Optimal. Leaf size=195 \[ \frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{6 x \sinh (a+b x)}{b^3}-\frac{6 \cosh (a+b x)}{b^4}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \sinh (a+b x)}{b} \]
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Rubi [A] time = 0.197753, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5449, 3296, 2638, 4180, 2531, 6609, 2282, 6589} \[ \frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{6 x \sinh (a+b x)}{b^3}-\frac{6 \cosh (a+b x)}{b^4}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \sinh (a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 5449
Rule 3296
Rule 2638
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^3 \sinh (a+b x) \tanh (a+b x) \, dx &=\int x^3 \cosh (a+b x) \, dx-\int x^3 \text{sech}(a+b x) \, dx\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \sinh (a+b x)}{b}+\frac{(3 i) \int x^2 \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{(3 i) \int x^2 \log \left (1+i e^{a+b x}\right ) \, dx}{b}-\frac{3 \int x^2 \sinh (a+b x) \, dx}{b}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{x^3 \sinh (a+b x)}{b}-\frac{(6 i) \int x \text{Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}+\frac{(6 i) \int x \text{Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}+\frac{6 \int x \cosh (a+b x) \, dx}{b^2}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac{6 x \sinh (a+b x)}{b^3}+\frac{x^3 \sinh (a+b x)}{b}+\frac{(6 i) \int \text{Li}_3\left (-i e^{a+b x}\right ) \, dx}{b^3}-\frac{(6 i) \int \text{Li}_3\left (i e^{a+b x}\right ) \, dx}{b^3}-\frac{6 \int \sinh (a+b x) \, dx}{b^3}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 \cosh (a+b x)}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac{6 x \sinh (a+b x)}{b^3}+\frac{x^3 \sinh (a+b x)}{b}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{(6 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 \cosh (a+b x)}{b^4}-\frac{3 x^2 \cosh (a+b x)}{b^2}+\frac{3 i x^2 \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac{6 i x \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{Li}_3\left (i e^{a+b x}\right )}{b^3}+\frac{6 i \text{Li}_4\left (-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{Li}_4\left (i e^{a+b x}\right )}{b^4}+\frac{6 x \sinh (a+b x)}{b^3}+\frac{x^3 \sinh (a+b x)}{b}\\ \end{align*}
Mathematica [A] time = 1.43698, size = 211, normalized size = 1.08 \[ -\frac{i \left (-3 b^2 x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )+3 b^2 x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )+6 b x \text{PolyLog}\left (3,-i e^{a+b x}\right )-6 b x \text{PolyLog}\left (3,i e^{a+b x}\right )-6 \text{PolyLog}\left (4,-i e^{a+b x}\right )+6 \text{PolyLog}\left (4,i e^{a+b x}\right )+b^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 x^3 \log \left (1+i e^{a+b x}\right )+i b^3 x^3 \sinh (a+b x)-3 i b^2 x^2 \cosh (a+b x)+6 i b x \sinh (a+b x)-6 i \cosh (a+b x)\right )}{b^4} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.125, size = 0, normalized size = 0. \begin{align*} \int{x}^{3}{\rm sech} \left (bx+a\right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (b^{3} x^{3} e^{\left (2 \, a\right )} - 3 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 6 \, e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} -{\left (b^{3} x^{3} + 3 \, b^{2} x^{2} + 6 \, b x + 6\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{4}} - 2 \, \int \frac{x^{3} e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.38427, size = 1704, normalized size = 8.74 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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