Optimal. Leaf size=180 \[ \frac {3 \text {Li}_4\left (e^{2 (a+b x)}\right )}{4 b^4}-\frac {3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 x}{8 b^3}+\frac {x^3}{4 b}-\frac {x^4}{4} \]
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Rubi [A] time = 0.24, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5450, 5372, 3311, 30, 2635, 8, 3716, 2190, 2531, 6609, 2282, 6589} \[ \frac {3 x^2 \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {PolyLog}\left (4,e^{2 (a+b x)}\right )}{4 b^4}-\frac {3 x^2 \sinh (a+b x) \cosh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}-\frac {3 \sinh (a+b x) \cosh (a+b x)}{8 b^4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 x}{8 b^3}+\frac {x^3}{4 b}-\frac {x^4}{4} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2190
Rule 2282
Rule 2531
Rule 2635
Rule 3311
Rule 3716
Rule 5372
Rule 5450
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \cosh ^2(a+b x) \coth (a+b x) \, dx &=\int x^3 \coth (a+b x) \, dx+\int x^3 \cosh (a+b x) \sinh (a+b x) \, dx\\ &=-\frac {x^4}{4}+\frac {x^3 \sinh ^2(a+b x)}{2 b}-2 \int \frac {e^{2 (a+b x)} x^3}{1-e^{2 (a+b x)}} \, dx-\frac {3 \int x^2 \sinh ^2(a+b x) \, dx}{2 b}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}-\frac {3 \int \sinh ^2(a+b x) \, dx}{4 b^3}+\frac {3 \int x^2 \, dx}{4 b}-\frac {3 \int x^2 \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=\frac {x^3}{4 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 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 \int 1 \, dx}{8 b^3}-\frac {3 \int x \text {Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=\frac {3 x}{8 b^3}+\frac {x^3}{4 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 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 \int \text {Li}_3\left (e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac {3 x}{8 b^3}+\frac {x^3}{4 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 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=\frac {3 x}{8 b^3}+\frac {x^3}{4 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 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}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b^4}-\frac {3 x^2 \cosh (a+b x) \sinh (a+b x)}{4 b^2}+\frac {3 x \sinh ^2(a+b x)}{4 b^3}+\frac {x^3 \sinh ^2(a+b x)}{2 b}\\ \end {align*}
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Mathematica [A] time = 2.61, size = 236, normalized size = 1.31 \[ \frac {\sinh (a) (\sinh (a)+\cosh (a)) \left (16 b^3 x^3 \log \left (1-e^{-a-b x}\right )+16 b^3 x^3 \log \left (e^{-a-b x}+1\right )+4 b^3 x^3 \cosh (2 (a+b x))-48 b^2 x^2 \text {Li}_2\left (-e^{-a-b x}\right )-48 b^2 x^2 \text {Li}_2\left (e^{-a-b x}\right )-6 b^2 x^2 \sinh (2 (a+b x))-96 b x \text {Li}_3\left (-e^{-a-b x}\right )-96 b x \text {Li}_3\left (e^{-a-b x}\right )-96 \text {Li}_4\left (-e^{-a-b x}\right )-96 \text {Li}_4\left (e^{-a-b x}\right )-3 \sinh (2 (a+b x))+6 b x \cosh (2 (a+b x))+4 b^4 x^4\right )}{8 \left (e^{2 a}-1\right ) b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.65, size = 876, normalized size = 4.87 \[ \text {result too large to display} \]
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 \[ \int x^{3} \cosh \left (b x + a\right )^{3} \operatorname {csch}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.60, size = 272, normalized size = 1.51 \[ -\frac {x^{4}}{4}+\frac {\left (4 x^{3} b^{3}-6 x^{2} b^{2}+6 b x -3\right ) {\mathrm e}^{2 b x +2 a}}{32 b^{4}}+\frac {\left (4 x^{3} b^{3}+6 x^{2} b^{2}+6 b x +3\right ) {\mathrm e}^{-2 b x -2 a}}{32 b^{4}}+\frac {6 \polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 a^{3} x}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {3 x^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {3 x^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {3 a^{4}}{2 b^{4}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}+\frac {6 \polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.58, size = 225, normalized size = 1.25 \[ -\frac {1}{2} \, x^{4} + \frac {{\left (8 \, b^{4} x^{4} e^{\left (2 \, a\right )} + {\left (4 \, b^{3} x^{3} e^{\left (4 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (4 \, a\right )} + 6 \, b x e^{\left (4 \, a\right )} - 3 \, e^{\left (4 \, a\right )}\right )} e^{\left (2 \, b x\right )} + {\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x\right )}\right )} e^{\left (-2 \, a\right )}}{32 \, b^{4}} + \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} \]
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 \[ \int \frac {x^3\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{\mathrm {sinh}\left (a+b\,x\right )} \,d x \]
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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