Optimal. Leaf size=89 \[ -\frac {\text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac {\sinh (a+b x) \cosh (a+b x)}{4 b^2}-\frac {x \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac {x \sinh ^2(a+b x)}{2 b}+\frac {x}{4 b}+\frac {x^2}{2} \]
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Rubi [A] time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 8
Rule 2190
Rule 2279
Rule 2391
Rule 2635
Rule 3718
Rule 5372
Rule 5449
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*}
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Mathematica [A] time = 0.23, size = 102, normalized size = 1.15 \[ -\frac {4 a^2-4 \text {Li}_2\left (-e^{-2 (a+b x)}\right )+8 a b x+8 a \log \left (e^{-2 (a+b x)}+1\right )+8 b x \log \left (e^{-2 (a+b x)}+1\right )+\sinh (2 (a+b x))-2 b x \cosh (2 (a+b x))-8 a \log (\cosh (a+b x))+4 b^2 x^2}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.73, size = 558, normalized size = 6.27 \[ \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 (i \, \cosh \left (b x + a\right ) + i \, \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 (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\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 ) + i\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 ) - i\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 (i \, \cosh \left (b x + a\right ) + i \, \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 (-i \, \cosh \left (b x + a\right ) - i \, \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 )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.47, size = 110, normalized size = 1.24 \[ \frac {x^{2}}{2}+\frac {\left (2 b x -1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{2}}+\frac {\left (2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{2}}+\frac {2 a x}{b}+\frac {a^{2}}{b^{2}}-\frac {x \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}-\frac {\polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 95, normalized size = 1.07 \[ x^{2} - \frac {{\left (8 \, b^{2} x^{2} e^{\left (2 \, a\right )} - {\left (2 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} - {\left (2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{16 \, b^{2}} - \frac {2 \, b x \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right )}{2 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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