Optimal. Leaf size=113 \[ \frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}-\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {sech}(a+b x)}{b} \]
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Rubi [A] time = 0.11, antiderivative size = 113, 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 = {5418, 4180, 2531, 2282, 6589} \[ -\frac {6 i x \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac {6 i \text {PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}-\frac {6 i \text {PolyLog}\left (3,i e^{a+b x}\right )}{b^4}+\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4180
Rule 5418
Rule 6589
Rubi steps
\begin {align*} \int x^3 \text {sech}(a+b x) \tanh (a+b x) \, dx &=-\frac {x^3 \text {sech}(a+b x)}{b}+\frac {3 \int x^2 \text {sech}(a+b x) \, dx}{b}\\ &=\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {x^3 \text {sech}(a+b x)}{b}-\frac {(6 i) \int x \log \left (1-i e^{a+b x}\right ) \, dx}{b^2}+\frac {(6 i) \int x \log \left (1+i e^{a+b x}\right ) \, dx}{b^2}\\ &=\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}-\frac {x^3 \text {sech}(a+b x)}{b}+\frac {(6 i) \int \text {Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^3}-\frac {(6 i) \int \text {Li}_2\left (i e^{a+b x}\right ) \, dx}{b^3}\\ &=\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}-\frac {x^3 \text {sech}(a+b x)}{b}+\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac {(6 i) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=\frac {6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac {6 i x \text {Li}_2\left (-i e^{a+b x}\right )}{b^3}+\frac {6 i x \text {Li}_2\left (i e^{a+b x}\right )}{b^3}+\frac {6 i \text {Li}_3\left (-i e^{a+b x}\right )}{b^4}-\frac {6 i \text {Li}_3\left (i e^{a+b x}\right )}{b^4}-\frac {x^3 \text {sech}(a+b x)}{b}\\ \end {align*}
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Mathematica [A] time = 2.28, size = 130, normalized size = 1.15 \[ -\frac {x^3 \text {sech}(a+b x)}{b}+\frac {3 i \left (b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )-2 b x \text {Li}_2\left (-i e^{a+b x}\right )+2 b x \text {Li}_2\left (i e^{a+b x}\right )+2 \text {Li}_3\left (-i e^{a+b x}\right )-2 \text {Li}_3\left (i e^{a+b x}\right )\right )}{b^4} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.77, size = 670, normalized size = 5.93 \[ -\frac {2 \, b^{3} x^{3} \cosh \left (b x + a\right ) + 2 \, b^{3} x^{3} \sinh \left (b x + a\right ) - {\left (6 i \, b x \cosh \left (b x + a\right )^{2} + 12 i \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 6 i \, b x \sinh \left (b x + a\right )^{2} + 6 i \, b x\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - {\left (-6 i \, b x \cosh \left (b x + a\right )^{2} - 12 i \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 6 i \, b x \sinh \left (b x + a\right )^{2} - 6 i \, b x\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - {\left (3 i \, a^{2} \cosh \left (b x + a\right )^{2} + 6 i \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 3 i \, a^{2} \sinh \left (b x + a\right )^{2} + 3 i \, a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - {\left (-3 i \, a^{2} \cosh \left (b x + a\right )^{2} - 6 i \, a^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 3 i \, a^{2} \sinh \left (b x + a\right )^{2} - 3 i \, a^{2}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - {\left (-3 i \, b^{2} x^{2} + {\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \cosh \left (b x + a\right )^{2} + {\left (-6 i \, b^{2} x^{2} + 6 i \, a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \sinh \left (b x + a\right )^{2} + 3 i \, a^{2}\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - {\left (3 i \, b^{2} x^{2} + {\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \cosh \left (b x + a\right )^{2} + {\left (6 i \, b^{2} x^{2} - 6 i \, a^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (3 i \, b^{2} x^{2} - 3 i \, a^{2}\right )} \sinh \left (b x + a\right )^{2} - 3 i \, a^{2}\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - {\left (-6 i \, \cosh \left (b x + a\right )^{2} - 12 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 6 i \, \sinh \left (b x + a\right )^{2} - 6 i\right )} {\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - {\left (6 i \, \cosh \left (b x + a\right )^{2} + 12 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 6 i \, \sinh \left (b x + a\right )^{2} + 6 i\right )} {\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \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 \[ \int x^{3} \operatorname {sech}\left (b x + a\right )^{2} \sinh \left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.56, size = 0, normalized size = 0.00 \[ \int x^{3} \mathrm {sech}\left (b x +a \right )^{2} \sinh \left (b x +a \right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} + 6 \, \int \frac {x^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b}\,{d x} \]
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 \[ \int \frac {x^3\,\mathrm {sinh}\left (a+b\,x\right )}{{\mathrm {cosh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x^{3} \sinh {\left (a + b x \right )} \operatorname {sech}^{2}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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