Optimal. Leaf size=148 \[ -\frac {3 \text {Li}_4\left (-e^{2 a+2 b x}\right )}{4 b^4}+\frac {3 \text {Li}_4\left (e^{2 a+2 b x}\right )}{4 b^4}+\frac {3 x \text {Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x \text {Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x^2 \text {Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \text {Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]
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Rubi [A] time = 0.15, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5461, 4182, 2531, 6609, 2282, 6589} \[ -\frac {3 x^2 \text {PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \text {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x \text {PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x \text {PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 \text {PolyLog}\left (4,-e^{2 a+2 b x}\right )}{4 b^4}+\frac {3 \text {PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4}-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4182
Rule 5461
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^3 \text {csch}(a+b x) \text {sech}(a+b x) \, dx &=2 \int x^3 \text {csch}(2 a+2 b x) \, dx\\ &=-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 \int x^2 \log \left (1-e^{2 a+2 b x}\right ) \, dx}{b}+\frac {3 \int x^2 \log \left (1+e^{2 a+2 b x}\right ) \, dx}{b}\\ &=-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \text {Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 \int x \text {Li}_2\left (-e^{2 a+2 b x}\right ) \, dx}{b^2}-\frac {3 \int x \text {Li}_2\left (e^{2 a+2 b x}\right ) \, dx}{b^2}\\ &=-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \text {Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x \text {Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 \int \text {Li}_3\left (-e^{2 a+2 b x}\right ) \, dx}{2 b^3}+\frac {3 \int \text {Li}_3\left (e^{2 a+2 b x}\right ) \, dx}{2 b^3}\\ &=-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \text {Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x \text {Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b^4}+\frac {3 \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{4 b^4}\\ &=-\frac {2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b}-\frac {3 x^2 \text {Li}_2\left (-e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x^2 \text {Li}_2\left (e^{2 a+2 b x}\right )}{2 b^2}+\frac {3 x \text {Li}_3\left (-e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 x \text {Li}_3\left (e^{2 a+2 b x}\right )}{2 b^3}-\frac {3 \text {Li}_4\left (-e^{2 a+2 b x}\right )}{4 b^4}+\frac {3 \text {Li}_4\left (e^{2 a+2 b x}\right )}{4 b^4}\\ \end {align*}
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Mathematica [A] time = 4.35, size = 150, normalized size = 1.01 \[ \frac {4 b^3 x^3 \log \left (1-e^{2 (a+b x)}\right )-4 b^3 x^3 \log \left (e^{2 (a+b x)}+1\right )-6 b^2 x^2 \text {Li}_2\left (-e^{2 (a+b x)}\right )+6 b^2 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )+6 b x \text {Li}_3\left (-e^{2 (a+b x)}\right )-6 b x \text {Li}_3\left (e^{2 (a+b x)}\right )-3 \text {Li}_4\left (-e^{2 (a+b x)}\right )+3 \text {Li}_4\left (e^{2 (a+b x)}\right )}{4 b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.44, size = 448, normalized size = 3.03 \[ \frac {b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 3 \, b^{2} x^{2} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - 6 \, b x {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, b x {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 6 \, b x {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 6 \, b x {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - {\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 ) - {\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 ) + {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) + 6 \, {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, {\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 6 \, {\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 6 \, {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{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 \[ \int x^{3} \operatorname {csch}\left (b x + a\right ) \operatorname {sech}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.45, size = 241, normalized size = 1.63 \[ -\frac {3 \polylog \left (4, -{\mathrm e}^{2 b x +2 a}\right )}{4 b^{4}}+\frac {6 \polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {6 \polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}-\frac {3 x^{2} \polylog \left (2, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{2}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {3 x^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {\ln \left (1+{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {3 x^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 x \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 x \polylog \left (3, -{\mathrm e}^{2 b x +2 a}\right )}{2 b^{3}}-\frac {x^{3} \ln \left (1+{\mathrm e}^{2 b x +2 a}\right )}{b}-\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.55, size = 203, normalized size = 1.37 \[ -\frac {4 \, b^{3} x^{3} \log \left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, b x + 2 \, a\right )})}{3 \, b^{4}} + \frac {b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})}{b^{4}} + \frac {b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})}{b^{4}} \]
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 \[ \int \frac {x^3}{\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {csch}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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