Integrand size = 135, antiderivative size = 29 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=\frac {x}{x+\frac {2 \left (-x^2+\log \left (-4+e^4-x+x^2\right )\right )}{x}} \]
[Out]
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 6820, 12, 6843, 32} \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=\frac {2}{2-\frac {x^2}{\log \left (x^2-x+e^4-4\right )}} \]
[In]
[Out]
Rule 6
Rule 12
Rule 32
Rule 6820
Rule 6843
Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{\left (-4+e^4\right ) x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx \\ & = \int \frac {2 x \left (-x+2 x^2-2 \left (-4+e^4-x+x^2\right ) \log \left (-4+e^4-x+x^2\right )\right )}{\left (4-e^4+x-x^2\right ) \left (x^2-2 \log \left (-4+e^4-x+x^2\right )\right )^2} \, dx \\ & = 2 \int \frac {x \left (-x+2 x^2-2 \left (-4+e^4-x+x^2\right ) \log \left (-4+e^4-x+x^2\right )\right )}{\left (4-e^4+x-x^2\right ) \left (x^2-2 \log \left (-4+e^4-x+x^2\right )\right )^2} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{(-2+x)^2} \, dx,x,\frac {x^2}{\log \left (-4+e^4-x+x^2\right )}\right ) \\ & = \frac {2}{2-\frac {x^2}{\log \left (-4+e^4-x+x^2\right )}} \\ \end{align*}
Time = 3.61 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=\frac {x^2}{-x^2+2 \log \left (-4+e^4-x+x^2\right )} \]
[In]
[Out]
Time = 4.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
risch | \(-\frac {x^{2}}{x^{2}-2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}\) | \(25\) |
parallelrisch | \(-\frac {x^{2}}{x^{2}-2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}\) | \(25\) |
norman | \(-\frac {2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}{x^{2}-2 \ln \left ({\mathrm e}^{4}+x^{2}-x -4\right )}\) | \(33\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=-\frac {x^{2}}{x^{2} - 2 \, \log \left (x^{2} - x + e^{4} - 4\right )} \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=\frac {x^{2}}{- x^{2} + 2 \log {\left (x^{2} - x - 4 + e^{4} \right )}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=-\frac {x^{2}}{x^{2} - 2 \, \log \left (x^{2} - x + e^{4} - 4\right )} \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=-\frac {x^{2}}{x^{2} - 2 \, \log \left (x^{2} - x + e^{4} - 4\right )} \]
[In]
[Out]
Time = 21.94 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {2 x^2-4 x^3+\left (-16 x+4 e^4 x-4 x^2+4 x^3\right ) \log \left (-4+e^4-x+x^2\right )}{-4 x^4+e^4 x^4-x^5+x^6+\left (16 x^2-4 e^4 x^2+4 x^3-4 x^4\right ) \log \left (-4+e^4-x+x^2\right )+\left (-16+4 e^4-4 x+4 x^2\right ) \log ^2\left (-4+e^4-x+x^2\right )} \, dx=\frac {x^2}{2\,\ln \left (x^2-x+{\mathrm {e}}^4-4\right )-x^2} \]
[In]
[Out]