Optimal. Leaf size=88 \[ \frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}-\frac{x^2}{2} \]
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Rubi [A] time = 0.120577, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5450, 5372, 2635, 8, 3716, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}-\frac{x^2}{2} \]
Antiderivative was successfully verified.
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Rule 5450
Rule 5372
Rule 2635
Rule 8
Rule 3716
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x \cosh ^2(a+b x) \coth (a+b x) \, dx &=\int x \coth (a+b x) \, dx+\int x \cosh (a+b x) \sinh (a+b x) \, dx\\ &=-\frac{x^2}{2}+\frac{x \sinh ^2(a+b x)}{2 b}-2 \int \frac{e^{2 (a+b x)} x}{1-e^{2 (a+b x)}} \, dx-\frac{\int \sinh ^2(a+b x) \, dx}{2 b}\\ &=-\frac{x^2}{2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{\int 1 \, dx}{4 b}-\frac{\int \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{x}{4 b}-\frac{x^2}{2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^2}\\ &=\frac{x}{4 b}-\frac{x^2}{2}+\frac{x \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac{\text{Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac{\cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac{x \sinh ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 0.233824, size = 82, normalized size = 0.93 \[ -\frac{4 \left (\text{PolyLog}\left (2,e^{-2 (a+b x)}\right )-(a+b x)^2\right )-8 (a+b x) \log \left (1-e^{-2 (a+b x)}\right )+\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))+8 a \log (\sinh (a+b x))}{8 b^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.07, size = 162, normalized size = 1.8 \begin{align*} -{\frac{{x}^{2}}{2}}+{\frac{ \left ( 2\,bx-1 \right ){{\rm e}^{2\,bx+2\,a}}}{16\,{b}^{2}}}+{\frac{ \left ( 2\,bx+1 \right ){{\rm e}^{-2\,bx-2\,a}}}{16\,{b}^{2}}}-2\,{\frac{ax}{b}}-{\frac{{a}^{2}}{{b}^{2}}}+{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{a\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{2}}}+2\,{\frac{a\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.36419, size = 153, normalized size = 1.74 \begin{align*} -x^{2} + \frac{{\left (8 \, b^{2} x^{2} e^{\left (2 \, a\right )} +{\left (2 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} +{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{16 \, b^{2}} + \frac{b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )}{b^{2}} + \frac{b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )}{b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.16096, size = 1355, normalized size = 15.4 \begin{align*} \frac{{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{4} + 4 \,{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} +{\left (2 \, b x - 1\right )} \sinh \left (b x + a\right )^{4} - 8 \,{\left (b^{2} x^{2} - 2 \, a^{2}\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (4 \, b^{2} x^{2} - 3 \,{\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} - 8 \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 2 \, b x + 16 \,{\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}\right )}{\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 16 \,{\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}\right )}{\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 16 \,{\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}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 16 \,{\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}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 16 \,{\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}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 4 \,{\left ({\left (2 \, b x - 1\right )} \cosh \left (b x + a\right )^{3} - 4 \,{\left (b^{2} x^{2} - 2 \, a^{2}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1}{16 \,{\left (b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2}\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 x \cosh \left (b x + a\right )^{3} \operatorname{csch}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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