Optimal. Leaf size=93 \[ \frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b} \]
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Rubi [A] time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5419, 4182, 2531, 2282, 6589} \[ -\frac {6 x \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac {6 x \text {PolyLog}\left (2,e^{a+b x}\right )}{b^3}+\frac {6 \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac {6 \text {PolyLog}\left (3,e^{a+b x}\right )}{b^4}-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4182
Rule 5419
Rule 6589
Rubi steps
\begin {align*} \int x^3 \coth (a+b x) \text {csch}(a+b x) \, dx &=-\frac {x^3 \text {csch}(a+b x)}{b}+\frac {3 \int x^2 \text {csch}(a+b x) \, dx}{b}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac {6 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \int \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^3}-\frac {6 \int \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {6 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {csch}(a+b x)}{b}-\frac {6 x \text {Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac {6 x \text {Li}_2\left (e^{a+b x}\right )}{b^3}+\frac {6 \text {Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac {6 \text {Li}_3\left (e^{a+b x}\right )}{b^4}\\ \end {align*}
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Mathematica [A] time = 7.21, size = 167, normalized size = 1.80 \[ -\frac {\text {csch}\left (\frac {1}{2} (a+b x)\right ) \text {sech}\left (\frac {1}{2} (a+b x)\right ) \left (6 b^2 x^2 \sinh (a+b x) \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))+6 b x \sinh (a+b x) \text {Li}_2(-\cosh (a+b x)-\sinh (a+b x))-6 b x \sinh (a+b x) \text {Li}_2(\cosh (a+b x)+\sinh (a+b x))-6 \sinh (a+b x) \text {Li}_3(-\cosh (a+b x)-\sinh (a+b x))+6 \sinh (a+b x) \text {Li}_3(\cosh (a+b x)+\sinh (a+b x))+b^3 x^3\right )}{2 b^4} \]
Warning: Unable to verify antiderivative.
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fricas [C] time = 0.43, size = 551, normalized size = 5.92 \[ -\frac {2 \, b^{3} x^{3} \cosh \left (b x + a\right ) + 2 \, b^{3} x^{3} \sinh \left (b x + a\right ) - 6 \, {\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 (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, {\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 (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 3 \, {\left (b^{2} x^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} x^{2} \sinh \left (b x + a\right )^{2} - b^{2} x^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 3 \, {\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 ) - 1\right ) + 3 \, {\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 (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\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, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\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, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{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}} \]
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 \[ \int x^{3} \cosh \left (b x + a\right ) \operatorname {csch}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.22, size = 174, normalized size = 1.87 \[ -\frac {2 x^{3} {\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {6 a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}-\frac {6 x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {6 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}-\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}+\frac {6 x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 121, normalized size = 1.30 \[ -\frac {2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} \]
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 \[ \int \frac {x^3\,\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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