Optimal. Leaf size=77 \[ \frac {i \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac {i \text {Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac {\cosh (a+b x)}{b^2}-\frac {2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}+\frac {x \sinh (a+b x)}{b} \]
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Rubi [A] time = 0.06, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 2279
Rule 2391
Rule 2638
Rule 3296
Rule 4180
Rule 5449
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*}
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Mathematica [B] time = 0.12, size = 213, normalized size = 2.77 \[ -\frac {-i \left (\text {Li}_2\left (-e^{i \left (-i a-i b x+\frac {\pi }{2}\right )}\right )-\text {Li}_2\left (e^{i \left (-i a-i b x+\frac {\pi }{2}\right )}\right )\right )-\left (\left (-i a-i b x+\frac {\pi }{2}\right ) \left (\log \left (1-e^{i \left (-i a-i b x+\frac {\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-i a-i b x+\frac {\pi }{2}\right )}\right )\right )\right )+\left (\frac {\pi }{2}-i a\right ) \log \left (\tan \left (\frac {1}{2} \left (-i a-i b x+\frac {\pi }{2}\right )\right )\right )}{b^2}+\frac {\cosh (b x) (b x \sinh (a)-\cosh (a))}{b^2}+\frac {\sinh (b x) (b x \cosh (a)-\sinh (a))}{b^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.68, size = 322, normalized size = 4.18 \[ \frac {{\left (b x - 1\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x - 1\right )} \sinh \left (b x + a\right )^{2} - b x + {\left (-2 i \, \cosh \left (b x + a\right ) - 2 i \, \sinh \left (b x + a\right )\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) + {\left (2 i \, \cosh \left (b x + a\right ) + 2 i \, \sinh \left (b x + a\right )\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (2 i \, a \cosh \left (b x + a\right ) + 2 i \, a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) + {\left (-2 i \, a \cosh \left (b x + a\right ) - 2 i \, a \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) + {\left ({\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right ) + {\left (2 i \, b x + 2 i \, a\right )} \sinh \left (b x + a\right )\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) + {\left ({\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right ) + {\left (-2 i \, b x - 2 i \, a\right )} \sinh \left (b x + a\right )\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right ) - 1}{2 \, {\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {sech}\left (b x + a\right ) \sinh \left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.39, size = 162, normalized size = 2.10 \[ \frac {\left (b x -1\right ) {\mathrm e}^{b x +a}}{2 b^{2}}-\frac {\left (b x +1\right ) {\mathrm e}^{-b x -a}}{2 b^{2}}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{b}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{b}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {2 a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left ({\left (b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (b x\right )} - {\left (b x + 1\right )} e^{\left (-b x\right )}\right )} e^{\left (-a\right )}}{2 \, b^{2}} - 2 \, \int \frac {x e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\mathrm {sinh}\left (a+b\,x\right )}^2}{\mathrm {cosh}\left (a+b\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x \sinh ^{2}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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