Optimal. Leaf size=157 \[ \frac{2 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{2 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{2 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^2 \text{csch}(a+b x)}{b} \]
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Rubi [A] time = 0.233922, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {2621, 321, 207, 5462, 14, 5205, 12, 4180, 2531, 2282, 6589, 4182, 2279, 2391} \[ \frac{2 i x \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{2 i x \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{2 \text{PolyLog}\left (2,-e^{a+b x}\right )}{b^3}+\frac{2 \text{PolyLog}\left (2,e^{a+b x}\right )}{b^3}-\frac{2 i \text{PolyLog}\left (3,-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{PolyLog}\left (3,i e^{a+b x}\right )}{b^3}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{2 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{x^2 \text{csch}(a+b x)}{b} \]
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 2282
Rule 6589
Rule 4182
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x^2 \text{csch}^2(a+b x) \text{sech}(a+b x) \, dx &=-\frac{x^2 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^2 \text{csch}(a+b x)}{b}-2 \int x \left (-\frac{\tan ^{-1}(\sinh (a+b x))}{b}-\frac{\text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^2 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^2 \text{csch}(a+b x)}{b}-2 \int \left (-\frac{x \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x \text{csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac{x^2 \tan ^{-1}(\sinh (a+b x))}{b}-\frac{x^2 \text{csch}(a+b x)}{b}+\frac{2 \int x \tan ^{-1}(\sinh (a+b x)) \, dx}{b}+\frac{2 \int x \text{csch}(a+b x) \, dx}{b}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \int \log \left (1-e^{a+b x}\right ) \, dx}{b^2}+\frac{2 \int \log \left (1+e^{a+b x}\right ) \, dx}{b^2}-\frac{\int b x^2 \text{sech}(a+b x) \, dx}{b}\\ &=-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac{2 \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}-\int x^2 \text{sech}(a+b x) \, dx\\ &=-\frac{2 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}+\frac{(2 i) \int x \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{(2 i) \int x \log \left (1+i e^{a+b x}\right ) \, dx}{b}\\ &=-\frac{2 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 i x \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{2 i x \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}-\frac{(2 i) \int \text{Li}_2\left (-i e^{a+b x}\right ) \, dx}{b^2}+\frac{(2 i) \int \text{Li}_2\left (i e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac{2 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 i x \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{2 i x \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{a+b x}\right )}{b^3}\\ &=-\frac{2 x^2 \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{4 x \tanh ^{-1}\left (e^{a+b x}\right )}{b^2}-\frac{x^2 \text{csch}(a+b x)}{b}-\frac{2 \text{Li}_2\left (-e^{a+b x}\right )}{b^3}+\frac{2 i x \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{2 i x \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{2 \text{Li}_2\left (e^{a+b x}\right )}{b^3}-\frac{2 i \text{Li}_3\left (-i e^{a+b x}\right )}{b^3}+\frac{2 i \text{Li}_3\left (i e^{a+b x}\right )}{b^3}\\ \end{align*}
Mathematica [A] time = 1.85044, size = 312, normalized size = 1.99 \[ \frac{4 i b x \text{PolyLog}\left (2,-i e^{a+b x}\right )-4 i b x \text{PolyLog}\left (2,i e^{a+b x}\right )+4 \text{PolyLog}\left (2,-e^{-a-b x}\right )-4 \text{PolyLog}\left (2,e^{-a-b x}\right )-4 i \text{PolyLog}\left (3,-i e^{a+b x}\right )+4 i \text{PolyLog}\left (3,i e^{a+b x}\right )-2 i b^2 x^2 \log \left (1-i e^{a+b x}\right )+2 i b^2 x^2 \log \left (1+i e^{a+b x}\right )-2 b^2 x^2 \text{csch}(a)+b^2 x^2 \text{csch}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{csch}\left (\frac{1}{2} (a+b x)\right )+b^2 x^2 \text{sech}\left (\frac{a}{2}\right ) \sinh \left (\frac{b x}{2}\right ) \text{sech}\left (\frac{1}{2} (a+b x)\right )+4 b x \log \left (1-e^{-a-b x}\right )-4 b x \log \left (e^{-a-b x}+1\right )+4 a \log \left (1-e^{-a-b x}\right )-4 a \log \left (e^{-a-b x}+1\right )-4 a \log \left (\tanh \left (\frac{1}{2} (a+b x)\right )\right )}{2 b^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.371, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \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^{2} e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac{2 \,{\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^{3}} + \frac{2 \,{\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^{3}} - 8 \, \int \frac{x^{2} 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 = 2.76475, size = 2685, normalized size = 17.1 \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^{2} \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(-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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