Optimal. Leaf size=148 \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 x \text{PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}-\frac{3 \text{PolyLog}\left (4,-e^{2 a+2 b x}\right )}{4 b^4}+\frac{3 \text{PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4}-\frac{2 x^3 \tanh ^{-1}\left (e^{2 a+2 b x}\right )}{b} \]
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Rubi [A] time = 0.14994, antiderivative size = 148, normalized size of antiderivative = 1., 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 5461
Rule 4182
Rule 2531
Rule 6609
Rule 2282
Rule 6589
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*}
Mathematica [A] time = 4.00892, size = 150, normalized size = 1.01 \[ \frac{-6 b^2 x^2 \text{PolyLog}\left (2,-e^{2 (a+b x)}\right )+6 b^2 x^2 \text{PolyLog}\left (2,e^{2 (a+b x)}\right )+6 b x \text{PolyLog}\left (3,-e^{2 (a+b x)}\right )-6 b x \text{PolyLog}\left (3,e^{2 (a+b x)}\right )-3 \text{PolyLog}\left (4,-e^{2 (a+b x)}\right )+3 \text{PolyLog}\left (4,e^{2 (a+b x)}\right )+4 b^3 x^3 \log \left (1-e^{2 (a+b x)}\right )-4 b^3 x^3 \log \left (e^{2 (a+b x)}+1\right )}{4 b^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.082, size = 241, normalized size = 1.6 \begin{align*}{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{3}}{b}}-{\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}}}+3\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-6\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{3}}{{b}^{4}}}+6\,{\frac{{\it polylog} \left ( 4,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+6\,{\frac{{\it polylog} \left ( 4,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-{\frac{3\,{\it polylog} \left ( 4,-{{\rm e}^{2\,bx+2\,a}} \right ) }{4\,{b}^{4}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{3}}{b}}+3\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ){x}^{2}}{{b}^{2}}}-6\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}-{\frac{{a}^{3}\ln \left ({{\rm e}^{bx+a}}-1 \right ) }{{b}^{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.16308, size = 274, normalized size = 1.85 \begin{align*} -\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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.63732, size = 1323, normalized size = 8.94 \begin{align*} \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}} \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} \operatorname{csch}{\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{csch}\left (b x + a\right ) \operatorname{sech}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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