Optimal. Leaf size=79 \[ \frac {i \text {Li}_2\left (-i e^{a+b x}\right )}{b^2}-\frac {i \text {Li}_2\left (i e^{a+b x}\right )}{b^2}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x \text {csch}(a+b x)}{b} \]
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Rubi [A] time = 0.11, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {2621, 321, 207, 5462, 5203, 12, 4180, 2279, 2391, 3770} \[ \frac {i \text {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}-\frac {i \text {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {x \text {csch}(a+b x)}{b} \]
Antiderivative was successfully verified.
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Rule 12
Rule 207
Rule 321
Rule 2279
Rule 2391
Rule 2621
Rule 3770
Rule 4180
Rule 5203
Rule 5462
Rubi steps
\begin {align*} \int x \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx &=-\frac {x \tan ^{-1}(\sinh (a+b x))}{b}-\frac {x \text {csch}(a+b x)}{b}-\int \left (-\frac {\tan ^{-1}(\sinh (a+b x))}{b}-\frac {\text {csch}(a+b x)}{b}\right ) \, dx\\ &=-\frac {x \tan ^{-1}(\sinh (a+b x))}{b}-\frac {x \text {csch}(a+b x)}{b}+\frac {\int \tan ^{-1}(\sinh (a+b x)) \, dx}{b}+\frac {\int \text {csch}(a+b x) \, dx}{b}\\ &=-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {x \text {csch}(a+b x)}{b}-\frac {\int b x \text {sech}(a+b x) \, dx}{b}\\ &=-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {x \text {csch}(a+b x)}{b}-\int x \text {sech}(a+b x) \, dx\\ &=-\frac {2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {x \text {csch}(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 {2 x \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {x \text {csch}(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 {\tanh ^{-1}(\cosh (a+b x))}{b^2}-\frac {x \text {csch}(a+b x)}{b}+\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}\\ \end {align*}
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Mathematica [A] time = 0.83, size = 112, normalized size = 1.42 \[ \frac {2 i \text {Li}_2(-i (\cosh (a+b x)+\sinh (a+b x)))-2 i \text {Li}_2(i (\cosh (a+b x)+\sinh (a+b x)))+b x \tanh \left (\frac {1}{2} (a+b x)\right )-b x \coth \left (\frac {1}{2} (a+b x)\right )+2 \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )-4 b x \tan ^{-1}(\sinh (a+b x)+\cosh (a+b x))}{2 b^2} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.44, size = 565, normalized size = 7.15 \[ -\frac {2 \, b x \cosh \left (b x + a\right ) + 2 \, b x \sinh \left (b x + a\right ) - {\left (-i \, \cosh \left (b x + a\right )^{2} - 2 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )^{2} + i\right )} {\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - {\left (i \, \cosh \left (b x + a\right )^{2} + 2 i \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )^{2} - i\right )} {\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) + {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - {\left (i \, a \cosh \left (b x + a\right )^{2} + 2 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + i \, a \sinh \left (b x + a\right )^{2} - i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) - {\left (-i \, a \cosh \left (b x + a\right )^{2} - 2 i \, a \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - i \, a \sinh \left (b x + a\right )^{2} + i \, a\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) - {\left ({\left (i \, b x + i \, a\right )} \cosh \left (b x + a\right )^{2} + {\left (2 i \, b x + 2 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (i \, b x + i \, a\right )} \sinh \left (b x + a\right )^{2} - i \, b x - i \, a\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) - {\left ({\left (-i \, b x - i \, a\right )} \cosh \left (b x + a\right )^{2} + {\left (-2 i \, b x - 2 i \, a\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (-i \, b x - i \, a\right )} \sinh \left (b x + a\right )^{2} + i \, b x + i \, a\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{2} \cosh \left (b x + a\right )^{2} + 2 \, b^{2} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )^{2} - b^{2}} \]
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 \[ \int x \operatorname {csch}\left (b x + a\right )^{2} \operatorname {sech}\left (b x + a\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.50, size = 179, normalized size = 2.27 \[ -\frac {2 \,{\mathrm e}^{b x +a} x}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {2 a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{b}+\frac {i \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {i \dilog \left (1+i {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{b}-\frac {i \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {i \dilog \left (1-i {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {\ln \left (1+{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{2}} \]
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 \[ -\frac {2 \, x e^{\left (b x + a\right )}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} - \frac {\log \left ({\left (e^{\left (b x + a\right )} + 1\right )} e^{\left (-a\right )}\right )}{b^{2}} + \frac {\log \left ({\left (e^{\left (b x + a\right )} - 1\right )} e^{\left (-a\right )}\right )}{b^{2}} - 8 \, \int \frac {x e^{\left (b x + a\right )}}{4 \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}\,{d x} \]
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 \[ \int \frac {x}{\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x \operatorname {csch}^{2}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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