Optimal. Leaf size=130 \[ \frac {\text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {\sinh ^2(a+b x)}{4 b^3}-\frac {x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^2}-\frac {x \sinh (a+b x) \cosh (a+b x)}{2 b^2}-\frac {x^2 \log \left (e^{2 (a+b x)}+1\right )}{b}+\frac {x^2 \sinh ^2(a+b x)}{2 b}+\frac {x^2}{4 b}+\frac {x^3}{3} \]
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Rubi [A] time = 0.19, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 30
Rule 2190
Rule 2282
Rule 2531
Rule 3310
Rule 3718
Rule 5372
Rule 5449
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*}
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Mathematica [A] time = 2.73, size = 154, normalized size = 1.18 \[ \frac {1}{24} \left (\frac {4 \left (2 b^2 x^2 \left (-\frac {2 b x}{e^{2 a}+1}-3 \log \left (e^{-2 (a+b x)}+1\right )\right )+6 b x \text {Li}_2\left (-e^{-2 (a+b x)}\right )+3 \text {Li}_3\left (-e^{-2 (a+b x)}\right )\right )}{b^3}+\frac {3 \cosh (2 b x) \left (\cosh (2 a) \left (2 b^2 x^2+1\right )-2 b x \sinh (2 a)\right )}{b^3}+\frac {3 \sinh (2 b x) \left (\sinh (2 a) \left (2 b^2 x^2+1\right )-2 b x \cosh (2 a)\right )}{b^3}-8 x^3 \tanh (a)\right ) \]
Antiderivative was successfully verified.
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fricas [C] time = 0.79, size = 789, normalized size = 6.07 \[ \text {result too large to display} \]
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 \[ \int x^{2} \operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.48, size = 152, normalized size = 1.17 \[ \frac {x^{3}}{3}+\frac {\left (2 x^{2} b^{2}-2 b x +1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{3}}+\frac {\left (2 x^{2} b^{2}+2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{3}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 a^{2} x}{b^{2}}-\frac {4 a^{3}}{3 b^{3}}-\frac {x^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}-\frac {x \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {\polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 138, normalized size = 1.06 \[ \frac {2}{3} \, x^{3} - \frac {{\left (16 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \, {\left (2 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 2 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} - 3 \, {\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{48 \, b^{3}} - \frac {2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})}{2 \, b^{3}} \]
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 \[ \int \frac {x^2\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
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 \[ \int x^{2} \sinh ^{3}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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