Integrand size = 3, antiderivative size = 39 \[ \int \log (\tanh (x)) \, dx=2 x \text {arctanh}\left (e^{2 x}\right )+x \log (\tanh (x))+\frac {1}{2} \operatorname {PolyLog}\left (2,-e^{2 x}\right )-\frac {\operatorname {PolyLog}\left (2,e^{2 x}\right )}{2} \]
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Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.667, Rules used = {2628, 5569, 4267, 2317, 2438} \[ \int \log (\tanh (x)) \, dx=2 x \text {arctanh}\left (e^{2 x}\right )+\frac {1}{2} \operatorname {PolyLog}\left (2,-e^{2 x}\right )-\frac {\operatorname {PolyLog}\left (2,e^{2 x}\right )}{2}+x \log (\tanh (x)) \]
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Rule 2317
Rule 2438
Rule 2628
Rule 4267
Rule 5569
Rubi steps \begin{align*} \text {integral}& = x \log (\tanh (x))-\int x \text {csch}(x) \text {sech}(x) \, dx \\ & = x \log (\tanh (x))-2 \int x \text {csch}(2 x) \, dx \\ & = 2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (\tanh (x))+\int \log \left (1-e^{2 x}\right ) \, dx-\int \log \left (1+e^{2 x}\right ) \, dx \\ & = 2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (\tanh (x))+\frac {1}{2} \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 x}\right )-\frac {1}{2} \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 x}\right ) \\ & = 2 x \tanh ^{-1}\left (e^{2 x}\right )+x \log (\tanh (x))+\frac {1}{2} \text {Li}_2\left (-e^{2 x}\right )-\frac {\text {Li}_2\left (e^{2 x}\right )}{2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \log (\tanh (x)) \, dx=\frac {1}{2} \log (\tanh (x)) \log (1+\tanh (x))+\frac {1}{2} \operatorname {PolyLog}(2,1-\tanh (x))+\frac {1}{2} \operatorname {PolyLog}(2,-\tanh (x)) \]
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Time = 0.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {\operatorname {dilog}\left (\tanh \left (x \right )\right )}{2}+\frac {\operatorname {dilog}\left (\tanh \left (x \right )+1\right )}{2}+\frac {\ln \left (\tanh \left (x \right )\right ) \ln \left (\tanh \left (x \right )+1\right )}{2}\) | \(24\) |
default | \(\frac {\operatorname {dilog}\left (\tanh \left (x \right )\right )}{2}+\frac {\operatorname {dilog}\left (\tanh \left (x \right )+1\right )}{2}+\frac {\ln \left (\tanh \left (x \right )\right ) \ln \left (\tanh \left (x \right )+1\right )}{2}\) | \(24\) |
risch | \(x \ln \left (-1+{\mathrm e}^{2 x}\right )+\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-1+{\mathrm e}^{2 x}\right )}{1+{\mathrm e}^{2 x}}\right )}^{2} x}{2}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) {\operatorname {csgn}\left (\frac {i \left (-1+{\mathrm e}^{2 x}\right )}{1+{\mathrm e}^{2 x}}\right )}^{2} x}{2}-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )-x \ln \left (1+{\mathrm e}^{x}\right )+\operatorname {dilog}\left ({\mathrm e}^{x}\right )-\frac {i \pi {\operatorname {csgn}\left (\frac {i \left (-1+{\mathrm e}^{2 x}\right )}{1+{\mathrm e}^{2 x}}\right )}^{3} x}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )\right ) \operatorname {csgn}\left (\frac {i}{1+{\mathrm e}^{2 x}}\right ) \operatorname {csgn}\left (\frac {i \left (-1+{\mathrm e}^{2 x}\right )}{1+{\mathrm e}^{2 x}}\right ) x}{2}-\ln \left ({\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{2 x}\right )+\ln \left ({\mathrm e}^{x}\right ) \ln \left (1+i {\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}\right ) \ln \left (1-i {\mathrm e}^{x}\right )+\operatorname {dilog}\left (1+i {\mathrm e}^{x}\right )+\operatorname {dilog}\left (1-i {\mathrm e}^{x}\right )\) | \(223\) |
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Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 2.59 \[ \int \log (\tanh (x)) \, dx=x \log \left (\frac {\sinh \left (x\right )}{\cosh \left (x\right )}\right ) - x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + x \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) + x \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) - x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) - {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + {\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + {\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) - {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) \]
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\[ \int \log (\tanh (x)) \, dx=\int \log {\left (\tanh {\left (x \right )} \right )}\, dx \]
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none
Time = 0.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \log (\tanh (x)) \, dx=x \log \left (e^{\left (2 \, x\right )} + 1\right ) - x \log \left (e^{x} + 1\right ) - x \log \left (-e^{x} + 1\right ) + x \log \left (\tanh \left (x\right )\right ) + \frac {1}{2} \, {\rm Li}_2\left (-e^{\left (2 \, x\right )}\right ) - {\rm Li}_2\left (-e^{x}\right ) - {\rm Li}_2\left (e^{x}\right ) \]
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\[ \int \log (\tanh (x)) \, dx=\int { \log \left (\tanh \left (x\right )\right ) \,d x } \]
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Time = 1.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.51 \[ \int \log (\tanh (x)) \, dx=x\,\ln \left (\mathrm {tanh}\left (x\right )\right )-\frac {\mathrm {polylog}\left (2,\mathrm {tanh}\left (x\right )\right )}{2}+\frac {\mathrm {polylog}\left (2,-\mathrm {tanh}\left (x\right )\right )}{2} \]
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