Optimal. Leaf size=91 \[ \frac{3 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}+\frac{x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^4}{4} \]
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Rubi [A] time = 0.159999, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3718, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )}{2 b^3}+\frac{3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )}{4 b^4}+\frac{x^3 \log \left (e^{2 (a+b x)}+1\right )}{b}-\frac{x^4}{4} \]
Antiderivative was successfully verified.
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Rule 3718
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^3 \tanh (a+b x) \, dx &=-\frac{x^4}{4}+2 \int \frac{e^{2 (a+b x)} x^3}{1+e^{2 (a+b x)}} \, dx\\ &=-\frac{x^4}{4}+\frac{x^3 \log \left (1+e^{2 (a+b x)}\right )}{b}-\frac{3 \int x^2 \log \left (1+e^{2 (a+b x)}\right ) \, dx}{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 \int x \text{Li}_2\left (-e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\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 \int \text{Li}_3\left (-e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\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 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=-\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}\\ \end{align*}
Mathematica [A] time = 2.23159, size = 88, normalized size = 0.97 \[ \frac{-6 b^2 x^2 \text{PolyLog}\left (2,-e^{-2 (a+b x)}\right )-6 b x \text{PolyLog}\left (3,-e^{-2 (a+b x)}\right )-3 \text{PolyLog}\left (4,-e^{-2 (a+b x)}\right )+4 b^3 x^3 \log \left (e^{-2 (a+b x)}+1\right )+b^4 x^4}{4 b^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.098, size = 116, normalized size = 1.3 \begin{align*} -{\frac{{x}^{4}}{4}}-2\,{\frac{{a}^{3}x}{{b}^{3}}}-{\frac{3\,{a}^{4}}{2\,{b}^{4}}}+{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,bx+2\,a}} \right ) }{b}}+{\frac{3\,{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{2}}}-{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,bx+2\,a}} \right ) }{2\,{b}^{3}}}+{\frac{3\,{\it polylog} \left ( 4,-{{\rm e}^{2\,bx+2\,a}} \right ) }{4\,{b}^{4}}}+2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.20261, size = 113, normalized size = 1.24 \begin{align*} -\frac{1}{4} \, x^{4} + \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.08174, size = 761, normalized size = 8.36 \begin{align*} -\frac{b^{4} x^{4} - 12 \, b^{2} x^{2}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 12 \, b^{2} x^{2}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + 24 \, b x{\rm polylog}\left (3, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + 24 \, b x{\rm polylog}\left (3, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) - 4 \,{\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 ) - 4 \,{\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 ) - 24 \,{\rm polylog}\left (4, i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - 24 \,{\rm polylog}\left (4, -i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right )}{4 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sinh{\left (a + b x \right )} \operatorname{sech}{\left (a + b x \right )}\, dx \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 )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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