Optimal. Leaf size=226 \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{6 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{6 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \text{sech}(a+b x)}{b} \]
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Rubi [A] time = 0.337148, antiderivative size = 226, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 13, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.722, Rules used = {2622, 321, 207, 5462, 14, 6273, 12, 4182, 2531, 6609, 2282, 6589, 4180} \[ -\frac{3 x^2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac{3 x^2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^2}+\frac{6 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac{6 x \text{PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac{6 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac{6 \text{PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x^3 \text{sech}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 2622
Rule 321
Rule 207
Rule 5462
Rule 14
Rule 6273
Rule 12
Rule 4182
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 4180
Rubi steps
\begin{align*} \int x^3 \text{csch}(a+b x) \text{sech}^2(a+b x) \, dx &=-\frac{x^3 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac{x^3 \text{sech}(a+b x)}{b}-3 \int x^2 \left (-\frac{\tanh ^{-1}(\cosh (a+b x))}{b}+\frac{\text{sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^3 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac{x^3 \text{sech}(a+b x)}{b}-3 \int \left (-\frac{x^2 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac{x^2 \text{sech}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^3 \tanh ^{-1}(\cosh (a+b x))}{b}+\frac{x^3 \text{sech}(a+b x)}{b}+\frac{3 \int x^2 \tanh ^{-1}(\cosh (a+b x)) \, dx}{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{\int b x^3 \text{csch}(a+b x) \, dx}{b}\\ &=-\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}+\int x^3 \text{csch}(a+b x) \, dx\\ &=-\frac{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}+\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{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^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\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{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{b^2}-\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}+\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{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\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{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_3\left (-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{6 x \text{Li}_3\left (e^{a+b x}\right )}{b^3}+\frac{x^3 \text{sech}(a+b x)}{b}-\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{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\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{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_3\left (-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{6 x \text{Li}_3\left (e^{a+b x}\right )}{b^3}+\frac{x^3 \text{sech}(a+b x)}{b}-\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{6 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac{3 x^2 \text{Li}_2\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{3 x^2 \text{Li}_2\left (e^{a+b x}\right )}{b^2}+\frac{6 x \text{Li}_3\left (-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{6 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{x^3 \text{sech}(a+b x)}{b}\\ \end{align*}
Mathematica [A] time = 0.377735, size = 282, normalized size = 1.25 \[ \frac{-3 i \left (-2 b x \text{PolyLog}\left (2,-i e^{a+b x}\right )+2 b x \text{PolyLog}\left (2,i e^{a+b x}\right )+2 \text{PolyLog}\left (3,-i e^{a+b x}\right )-2 \text{PolyLog}\left (3,i e^{a+b x}\right )+b^2 x^2 \log \left (1-i e^{a+b x}\right )-b^2 x^2 \log \left (1+i e^{a+b x}\right )\right )-3 b^2 x^2 \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))+3 b^2 x^2 \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))+6 b x \text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))-6 b x \text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))-6 \text{PolyLog}(4,-\sinh (a+b x)-\cosh (a+b x))+6 \text{PolyLog}(4,\sinh (a+b x)+\cosh (a+b x))+b^3 x^3 \text{sech}(a+b x)-2 b^3 x^3 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))}{b^4} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.521, size = 0, normalized size = 0. \begin{align*} \int{x}^{3}{\rm csch} \left (bx+a\right ) \left ({\rm sech} \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{2 \, x^{3} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} + b} - \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}} - 24 \, \int \frac{x^{2} e^{\left (b x + a\right )}}{4 \,{\left (b e^{\left (2 \, b x + 2 \, a\right )} + b\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.89573, size = 3553, normalized size = 15.72 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{csch}{\left (a + b x \right )} \operatorname{sech}^{2}{\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*} \mathit{sage}_{0} x \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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