Integrand size = 13, antiderivative size = 25 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{2}+\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \]
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Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {720, 31, 649, 209, 266} \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{2}-\frac {1}{4} \log \left (x^2+1\right )+\frac {1}{2} \log (x+1) \]
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Rule 31
Rule 209
Rule 266
Rule 649
Rule 720
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {1}{1+x} \, dx+\frac {1}{2} \int \frac {1-x}{1+x^2} \, dx \\ & = \frac {1}{2} \log (1+x)+\frac {1}{2} \int \frac {1}{1+x^2} \, dx-\frac {1}{2} \int \frac {x}{1+x^2} \, dx \\ & = \frac {1}{2} \tan ^{-1}(x)+\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {\arctan (x)}{2}+\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \]
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Time = 0.80 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
default | \(\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}\) | \(20\) |
risch | \(\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}\) | \(20\) |
parallelrisch | \(\frac {\ln \left (x +1\right )}{2}-\frac {\ln \left (x -i\right )}{4}-\frac {i \ln \left (x -i\right )}{4}-\frac {\ln \left (x +i\right )}{4}+\frac {i \ln \left (x +i\right )}{4}\) | \(38\) |
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Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x + 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {\log {\left (x + 1 \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{4} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x + 1\right ) \]
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Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(1+x) \left (1+x^2\right )} \, dx=\frac {\ln \left (x+1\right )}{2}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]
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