Optimal. Leaf size=87 \[ -\frac {3 \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x \text {Li}_2\left (e^{2 (a+b x)}\right )}{b^3}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac {x^3 \coth (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \]
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Rubi [A] time = 0.19, antiderivative size = 87, 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, 3716, 2190, 2531, 2282, 6589, 30} \[ \frac {3 x \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{b^3}-\frac {3 \text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^4}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\frac {x^3 \coth (a+b x)}{b}-\frac {x^3}{b}+\frac {x^4}{4} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2190
Rule 2282
Rule 2531
Rule 3716
Rule 3720
Rule 6589
Rubi steps
\begin {align*} \int x^3 \coth ^2(a+b x) \, dx &=-\frac {x^3 \coth (a+b x)}{b}+\frac {3 \int x^2 \coth (a+b x) \, dx}{b}+\int x^3 \, dx\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (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 {x^3 \coth (a+b x)}{b}+\frac {3 x^2 \log \left (1-e^{2 (a+b x)}\right )}{b^2}-\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 {x^3 \coth (a+b x)}{b}+\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 \int \text {Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^3}\\ &=-\frac {x^3}{b}+\frac {x^4}{4}-\frac {x^3 \coth (a+b x)}{b}+\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 \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 {x^3 \coth (a+b x)}{b}+\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}\\ \end {align*}
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Mathematica [B] time = 0.60, size = 204, normalized size = 2.34 \[ -\frac {e^{2 a} \left (2 e^{-2 a} b^3 x^3-3 \left (1-e^{-2 a}\right ) b^2 x^2 \log \left (1-e^{-a-b x}\right )-3 \left (1-e^{-2 a}\right ) b^2 x^2 \log \left (e^{-a-b x}+1\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text {Li}_2\left (-e^{-a-b x}\right )+\text {Li}_3\left (-e^{-a-b x}\right )\right )+6 \left (1-e^{-2 a}\right ) \left (b x \text {Li}_2\left (e^{-a-b x}\right )+\text {Li}_3\left (e^{-a-b x}\right )\right )\right )}{\left (e^{2 a}-1\right ) b^4}+\frac {x^3 \text {csch}(a) \sinh (b x) \text {csch}(a+b x)}{b}+\frac {x^4}{4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.50, size = 632, normalized size = 7.26 \[ -\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 (\cosh \left (b x + a\right ) + \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 (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 12 \, {\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 ) - 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 ) - 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 (-\cosh \left (b x + a\right ) - \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, \cosh \left (b x + a\right ) + \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, -\cosh \left (b x + a\right ) - \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 \[ \int x^{3} \cosh \left (b x + a\right )^{2} \operatorname {csch}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.58, size = 198, normalized size = 2.28 \[ \frac {x^{4}}{4}-\frac {2 x^{3}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {3 a^{2} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}-\frac {6 a^{2} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 x^{3}}{b}+\frac {6 a^{2} x}{b^{3}}+\frac {4 a^{3}}{b^{4}}+\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}-\frac {3 \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{b^{4}}+\frac {6 x \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {6 x \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 146, normalized size = 1.68 \[ -\frac {2 \, x^{3}}{b} + \frac {b x^{4} e^{\left (2 \, b x + 2 \, a\right )} - b x^{4} - 8 \, x^{3}}{4 \, {\left (b e^{\left (2 \, b x + 2 \, a\right )} - b\right )}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} \]
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 \[ \int \frac {x^3\,{\mathrm {cosh}\left (a+b\,x\right )}^2}{{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]
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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