Integrand size = 10, antiderivative size = 36 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {x}{2}+\frac {x^2}{4}-\frac {1}{2} \log (1+x)+\frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right ) \]
[Out]
Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2581, 30, 45} \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {x^2}{4}+\frac {1}{2} x^2 \log \left (\frac {x+1}{x^2}\right )+\frac {x}{2}-\frac {1}{2} \log (x+1) \]
[In]
[Out]
Rule 30
Rule 45
Rule 2581
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right )-\frac {1}{2} \int \frac {x^2}{1+x} \, dx+\int x \, dx \\ & = \frac {x^2}{2}+\frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right )-\frac {1}{2} \int \left (-1+x+\frac {1}{1+x}\right ) \, dx \\ & = \frac {x}{2}+\frac {x^2}{4}-\frac {1}{2} \log (1+x)+\frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {1}{4} \left (-2 \log (1+x)+x \left (2+x+2 x \log \left (\frac {1+x}{x^2}\right )\right )\right ) \]
[In]
[Out]
Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {x}{2}+\frac {x^{2}}{4}-\frac {\ln \left (x +1\right )}{2}+\frac {x^{2} \ln \left (\frac {x +1}{x^{2}}\right )}{2}\) | \(29\) |
parts | \(\frac {x}{2}+\frac {x^{2}}{4}-\frac {\ln \left (x +1\right )}{2}+\frac {x^{2} \ln \left (\frac {x +1}{x^{2}}\right )}{2}\) | \(29\) |
parallelrisch | \(\frac {x^{2} \ln \left (\frac {x +1}{x^{2}}\right )}{2}-\frac {1}{2}+\frac {x^{2}}{4}-\ln \left (x \right )+\frac {x}{2}-\frac {\ln \left (\frac {x +1}{x^{2}}\right )}{2}\) | \(38\) |
derivativedivides | \(\frac {x^{2} \ln \left (\frac {1+\frac {1}{x}}{x}\right )}{2}+\frac {x^{2}}{4}+\frac {x}{2}+\frac {\ln \left (\frac {1}{x}\right )}{2}-\frac {\ln \left (1+\frac {1}{x}\right )}{2}\) | \(39\) |
default | \(\frac {x^{2} \ln \left (\frac {1+\frac {1}{x}}{x}\right )}{2}+\frac {x^{2}}{4}+\frac {x}{2}+\frac {\ln \left (\frac {1}{x}\right )}{2}-\frac {\ln \left (1+\frac {1}{x}\right )}{2}\) | \(39\) |
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (\frac {x + 1}{x^{2}}\right ) + \frac {1}{4} \, x^{2} + \frac {1}{2} \, x - \frac {1}{2} \, \log \left (x + 1\right ) \]
[In]
[Out]
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.75 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {x^{2} \log {\left (\frac {x + 1}{x^{2}} \right )}}{2} + \frac {x^{2}}{4} + \frac {x}{2} - \frac {\log {\left (x + 1 \right )}}{2} \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (\frac {x + 1}{x^{2}}\right ) + \frac {1}{4} \, x^{2} + \frac {1}{2} \, x - \frac {1}{2} \, \log \left (x + 1\right ) \]
[In]
[Out]
none
Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {1}{2} \, x^{2} \log \left (\frac {x + 1}{x^{2}}\right ) + \frac {1}{4} \, x^{2} + \frac {1}{2} \, x - \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \]
[In]
[Out]
Time = 1.63 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int x \log \left (\frac {1+x}{x^2}\right ) \, dx=\frac {x}{2}-\frac {\ln \left (x\,\left (x+1\right )\right )}{3}-\frac {\ln \left (\frac {x+1}{x^2}\right )}{6}+\frac {x^2\,\ln \left (\frac {x+1}{x^2}\right )}{2}+\frac {x^2}{4} \]
[In]
[Out]