Optimal. Leaf size=201 \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{3 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{3 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^4}+\frac{3 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^4}-\frac{3 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{3 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b} \]
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Rubi [A] time = 0.35973, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5457, 4182, 2531, 6609, 2282, 6589, 4186, 2279, 2391} \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{3 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{3 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^4}+\frac{3 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^4}-\frac{3 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{3 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b} \]
Antiderivative was successfully verified.
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Rule 5457
Rule 4182
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 4186
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x^3 \coth ^2(a+b x) \text{csch}(a+b x) \, dx &=\int x^3 \text{csch}(a+b x) \, dx+\int x^3 \text{csch}^3(a+b x) \, dx\\ &=-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{1}{2} \int x^3 \text{csch}(a+b x) \, dx+\frac{3 \int x \text{csch}(a+b x) \, dx}{b^2}-\frac{3 \int x^2 \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac{3 \int x^2 \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{3 x^2 \text{Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{b^2}-\frac{3 \int \log \left (1-e^{a+b x}\right ) \, dx}{b^3}+\frac{3 \int \log \left (1+e^{a+b x}\right ) \, dx}{b^3}+\frac{6 \int x \text{Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}-\frac{6 \int x \text{Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}+\frac{3 \int x^2 \log \left (1-e^{a+b x}\right ) \, dx}{2 b}-\frac{3 \int x^2 \log \left (1+e^{a+b x}\right ) \, dx}{2 b}\\ &=-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{3 x^2 \text{Li}_2\left (-e^{a+b x}\right )}{2 b^2}+\frac{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{2 b^2}+\frac{6 x \text{Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac{6 x \text{Li}_3\left (e^{a+b x}\right )}{b^3}-\frac{3 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{6 \int \text{Li}_3\left (-e^{a+b x}\right ) \, dx}{b^3}+\frac{6 \int \text{Li}_3\left (e^{a+b x}\right ) \, dx}{b^3}-\frac{3 \int x \text{Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}+\frac{3 \int x \text{Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{3 \text{Li}_2\left (-e^{a+b x}\right )}{b^4}-\frac{3 x^2 \text{Li}_2\left (-e^{a+b x}\right )}{2 b^2}+\frac{3 \text{Li}_2\left (e^{a+b x}\right )}{b^4}+\frac{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac{3 x \text{Li}_3\left (e^{a+b x}\right )}{b^3}-\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac{3 \int \text{Li}_3\left (-e^{a+b x}\right ) \, dx}{b^3}-\frac{3 \int \text{Li}_3\left (e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{3 \text{Li}_2\left (-e^{a+b x}\right )}{b^4}-\frac{3 x^2 \text{Li}_2\left (-e^{a+b x}\right )}{2 b^2}+\frac{3 \text{Li}_2\left (e^{a+b x}\right )}{b^4}+\frac{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac{3 x \text{Li}_3\left (e^{a+b x}\right )}{b^3}-\frac{6 \text{Li}_4\left (-e^{a+b x}\right )}{b^4}+\frac{6 \text{Li}_4\left (e^{a+b x}\right )}{b^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac{6 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac{x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{csch}(a+b x)}{2 b^2}-\frac{x^3 \coth (a+b x) \text{csch}(a+b x)}{2 b}-\frac{3 \text{Li}_2\left (-e^{a+b x}\right )}{b^4}-\frac{3 x^2 \text{Li}_2\left (-e^{a+b x}\right )}{2 b^2}+\frac{3 \text{Li}_2\left (e^{a+b x}\right )}{b^4}+\frac{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac{3 x \text{Li}_3\left (e^{a+b x}\right )}{b^3}-\frac{3 \text{Li}_4\left (-e^{a+b x}\right )}{b^4}+\frac{3 \text{Li}_4\left (e^{a+b x}\right )}{b^4}\\ \end{align*}
Mathematica [A] time = 6.84958, size = 280, normalized size = 1.39 \[ -\frac{12 \left (b^2 x^2+2\right ) \text{PolyLog}\left (2,-e^{a+b x}\right )-12 \left (b^2 x^2+2\right ) \text{PolyLog}\left (2,e^{a+b x}\right )-24 b x \text{PolyLog}\left (3,-e^{a+b x}\right )+24 b x \text{PolyLog}\left (3,e^{a+b x}\right )+24 \text{PolyLog}\left (4,-e^{a+b x}\right )-24 \text{PolyLog}\left (4,e^{a+b x}\right )-4 b^3 x^3 \log \left (1-e^{a+b x}\right )+4 b^3 x^3 \log \left (e^{a+b x}+1\right )+b^3 x^3 \text{csch}^2\left (\frac{1}{2} (a+b x)\right )+12 b^2 x^2 \text{csch}(a)+b^3 x^3 \text{sech}^2\left (\frac{1}{2} (a+b x)\right )-6 b^2 x^2 \text{csch}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{csch}\left (\frac{1}{2} (a+b x)\right )-6 b^2 x^2 \text{sech}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right )-24 b x \log \left (1-e^{a+b x}\right )+24 b x \log \left (e^{a+b x}+1\right )}{8 b^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.078, size = 340, normalized size = 1.7 \begin{align*} -{\frac{{x}^{2}{{\rm e}^{bx+a}} \left ( bx{{\rm e}^{2\,bx+2\,a}}+bx+3\,{{\rm e}^{2\,bx+2\,a}}-3 \right ) }{{b}^{2} \left ({{\rm e}^{2\,bx+2\,a}}-1 \right ) ^{2}}}-3\,{\frac{{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-3\,{\frac{{\it polylog} \left ( 4,-{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+6\,{\frac{a{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+{\frac{{a}^{3}{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{{\it polylog} \left ( 4,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}-3\,{\frac{a\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{{b}^{4}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) a}{{b}^{4}}}-{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ){x}^{3}}{2\,b}}-{\frac{3\,{\it polylog} \left ( 2,-{{\rm e}^{bx+a}} \right ){x}^{2}}{2\,{b}^{2}}}+3\,{\frac{{\it polylog} \left ( 3,-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){x}^{3}}{2\,b}}+{\frac{3\,{\it polylog} \left ( 2,{{\rm e}^{bx+a}} \right ){x}^{2}}{2\,{b}^{2}}}-3\,{\frac{{\it polylog} \left ( 3,{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}-3\,{\frac{\ln \left ( 1+{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}+3\,{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ) x}{{b}^{3}}}-{\frac{{a}^{3}\ln \left ( 1+{{\rm e}^{bx+a}} \right ) }{2\,{b}^{4}}}+{\frac{\ln \left ( 1-{{\rm e}^{bx+a}} \right ){a}^{3}}{2\,{b}^{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.62837, size = 354, normalized size = 1.76 \begin{align*} -\frac{{\left (b x^{3} e^{\left (3 \, a\right )} + 3 \, x^{2} e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} +{\left (b x^{3} e^{a} - 3 \, x^{2} e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} - \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 )})}{2 \, 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 )})}{2 \, b^{4}} - \frac{3 \,{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )}}{b^{4}} + \frac{3 \,{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )}}{b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.48115, size = 4602, normalized size = 22.9 \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} \cosh \left (b x + a\right )^{2} \operatorname{csch}\left (b x + a\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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