Optimal. Leaf size=119 \[ \frac{x \text{PolyLog}\left (3,-e^{-a-b x}\right )}{b^2}-\frac{x \text{PolyLog}\left (3,e^{-a-b x}\right )}{b^2}+\frac{\text{PolyLog}\left (4,-e^{-a-b x}\right )}{b^3}-\frac{\text{PolyLog}\left (4,e^{-a-b x}\right )}{b^3}+\frac{x^2 \text{PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac{x^2 \text{PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]
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Rubi [A] time = 0.0994868, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6214, 2531, 6609, 2282, 6589} \[ \frac{x \text{PolyLog}\left (3,-e^{-a-b x}\right )}{b^2}-\frac{x \text{PolyLog}\left (3,e^{-a-b x}\right )}{b^2}+\frac{\text{PolyLog}\left (4,-e^{-a-b x}\right )}{b^3}-\frac{\text{PolyLog}\left (4,e^{-a-b x}\right )}{b^3}+\frac{x^2 \text{PolyLog}\left (2,-e^{-a-b x}\right )}{2 b}-\frac{x^2 \text{PolyLog}\left (2,e^{-a-b x}\right )}{2 b} \]
Antiderivative was successfully verified.
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Rule 6214
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \coth ^{-1}\left (e^{a+b x}\right ) \, dx &=-\left (\frac{1}{2} \int x^2 \log \left (1-e^{-a-b x}\right ) \, dx\right )+\frac{1}{2} \int x^2 \log \left (1+e^{-a-b x}\right ) \, dx\\ &=\frac{x^2 \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x^2 \text{Li}_2\left (e^{-a-b x}\right )}{2 b}-\frac{\int x \text{Li}_2\left (-e^{-a-b x}\right ) \, dx}{b}+\frac{\int x \text{Li}_2\left (e^{-a-b x}\right ) \, dx}{b}\\ &=\frac{x^2 \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x^2 \text{Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac{x \text{Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac{x \text{Li}_3\left (e^{-a-b x}\right )}{b^2}-\frac{\int \text{Li}_3\left (-e^{-a-b x}\right ) \, dx}{b^2}+\frac{\int \text{Li}_3\left (e^{-a-b x}\right ) \, dx}{b^2}\\ &=\frac{x^2 \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x^2 \text{Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac{x \text{Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac{x \text{Li}_3\left (e^{-a-b x}\right )}{b^2}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{-a-b x}\right )}{b^3}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{-a-b x}\right )}{b^3}\\ &=\frac{x^2 \text{Li}_2\left (-e^{-a-b x}\right )}{2 b}-\frac{x^2 \text{Li}_2\left (e^{-a-b x}\right )}{2 b}+\frac{x \text{Li}_3\left (-e^{-a-b x}\right )}{b^2}-\frac{x \text{Li}_3\left (e^{-a-b x}\right )}{b^2}+\frac{\text{Li}_4\left (-e^{-a-b x}\right )}{b^3}-\frac{\text{Li}_4\left (e^{-a-b x}\right )}{b^3}\\ \end{align*}
Mathematica [A] time = 0.0298635, size = 149, normalized size = 1.25 \[ \frac{-3 b^2 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )+3 b^2 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )+6 b x \text{PolyLog}\left (3,-e^{a+b x}\right )-6 b x \text{PolyLog}\left (3,e^{a+b x}\right )-6 \text{PolyLog}\left (4,-e^{a+b x}\right )+6 \text{PolyLog}\left (4,e^{a+b x}\right )+b^3 x^3 \log \left (1-e^{a+b x}\right )-b^3 x^3 \log \left (e^{a+b x}+1\right )+2 b^3 x^3 \coth ^{-1}\left (e^{a+b x}\right )}{6 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.076, size = 185, normalized size = 1.6 \begin{align*}{\frac{{x}^{3}{\rm arccoth} \left ({{\rm e}^{bx+a}}\right )}{3}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ){x}^{2}}{2\,b}}-{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}-{\frac{\ln \left ({{\rm e}^{bx+a}}+1 \right ){x}^{3}}{6}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ){x}^{2}}{2\,b}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) x}{{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{3}}{6}}-{\frac{\ln \left ({{\rm e}^{bx+a}}+1 \right ){a}^{3}}{6\,{b}^{3}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{3}}{6\,{b}^{3}}}+{\frac{{a}^{3}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{3\,{b}^{3}}}-{\frac{{\it polylog} \left ( 4,-{{\rm e}^{bx+a}} \right ) }{{b}^{3}}}+{\frac{{\it polylog} \left ( 4,{{\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 = 1.12452, size = 192, normalized size = 1.61 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arcoth}\left (e^{\left (b x + a\right )}\right ) - \frac{1}{6} \, b{\left (\frac{b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \,{\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} - \frac{b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x{\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \,{\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.75526, size = 730, normalized size = 6.13 \begin{align*} \frac{b^{3} x^{3} \log \left (\frac{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1}{\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1}\right ) - b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, b^{2} x^{2}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, b x{\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, b x{\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) +{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \,{\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \,{\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{6 \, 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} \operatorname{arcoth}\left (e^{\left (b x + a\right )}\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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