text{PolyLog}\left (2,-e^{-a-b x}\right )+2 \text{PolyLog}\left (2,e^{-a-b x}\right )+b^2 x^2 \text{csch}(a+b x)-2 b x \log \left (1-e^{-a-b x}\right )+2 b x \log \left (e^{-a-b x}+1\right )-2 a \log \left (1-e^{-a-b x}\right )+2 a \log \left (e^{-a-b x}+1\right )+2 a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.027, size = 134, normalized size = 2.3 \begin{align*} -2\,{\frac{{x}^{2}{{\rm e}^{bx+a}}}{b \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) }}-2\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-2\,{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+2\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{3}}}+2\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+4\,{\frac{a{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.36937, size = 112, normalized size = 1.9 \begin{align*} -\frac{2 \, x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac{2 \,{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{3}} + \frac{2 \,{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.13463, size = 1022, normalized size = 17.32 \begin{align*} -\frac{2 \,{\left (b^{2} x^{2} \cosh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right ) -{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) +{\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) +{\left (b x \cosh \left (b x + a\right )^{2} + 2 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b x \sinh \left (b x + a\right )^{2} - b x\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) +{\left (a \cosh \left (b x + a\right )^{2} + 2 \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a \sinh \left (b x + a\right )^{2} - a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) -{\left ({\left (b x + a\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x + a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x + a\right )} \sinh \left (b x + a\right )^{2} - b x - a\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )\right )}}{b^{3} \cosh \left (b x + a\right )^{2} + 2 \, b^{3} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )^{2} - b^{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 x^{2} \cosh \left (b x + a\right ) \operatorname{csch}\left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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