Integrand size = 16, antiderivative size = 27 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{2}+\log (x)-\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \]
[Out]
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {908, 649, 209, 266} \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{2}-\frac {1}{4} \log \left (x^2+1\right )+\log (x)-\frac {1}{2} \log (x+1) \]
[In]
[Out]
Rule 209
Rule 266
Rule 649
Rule 908
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x}-\frac {1}{2 (1+x)}+\frac {-1-x}{2 \left (1+x^2\right )}\right ) \, dx \\ & = \log (x)-\frac {1}{2} \log (1+x)+\frac {1}{2} \int \frac {-1-x}{1+x^2} \, dx \\ & = \log (x)-\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 {\arctan (x)}{2}+\log (x)-\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{2}+\log (x)-\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \]
[In]
[Out]
Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\frac {\arctan \left (x \right )}{2}+\ln \left (x \right )-\frac {\ln \left (1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}\) | \(22\) |
risch | \(-\frac {\arctan \left (x \right )}{2}+\ln \left (x \right )-\frac {\ln \left (1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}\) | \(22\) |
parallelrisch | \(\ln \left (x \right )-\frac {\ln \left (1+x \right )}{2}-\frac {\ln \left (x -i\right )}{4}+\frac {i \ln \left (x -i\right )}{4}-\frac {\ln \left (x +i\right )}{4}-\frac {i \ln \left (x +i\right )}{4}\) | \(40\) |
[In]
[Out]
none
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) + \log \left (x\right ) \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x + 1 \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{4} - \frac {\operatorname {atan}{\left (x \right )}}{2} \]
[In]
[Out]
none
Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) + \log \left (x\right ) \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x+1\right )}{2}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]
[In]
[Out]