Optimal. Leaf size=130 \[ -\frac{x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}-\frac{x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}+\frac{x^3}{3} \]
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Rubi [A] time = 0.18999, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5449, 5372, 3310, 30, 3718, 2190, 2531, 2282, 6589} \[ -\frac{x \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{\sinh ^2(a+b x)}{4 b^3}-\frac{x \sinh (a+b x) \cosh (a+b x)}{2 b^2}-\frac{x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{x^2}{4 b}+\frac{x^3}{3} \]
Antiderivative was successfully verified.
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Rule 5449
Rule 5372
Rule 3310
Rule 30
Rule 3718
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \sinh ^2(a+b x) \tanh (a+b x) \, dx &=\int x^2 \cosh (a+b x) \sinh (a+b x) \, dx-\int x^2 \tanh (a+b x) \, dx\\ &=\frac{x^3}{3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}-2 \int \frac{e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}} \, dx-\frac{\int x \sinh ^2(a+b x) \, dx}{b}\\ &=\frac{x^3}{3}-\frac{x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{\int x \, dx}{2 b}+\frac{2 \int x \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac{x^2}{4 b}+\frac{x^3}{3}-\frac{x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{x \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{\int \text{Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{x^2}{4 b}+\frac{x^3}{3}-\frac{x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{x \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^3}\\ &=\frac{x^2}{4 b}+\frac{x^3}{3}-\frac{x^2 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{x \text{Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}+\frac{\text{Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}-\frac{x \cosh (a+b x) \sinh (a+b x)}{2 b^2}+\frac{\sinh ^2(a+b x)}{4 b^3}+\frac{x^2 \sinh ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 2.63895, size = 154, normalized size = 1.18 \[ \frac{1}{24} \left (\frac{4 \left (6 b x \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )+3 \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )+2 b^2 x^2 \left (-\frac{2 b x}{e^{2 a}+1}-3 \log \left (e^{-2 (a+b x)}+1\right )\right )\right )}{b^3}+\frac{3 \cosh (2 b x) \left (\cosh (2 a) \left (2 b^2 x^2+1\right )-2 b x \sinh (2 a)\right )}{b^3}+\frac{3 \sinh (2 b x) \left (\sinh (2 a) \left (2 b^2 x^2+1\right )-2 b x \cosh (2 a)\right )}{b^3}-8 x^3 \tanh (a)\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.069, size = 152, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}}{3}}+{\frac{ \left ( 2\,{x}^{2}{b}^{2}-2\,bx+1 \right ){{\rm e}^{2\,bx+2\,a}}}{16\,{b}^{3}}}+{\frac{ \left ( 2\,{x}^{2}{b}^{2}+2\,bx+1 \right ){{\rm e}^{-2\,bx-2\,a}}}{16\,{b}^{3}}}+2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{3}}}-2\,{\frac{{a}^{2}x}{{b}^{2}}}-{\frac{4\,{a}^{3}}{3\,{b}^{3}}}-{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}-{\frac{x{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{{b}^{2}}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.38972, size = 186, normalized size = 1.43 \begin{align*} \frac{2}{3} \, x^{3} - \frac{{\left (16 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \,{\left (2 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} - 3 \,{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{48 \, b^{3}} - \frac{2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) -{\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})}{2 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.31393, size = 2090, normalized size = 16.08 \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(-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{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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