Optimal. Leaf size=237 \[ \frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 x \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^3 \text{csch}(a+b x)}{b} \]
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Rubi [A] time = 0.343965, antiderivative size = 237, 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 = {2621, 321, 207, 5462, 14, 5205, 12, 4180, 2531, 6609, 2282, 6589, 4182} \[ \frac{3 i x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{3 i x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{6 x \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{6 x \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{6 i x \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{6 i x \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}+\frac{6 \text{PolyLog}\left (3,-e^{a+b x}\right )}{b^4}-\frac{6 \text{PolyLog}\left (3,e^{a+b x}\right )}{b^4}+\frac{6 i \text{PolyLog}\left (4,-i e^{a+b x}\right )}{b^4}-\frac{6 i \text{PolyLog}\left (4,i e^{a+b x}\right )}{b^4}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^3 \text{csch}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 2621
Rule 321
Rule 207
Rule 5462
Rule 14
Rule 5205
Rule 12
Rule 4180
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rule 4182
Rubi steps
\begin{align*} \int x^3 \text{csch}^2(a+b x) \text{sech}(a+b x) \, dx &=-\frac{x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^3 \text{csch}(a+b x)}{b}-3 \int x^2 \left (-\frac{\tan ^{-1}(\sinh (a+b x))}{b}-\frac{\text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^3 \text{csch}(a+b x)}{b}-3 \int \left (-\frac{x^2 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^2 \text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^3 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^3 \text{csch}(a+b x)}{b}+\frac{3 \int x^2 \tan ^{-1}(\sinh (a+b x)) \, dx}{b}+\frac{3 \int x^2 \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 \int x \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac{6 \int x \log \left (1+e^{a+b x}\right ) \, dx}{b^2}-\frac{\int b x^3 \text{sech}(a+b x) \, dx}{b}\\ &=-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \int \text{Li}_2\left (-e^{a+b x}\right ) \, dx}{b^3}-\frac{6 \int \text{Li}_2\left (e^{a+b x}\right ) \, dx}{b^3}-\int x^3 \text{sech}(a+b x) \, dx\\ &=-\frac{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{6 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}-\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\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{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\frac{6 \text{Li}_3\left (e^{a+b x}\right )}{b^4}-\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{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\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 \text{Li}_3\left (e^{a+b x}\right )}{b^4}+\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{2 x^3 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{6 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\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 \text{Li}_3\left (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{(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 x^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^3 \text{csch}(a+b x)}{b}-\frac{6 x \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\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 x \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{6 \text{Li}_3\left (-e^{a+b x}\right )}{b^4}-\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 \text{Li}_3\left (e^{a+b x}\right )}{b^4}+\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}\\ \end{align*}
Mathematica [A] time = 1.76773, size = 333, normalized size = 1.41 \[ \frac{-2 i \left (-3 b^2 x^2 \text{PolyLog}\left (2,-i e^{a+b x}\right )+3 b^2 x^2 \text{PolyLog}\left (2,i e^{a+b x}\right )+6 b x \text{PolyLog}\left (3,-i e^{a+b x}\right )-6 b x \text{PolyLog}\left (3,i e^{a+b x}\right )-6 \text{PolyLog}\left (4,-i e^{a+b x}\right )+6 \text{PolyLog}\left (4,i e^{a+b x}\right )+b^3 x^3 \log \left (1-i e^{a+b x}\right )-b^3 x^3 \log \left (1+i e^{a+b x}\right )\right )-12 \left (b x \text{PolyLog}(2,-\sinh (a+b x)-\cosh (a+b x))-b x \text{PolyLog}(2,\sinh (a+b x)+\cosh (a+b x))-\text{PolyLog}(3,-\sinh (a+b x)-\cosh (a+b x))+\text{PolyLog}(3,\sinh (a+b x)+\cosh (a+b x))+b^2 x^2 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))\right )-2 b^3 x^3 \text{csch}(a)+b^3 x^3 \text{csch}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{csch}\left (\frac{1}{2} (a+b x)\right )+b^3 x^3 \text{sech}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right )}{2 b^4} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.454, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ({\rm csch} \left (bx+a\right ) \right ) ^{2}{\rm sech} \left (bx+a\right )\, 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{3 \,{\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(-e^{\left (b x + a\right )})\right )}}{b^{4}} + \frac{3 \,{\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x{\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \,{\rm Li}_{3}(e^{\left (b x + a\right )})\right )}}{b^{4}} - 8 \, \int \frac{x^{3} e^{\left (b x + a\right )}}{4 \,{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.80188, size = 3559, normalized size = 15.02 \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}^{2}{\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 )^{2} \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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