Optimal. Leaf size=89 \[ -\frac {3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}+\frac {3 x^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac {x^3 \tanh (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \]
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Rubi [A] time = 0.18, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3720, 3718, 2190, 2531, 2282, 6589, 30} \[ \frac {3 x \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \log \left (e^{2 (a+b x)}+1\right )}{b^2}-\frac {x^3 \tanh (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 3720
Rule 6589
Rubi steps
\begin {align*} \int x^3 \tanh ^2(a+b x) \, dx &=-\frac {x^3 \tanh (a+b x)}{b}+\frac {3 \int x^2 \tanh (a+b x) \, dx}{b}+\int x^3 \, dx\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \tanh (a+b x)}{b}+\frac {6 \int \frac {e^{2 (a+b x)} x^2}{1+e^{2 (a+b x)}} \, dx}{b}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}-\frac {x^3 \tanh (a+b x)}{b}-\frac {6 \int x \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {x^3 \tanh (a+b x)}{b}-\frac {3 \int \text {Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {x^3 \tanh (a+b x)}{b}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{2 b^4}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}+\frac {3 x^2 \log \left (1+e^{2 (a+b x)}\right )}{b^2}+\frac {3 x \text {Li}_2\left (-e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^4}-\frac {x^3 \tanh (a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 4.65, size = 104, normalized size = 1.17 \[ \frac {2 b^2 x^2 \left (\frac {2 b x}{e^{2 a}+1}+3 \log \left (e^{-2 (a+b x)}+1\right )\right )-6 b x \text {Li}_2\left (-e^{-2 (a+b x)}\right )-3 \text {Li}_3\left (-e^{-2 (a+b x)}\right )}{2 b^4}-\frac {x^3 \text {sech}(a) \sinh (b x) \text {sech}(a+b x)}{b}+\frac {x^4}{4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.71, size = 721, normalized size = 8.10 \[ \frac {b^{4} x^{4} - 8 \, a^{3} + {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{4} x^{4} - 8 \, b^{3} x^{3} - 8 \, a^{3}\right )} \sinh \left (b x + a\right )^{2} + 24 \, {\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} + b x\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 24 \, {\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} + b x\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 12 \, {\left (a^{2} \cosh \left (b x + a\right )^{2} + 2 \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )^{2} + a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 12 \, {\left (a^{2} \cosh \left (b x + a\right )^{2} + 2 \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + a^{2} \sinh \left (b x + a\right )^{2} + a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 12 \, {\left (b^{2} x^{2} + {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + 12 \, {\left (b^{2} x^{2} + {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b^{2} x^{2} - a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b^{2} x^{2} - a^{2}\right )} \sinh \left (b x + a\right )^{2} - a^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 24 \, {\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} + 1\right )} {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 24 \, {\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} + 1\right )} {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{4 \, {\left (b^{4} \cosh \left (b x + a\right )^{2} + 2 \, b^{4} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{4} \sinh \left (b x + a\right )^{2} + b^{4}\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 \[ \int x^{3} \operatorname {sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.57, size = 125, normalized size = 1.40 \[ \frac {x^{4}}{4}+\frac {2 x^{3}}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )}-\frac {6 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 x^{3}}{b}+\frac {6 a^{2} x}{b^{3}}+\frac {4 a^{3}}{b^{4}}+\frac {3 x^{2} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b^{2}}+\frac {3 x \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{b^{3}}-\frac {3 \polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.42, size = 108, normalized size = 1.21 \[ -\frac {2 \, x^{3}}{b} + \frac {b x^{4} e^{\left (2 \, b x + 2 \, a\right )} + b x^{4} + 8 \, x^{3}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}} + \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )})\right )}}{2 \, b^{4}} \]
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 \[ \int \frac {x^3\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x^{3} \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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