Optimal. Leaf size=91 \[ \frac {3 \text {Li}_4\left (-e^{2 (a+b x)}\right )}{4 b^4}-\frac {3 x \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 x^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}+\frac {x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^4}{4} \]
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Rubi [A] time = 0.16, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3718, 2190, 2531, 6609, 2282, 6589} \[ \frac {3 x^2 \text {PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}+\frac {x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac {x^4}{4} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 3718
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \tanh (a+b x) \, dx &=-\frac {x^4}{4}+2 \int \frac {e^{2 (a+b x)} x^3}{1+e^{2 (a+b x)}} \, dx\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac {3 \int x^2 \log \left (1+e^{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 \int x \text {Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \int \text {Li}_3\left (-e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (-e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {Li}_4\left (-e^{2 (a+b x)}\right )}{4 b^4}\\ \end {align*}
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Mathematica [A] time = 2.31, size = 88, normalized size = 0.97 \[ \frac {4 b^3 x^3 \log \left (e^{-2 (a+b x)}+1\right )-6 b^2 x^2 \text {Li}_2\left (-e^{-2 (a+b x)}\right )-6 b x \text {Li}_3\left (-e^{-2 (a+b x)}\right )-3 \text {Li}_4\left (-e^{-2 (a+b x)}\right )+b^4 x^4}{4 b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.64, size = 257, normalized size = 2.82 \[ -\frac {b^{4} x^{4} - 12 \, b^{2} x^{2} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 12 \, b^{2} x^{2} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 24 \, b x {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 24 \, b x {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 24 \, {\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 24 \, {\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{4 \, b^{4}} \]
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 \[ \int x^{3} \operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.32, size = 116, normalized size = 1.27 \[ -\frac {x^{4}}{4}-\frac {2 a^{3} x}{b^{3}}-\frac {3 a^{4}}{2 b^{4}}+\frac {x^{3} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}+\frac {3 x^{2} \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}-\frac {3 x \polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}}+\frac {3 \polylog \left (4, -{\mathrm e}^{2 b x +2 a}\right )}{4 b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 84, normalized size = 0.92 \[ -\frac {1}{4} \, x^{4} + \frac {4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} \]
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 \[ \int \frac {x^3\,\mathrm {sinh}\left (a+b\,x\right )}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
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 \[ \int x^{3} \sinh {\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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