Integrand size = 13, antiderivative size = 38 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {\arctan (x)}{2}+\frac {1}{6} \log (1+x)+\frac {1}{4} \log \left (1+x^2\right )-\frac {1}{3} \log \left (1-x+x^2\right ) \]
[Out]
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2083, 649, 209, 266, 642} \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {\arctan (x)}{2}+\frac {1}{4} \log \left (x^2+1\right )-\frac {1}{3} \log \left (x^2-x+1\right )+\frac {1}{6} \log (x+1) \]
[In]
[Out]
Rule 209
Rule 266
Rule 642
Rule 649
Rule 2083
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{6 (1+x)}+\frac {1+x}{2 \left (1+x^2\right )}+\frac {1-2 x}{3 \left (1-x+x^2\right )}\right ) \, dx \\ & = \frac {1}{6} \log (1+x)+\frac {1}{3} \int \frac {1-2 x}{1-x+x^2} \, dx+\frac {1}{2} \int \frac {1+x}{1+x^2} \, dx \\ & = \frac {1}{6} \log (1+x)-\frac {1}{3} \log \left (1-x+x^2\right )+\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}{6} \log (1+x)+\frac {1}{4} \log \left (1+x^2\right )-\frac {1}{3} \log \left (1-x+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {\arctan (x)}{2}+\frac {1}{6} \log (1+x)+\frac {1}{4} \log \left (1+x^2\right )-\frac {1}{3} \log \left (1-x+x^2\right ) \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{2}+1\right )}{4}-\frac {\ln \left (x^{2}-x +1\right )}{3}\) | \(31\) |
risch | \(\frac {\arctan \left (x \right )}{2}+\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{2}+1\right )}{4}-\frac {\ln \left (x^{2}-x +1\right )}{3}\) | \(31\) |
parallelrisch | \(\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x -i\right )}{4}-\frac {i \ln \left (x -i\right )}{4}+\frac {\ln \left (x +i\right )}{4}+\frac {i \ln \left (x +i\right )}{4}-\frac {\ln \left (x^{2}-x +1\right )}{3}\) | \(49\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]
[In]
[Out]
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {\log {\left (x + 1 \right )}}{6} + \frac {\log {\left (x^{2} + 1 \right )}}{4} - \frac {\log {\left (x^{2} - x + 1 \right )}}{3} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) \]
[In]
[Out]
Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {1}{1+x^2+x^3+x^5} \, dx=\frac {\ln \left (x+1\right )}{6}-\frac {\ln \left (x^2-x+1\right )}{3}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]
[In]
[Out]