Optimal. Leaf size=89 \[ -\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}-\frac{x \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}+\frac{x^2}{2} \]
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Rubi [A] time = 0.116242, antiderivative size = 89, 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 = {5449, 5372, 2635, 8, 3718, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{\sinh (a+b x) \cosh (a+b x)}{4 b^2}-\frac{x \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x \sinh ^2(a+b x)}{2 b}+\frac{x}{4 b}+\frac{x^2}{2} \]
Antiderivative was successfully verified.
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Rule 5449
Rule 5372
Rule 2635
Rule 8
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x \sinh ^2(a+b x) \tanh (a+b x) \, dx &=\int x \cosh (a+b x) \sinh (a+b x) \, dx-\int x \tanh (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 = 0.221195, size = 102, normalized size = 1.15 \[ -\frac{-4 \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+4 a^2+8 a b x+8 a \log \left (e^{-2 (a+b x)}+1\right )+8 b x \log \left (e^{-2 (a+b x)}+1\right )+\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))-8 a \log (\cosh (a+b x))+4 b^2 x^2}{8 b^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.065, size = 110, normalized size = 1.2 \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{x\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{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 = 1.36182, size = 128, normalized size = 1.44 \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{2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.28533, size = 1561, normalized size = 17.54 \begin{align*} \text{result too large to display} \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 \sinh ^{3}{\left (a + b x \right )} \operatorname{sech}{\left (a + b 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{sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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