Optimal. Leaf size=77 \[ \frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{\cosh (a+b x)}{b^2}-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \sinh (a+b x)}{b} \]
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Rubi [A] time = 0.0637488, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5449, 3296, 2638, 4180, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac{i \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac{\cosh (a+b x)}{b^2}-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \sinh (a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 5449
Rule 3296
Rule 2638
Rule 4180
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x \sinh (a+b x) \tanh (a+b x) \, dx &=\int x \cosh (a+b x) \, dx-\int x \text{sech}(a+b x) \, dx\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac{x \sinh (a+b x)}{b}+\frac{i \int \log \left (1-i e^{a+b x}\right ) \, dx}{b}-\frac{i \int \log \left (1+i e^{a+b x}\right ) \, dx}{b}-\frac{\int \sinh (a+b x) \, dx}{b}\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\cosh (a+b x)}{b^2}+\frac{x \sinh (a+b x)}{b}+\frac{i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}\\ &=-\frac{2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{\cosh (a+b x)}{b^2}+\frac{i \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac{i \text{Li}_2\left (i e^{a+b x}\right )}{b^2}+\frac{x \sinh (a+b x)}{b}\\ \end{align*}
Mathematica [B] time = 0.109484, size = 212, normalized size = 2.75 \[ \frac{i \left (\text{PolyLog}\left (2,-e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )-\text{PolyLog}\left (2,e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )\right )+\left (-i a-i b x+\frac{\pi }{2}\right ) \left (\log \left (1-e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-i a-i b x+\frac{\pi }{2}\right )}\right )\right )-\left (\frac{\pi }{2}-i a\right ) \log \left (\tan \left (\frac{1}{2} \left (-i a-i b x+\frac{\pi }{2}\right )\right )\right )}{b^2}+\frac{\cosh (b x) (b x \sinh (a)-\cosh (a))}{b^2}+\frac{\sinh (b x) (b x \cosh (a)-\sinh (a))}{b^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.057, size = 162, normalized size = 2.1 \begin{align*}{\frac{ \left ( bx-1 \right ){{\rm e}^{bx+a}}}{2\,{b}^{2}}}-{\frac{ \left ( bx+1 \right ){{\rm e}^{-bx-a}}}{2\,{b}^{2}}}+{\frac{i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) x}{b}}+{\frac{i\ln \left ( 1+i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}-{\frac{i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) x}{b}}-{\frac{i\ln \left ( 1-i{{\rm e}^{bx+a}} \right ) a}{{b}^{2}}}+{\frac{i{\it dilog} \left ( 1+i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}-{\frac{i{\it dilog} \left ( 1-i{{\rm e}^{bx+a}} \right ) }{{b}^{2}}}+2\,{\frac{a\arctan \left ({{\rm e}^{bx+a}} \right ) }{{b}^{2}}} \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{{\left ({\left (b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} -{\left (b x + 1\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{2}} - 2 \, \int \frac{x e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.23961, size = 973, normalized size = 12.64 \begin{align*} \frac{{\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x - 1\right )} \sinh \left (b x + a\right )^{2} - b x +{\left (-2 i \, \cosh \left (b x + a\right ) - 2 i \, \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) +{\left (2 i \, \cosh \left (b x + a\right ) + 2 i \, \sinh \left (b x + a\right )\right )}{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) +{\left (2 i \, a \cosh \left (b x + a\right ) + 2 i \, a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) +{\left (-2 i \, a \cosh \left (b x + a\right ) - 2 i \, a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) +{\left ({\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right ) +{\left (2 i \, b x + 2 i \, a\right )} \sinh \left (b x + a\right )\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) +{\left ({\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right ) +{\left (-2 i \, b x - 2 i \, a\right )} \sinh \left (b x + a\right )\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 1}{2 \,{\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \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 \sinh ^{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 \operatorname{sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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