Integrand size = 10, antiderivative size = 32 \[ \int \frac {\log (x)}{1+x^2} \, dx=\arctan (x) \log (x)-\frac {1}{2} i \operatorname {PolyLog}(2,-i x)+\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
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Time = 0.03 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {209, 2361, 4940, 2438} \[ \int \frac {\log (x)}{1+x^2} \, dx=\arctan (x) \log (x)-\frac {1}{2} i \operatorname {PolyLog}(2,-i x)+\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
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Rule 209
Rule 2361
Rule 2438
Rule 4940
Rubi steps \begin{align*} \text {integral}& = \tan ^{-1}(x) \log (x)-\int \frac {\tan ^{-1}(x)}{x} \, dx \\ & = \tan ^{-1}(x) \log (x)-\frac {1}{2} i \int \frac {\log (1-i x)}{x} \, dx+\frac {1}{2} i \int \frac {\log (1+i x)}{x} \, dx \\ & = \tan ^{-1}(x) \log (x)-\frac {1}{2} i \text {Li}_2(-i x)+\frac {1}{2} i \text {Li}_2(i x) \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(65\) vs. \(2(32)=64\).
Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {\log (x)}{1+x^2} \, dx=-\frac {1}{2} i \log (-i (i-x)) \log (x)+\frac {1}{2} i \log (x) \log (-i (i+x))-\frac {1}{2} i \operatorname {PolyLog}(2,-i x)+\frac {1}{2} i \operatorname {PolyLog}(2,i x) \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.62 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81
method | result | size |
meijerg | \(\left (\frac {\ln \left (x \right ) \Phi \left (-x^{2}, 1, \frac {1}{2}\right )}{2}-\frac {\Phi \left (-x^{2}, 2, \frac {1}{2}\right )}{4}\right ) x\) | \(26\) |
default | \(-\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}+\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}\) | \(46\) |
risch | \(-\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}+\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}\) | \(46\) |
parts | \(-\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}+\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}-\frac {i \operatorname {dilog}\left (i x +1\right )}{2}+\frac {i \operatorname {dilog}\left (-i x +1\right )}{2}\) | \(46\) |
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\[ \int \frac {\log (x)}{1+x^2} \, dx=\int { \frac {\log \left (x\right )}{x^{2} + 1} \,d x } \]
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\[ \int \frac {\log (x)}{1+x^2} \, dx=\int \frac {\log {\left (x \right )}}{x^{2} + 1}\, dx \]
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none
Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {\log (x)}{1+x^2} \, dx=\frac {1}{4} \, \pi \log \left (x^{2} + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (i \, x + 1\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, x + 1\right ) \]
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\[ \int \frac {\log (x)}{1+x^2} \, dx=\int { \frac {\log \left (x\right )}{x^{2} + 1} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {\log (x)}{1+x^2} \, dx=\mathrm {atan}\left (x\right )\,\ln \left (x\right )-\frac {\mathrm {polylog}\left (2,-x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {\mathrm {polylog}\left (2,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \]
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