Integrand size = 7, antiderivative size = 79 \[ \int \coth ^{-1}(\tan (a+b x)) \, dx=x \coth ^{-1}(\tan (a+b x))+i x \arctan \left (e^{2 i (a+b x)}\right )-\frac {i \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b} \]
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Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6383, 4266, 2317, 2438} \[ \int \coth ^{-1}(\tan (a+b x)) \, dx=i x \arctan \left (e^{2 i (a+b x)}\right )-\frac {i \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+x \coth ^{-1}(\tan (a+b x)) \]
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Rule 2317
Rule 2438
Rule 4266
Rule 6383
Rubi steps \begin{align*} \text {integral}& = x \coth ^{-1}(\tan (a+b x))-b \int x \sec (2 a+2 b x) \, dx \\ & = x \coth ^{-1}(\tan (a+b x))+i x \arctan \left (e^{2 i (a+b x)}\right )+\frac {1}{2} \int \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac {1}{2} \int \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx \\ & = x \coth ^{-1}(\tan (a+b x))+i x \arctan \left (e^{2 i (a+b x)}\right )-\frac {i \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{4 b}+\frac {i \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{4 b} \\ & = x \coth ^{-1}(\tan (a+b x))+i x \arctan \left (e^{2 i (a+b x)}\right )-\frac {i \operatorname {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i \operatorname {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.61 \[ \int \coth ^{-1}(\tan (a+b x)) \, dx=x \coth ^{-1}(\tan (a+b x))-\frac {(-4 a+\pi -4 b x) \left (\log \left (1-i e^{-2 i (a+b x)}\right )-\log \left (1+i e^{-2 i (a+b x)}\right )\right )-(-4 a+\pi ) \log \left (\cot \left (a+\frac {\pi }{4}+b x\right )\right )+2 i \left (\operatorname {PolyLog}\left (2,-i e^{-2 i (a+b x)}\right )-\operatorname {PolyLog}\left (2,i e^{-2 i (a+b x)}\right )\right )}{8 b} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (64 ) = 128\).
Time = 1.06 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.14
method | result | size |
derivativedivides | \(\frac {\arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arccoth}\left (\tan \left (b x +a \right )\right )+\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}}{b}\) | \(169\) |
default | \(\frac {\arctan \left (\tan \left (b x +a \right )\right ) \operatorname {arccoth}\left (\tan \left (b x +a \right )\right )+\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {\arctan \left (\tan \left (b x +a \right )\right ) \ln \left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{2}-\frac {i \operatorname {dilog}\left (1+\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}+\frac {i \operatorname {dilog}\left (1-\frac {i \left (1+i \tan \left (b x +a \right )\right )^{2}}{1+\tan \left (b x +a \right )^{2}}\right )}{4}}{b}\) | \(169\) |
risch | \(\text {Expression too large to display}\) | \(1161\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (57) = 114\).
Time = 0.28 (sec) , antiderivative size = 498, normalized size of antiderivative = 6.30 \[ \int \coth ^{-1}(\tan (a+b x)) \, dx=\frac {4 \, b x \log \left (\frac {\tan \left (b x + a\right ) + 1}{\tan \left (b x + a\right ) - 1}\right ) - 2 \, {\left (b x + a\right )} \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, a \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) + i - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, a \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) + i - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b x + a\right )} \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, {\left (b x + a\right )} \log \left (\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b x + a\right )} \log \left (\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, a \log \left (\frac {\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, a \log \left (\frac {\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + i \, {\rm Li}_2\left (-\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + i \, {\rm Li}_2\left (-\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \, {\rm Li}_2\left (-\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \, {\rm Li}_2\left (-\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right )}{8 \, b} \]
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\[ \int \coth ^{-1}(\tan (a+b x)) \, dx=\int \operatorname {acoth}{\left (\tan {\left (a + b x \right )} \right )}\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (57) = 114\).
Time = 0.34 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.30 \[ \int \coth ^{-1}(\tan (a+b x)) \, dx=\frac {4 \, {\left (b x + a\right )} \operatorname {arcoth}\left (\tan \left (b x + a\right )\right ) + {\left (\arctan \left (\frac {1}{2} \, \tan \left (b x + a\right ) + \frac {1}{2}, \frac {1}{2} \, \tan \left (b x + a\right ) + \frac {1}{2}\right ) - \arctan \left (\frac {1}{2} \, \tan \left (b x + a\right ) - \frac {1}{2}, -\frac {1}{2} \, \tan \left (b x + a\right ) + \frac {1}{2}\right )\right )} \log \left (\tan \left (b x + a\right )^{2} + 1\right ) - {\left (b x + a\right )} \log \left (\frac {1}{2} \, \tan \left (b x + a\right )^{2} + \tan \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b x + a\right )} \log \left (\frac {1}{2} \, \tan \left (b x + a\right )^{2} - \tan \left (b x + a\right ) + \frac {1}{2}\right ) - i \, {\rm Li}_2\left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \tan \left (b x + a\right ) - \frac {1}{2} i + \frac {1}{2}\right ) + i \, {\rm Li}_2\left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \tan \left (b x + a\right ) + \frac {1}{2} i + \frac {1}{2}\right ) + i \, {\rm Li}_2\left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \tan \left (b x + a\right ) + \frac {1}{2} i + \frac {1}{2}\right ) - i \, {\rm Li}_2\left (-\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \tan \left (b x + a\right ) - \frac {1}{2} i + \frac {1}{2}\right )}{4 \, b} \]
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\[ \int \coth ^{-1}(\tan (a+b x)) \, dx=\int { \operatorname {arcoth}\left (\tan \left (b x + a\right )\right ) \,d x } \]
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Timed out. \[ \int \coth ^{-1}(\tan (a+b x)) \, dx=\int \mathrm {acoth}\left (\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]
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