Integrand size = 154, antiderivative size = 24 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {1}{-e^x+\frac {\log \left (x^2\right )}{2}+\log \left (-e^x+x\right )} \]
[Out]
Time = 0.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6820, 6818} \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {2}{-\log \left (x^2\right )+2 e^x-2 \log \left (x-e^x\right )} \]
[In]
[Out]
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {8 x+4 e^{2 x} x-4 e^x \left (1+x+x^2\right )}{\left (e^x-x\right ) x \left (2 e^x-\log \left (x^2\right )-2 \log \left (-e^x+x\right )\right )^2} \, dx \\ & = -\frac {2}{2 e^x-\log \left (x^2\right )-2 \log \left (-e^x+x\right )} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {2}{-2 e^x+\log \left (x^2\right )+2 \log \left (-e^x+x\right )} \]
[In]
[Out]
Time = 4.72 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(-\frac {2}{2 \,{\mathrm e}^{x}-\ln \left (x^{2}\right )-2 \ln \left (x -{\mathrm e}^{x}\right )}\) | \(25\) |
risch | \(\frac {4 i}{\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}-4 i {\mathrm e}^{x}+4 i \ln \left (x \right )+4 i \ln \left (x -{\mathrm e}^{x}\right )}\) | \(71\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {2}{2 \, e^{x} - \log \left (x^{2}\right ) - 2 \, \log \left (x - e^{x}\right )} \]
[In]
[Out]
Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {2}{- 2 e^{x} + \log {\left (x^{2} \right )} + 2 \log {\left (x - e^{x} \right )}} \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {1}{e^{x} - \log \left (x - e^{x}\right ) - \log \left (x\right )} \]
[In]
[Out]
none
Time = 0.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=-\frac {2}{2 \, e^{x} - \log \left (x^{2}\right ) - 2 \, \log \left (x - e^{x}\right )} \]
[In]
[Out]
Time = 9.90 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {8 x+4 e^{2 x} x+e^x \left (-4-4 x-4 x^2\right )}{4 e^{3 x} x-4 e^{2 x} x^2+\left (-4 e^{2 x} x+4 e^x x^2\right ) \log \left (x^2\right )+\left (e^x x-x^2\right ) \log ^2\left (x^2\right )+\left (-8 e^{2 x} x+8 e^x x^2+\left (4 e^x x-4 x^2\right ) \log \left (x^2\right )\right ) \log \left (-e^x+x\right )+\left (4 e^x x-4 x^2\right ) \log ^2\left (-e^x+x\right )} \, dx=\frac {2}{\ln \left (x^2\right )+2\,\ln \left (x-{\mathrm {e}}^x\right )-2\,{\mathrm {e}}^x} \]
[In]
[Out]