Optimal. Leaf size=51 \[ \frac{1}{2} x \text{PolyLog}\left (2,-e^{-x}\right )-\frac{1}{2} x \text{PolyLog}\left (2,e^{-x}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-x}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-x}\right ) \]
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Rubi [A] time = 0.0474447, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {6214, 2531, 2282, 6589} \[ \frac{1}{2} x \text{PolyLog}\left (2,-e^{-x}\right )-\frac{1}{2} x \text{PolyLog}\left (2,e^{-x}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-x}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-x}\right ) \]
Antiderivative was successfully verified.
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Rule 6214
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x \coth ^{-1}\left (e^x\right ) \, dx &=-\left (\frac{1}{2} \int x \log \left (1-e^{-x}\right ) \, dx\right )+\frac{1}{2} \int x \log \left (1+e^{-x}\right ) \, dx\\ &=\frac{1}{2} x \text{Li}_2\left (-e^{-x}\right )-\frac{1}{2} x \text{Li}_2\left (e^{-x}\right )-\frac{1}{2} \int \text{Li}_2\left (-e^{-x}\right ) \, dx+\frac{1}{2} \int \text{Li}_2\left (e^{-x}\right ) \, dx\\ &=\frac{1}{2} x \text{Li}_2\left (-e^{-x}\right )-\frac{1}{2} x \text{Li}_2\left (e^{-x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{-x}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{-x}\right )\\ &=\frac{1}{2} x \text{Li}_2\left (-e^{-x}\right )-\frac{1}{2} x \text{Li}_2\left (e^{-x}\right )+\frac{\text{Li}_3\left (-e^{-x}\right )}{2}-\frac{\text{Li}_3\left (e^{-x}\right )}{2}\\ \end{align*}
Mathematica [A] time = 0.0223064, size = 71, normalized size = 1.39 \[ \frac{1}{4} \left (-2 x \text{PolyLog}\left (2,-e^x\right )+2 x \text{PolyLog}\left (2,e^x\right )+2 \text{PolyLog}\left (3,-e^x\right )-2 \text{PolyLog}\left (3,e^x\right )+x^2 \log \left (1-e^x\right )-x^2 \log \left (e^x+1\right )+2 x^2 \coth ^{-1}\left (e^x\right )\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.049, size = 62, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}{\rm arccoth} \left ({{\rm e}^{x}}\right )}{2}}-{\frac{{x}^{2}\ln \left ({{\rm e}^{x}}+1 \right ) }{4}}-{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{x}} \right ) }{2}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{x}} \right ) }{2}}+{\frac{{x}^{2}\ln \left ( 1-{{\rm e}^{x}} \right ) }{4}}+{\frac{x{\it polylog} \left ( 2,{{\rm e}^{x}} \right ) }{2}}-{\frac{{\it polylog} \left ( 3,{{\rm e}^{x}} \right ) }{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.1231, size = 80, normalized size = 1.57 \begin{align*} \frac{1}{2} \, x^{2} \operatorname{arcoth}\left (e^{x}\right ) - \frac{1}{4} \, x^{2} \log \left (e^{x} + 1\right ) + \frac{1}{4} \, x^{2} \log \left (-e^{x} + 1\right ) - \frac{1}{2} \, x{\rm Li}_2\left (-e^{x}\right ) + \frac{1}{2} \, x{\rm Li}_2\left (e^{x}\right ) + \frac{1}{2} \,{\rm Li}_{3}(-e^{x}) - \frac{1}{2} \,{\rm Li}_{3}(e^{x}) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.6609, size = 374, normalized size = 7.33 \begin{align*} \frac{1}{4} \, x^{2} \log \left (\frac{\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) + \sinh \left (x\right ) - 1}\right ) - \frac{1}{4} \, x^{2} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac{1}{4} \, x^{2} \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) + \frac{1}{2} \, x{\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - \frac{1}{2} \, x{\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) - \frac{1}{2} \,{\rm polylog}\left (3, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + \frac{1}{2} \,{\rm polylog}\left (3, -\cosh \left (x\right ) - \sinh \left (x\right )\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{acoth}{\left (e^{x} \right )}\, dx \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 \operatorname{arcoth}\left (e^{x}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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