Integrand size = 80, antiderivative size = 22 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{-e^{e^{20}}-2 x+x^2-\log (x)} \]
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\[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {1-e^{e^{20}}-x^2-\log (x)}{\left (e^{e^{20}}-(-2+x) x+\log (x)\right )^2} \, dx \\ & = \int \left (-\frac {-1-2 x+2 x^2}{\left (-e^{e^{20}}-2 x+x^2-\log (x)\right )^2}-\frac {1}{e^{e^{20}}+2 x-x^2+\log (x)}\right ) \, dx \\ & = -\int \frac {-1-2 x+2 x^2}{\left (-e^{e^{20}}-2 x+x^2-\log (x)\right )^2} \, dx-\int \frac {1}{e^{e^{20}}+2 x-x^2+\log (x)} \, dx \\ & = -\int \frac {1}{e^{e^{20}}+2 x-x^2+\log (x)} \, dx-\int \left (-\frac {2 x}{\left (-e^{e^{20}}-2 x+x^2-\log (x)\right )^2}+\frac {2 x^2}{\left (-e^{e^{20}}-2 x+x^2-\log (x)\right )^2}-\frac {1}{\left (e^{e^{20}}+2 x-x^2+\log (x)\right )^2}\right ) \, dx \\ & = 2 \int \frac {x}{\left (-e^{e^{20}}-2 x+x^2-\log (x)\right )^2} \, dx-2 \int \frac {x^2}{\left (-e^{e^{20}}-2 x+x^2-\log (x)\right )^2} \, dx+\int \frac {1}{\left (e^{e^{20}}+2 x-x^2+\log (x)\right )^2} \, dx-\int \frac {1}{e^{e^{20}}+2 x-x^2+\log (x)} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=-\frac {x}{e^{e^{20}}+2 x-x^2+\log (x)} \]
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Time = 0.92 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
default | \(-\frac {x}{-x^{2}+{\mathrm e}^{{\mathrm e}^{20}}+\ln \left (x \right )+2 x}\) | \(20\) |
norman | \(-\frac {x}{-x^{2}+{\mathrm e}^{{\mathrm e}^{20}}+\ln \left (x \right )+2 x}\) | \(20\) |
risch | \(-\frac {x}{-x^{2}+{\mathrm e}^{{\mathrm e}^{20}}+\ln \left (x \right )+2 x}\) | \(20\) |
parallelrisch | \(-\frac {x}{-x^{2}+{\mathrm e}^{{\mathrm e}^{20}}+\ln \left (x \right )+2 x}\) | \(20\) |
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{x^{2} - 2 \, x - e^{\left (e^{20}\right )} - \log \left (x\right )} \]
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Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=- \frac {x}{- x^{2} + 2 x + \log {\left (x \right )} + e^{e^{20}}} \]
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Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{x^{2} - 2 \, x - e^{\left (e^{20}\right )} - \log \left (x\right )} \]
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Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=\frac {x}{x^{2} - 2 \, x - e^{\left (e^{20}\right )} - \log \left (x\right )} \]
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Time = 11.71 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1-e^{e^{20}}-x^2-\log (x)}{e^{2 e^{20}}+4 x^2-4 x^3+x^4+e^{e^{20}} \left (4 x-2 x^2\right )+\left (2 e^{e^{20}}+4 x-2 x^2\right ) \log (x)+\log ^2(x)} \, dx=-\frac {x}{2\,x+{\mathrm {e}}^{{\mathrm {e}}^{20}}+\ln \left (x\right )-x^2} \]
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