Integrand size = 56, antiderivative size = 23 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\log \left (1-10 e^{e^x+x} (5-5 (2-x))\right )} \]
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\[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {50 e^{e^x+x} \left (-e^x+x+e^x x\right )}{\left (1-e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx \\ & = 50 \int \frac {e^{e^x+x} \left (-e^x+x+e^x x\right )}{\left (1-e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx \\ & = 50 \int \left (-\frac {e^x}{50 \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}-\frac {e^x \left (1+50 e^{e^x} x\right )}{50 \left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}\right ) \, dx \\ & = -\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx-\int \frac {e^x \left (1+50 e^{e^x} x\right )}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx \\ & = -\int \left (\frac {e^x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}+\frac {50 e^{e^x+x} x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )}\right ) \, dx-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx \\ & = -\left (50 \int \frac {e^{e^x+x} x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\right )-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx-\int \frac {e^x}{\left (-1-50 e^{e^x+x}+50 e^{e^x+x} x\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx \\ & = -\left (50 \int \frac {e^{e^x+x} x}{\left (-1+50 e^{e^x+x} (-1+x)\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx\right )-\int \frac {e^x}{\log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx-\int \frac {e^x}{\left (-1+50 e^{e^x+x} (-1+x)\right ) \log ^2\left (1-50 e^{e^x+x} (-1+x)\right )} \, dx \\ \end{align*}
Time = 0.69 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\log \left (1-50 e^{e^x+x} (-1+x)\right )} \]
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Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {1}{\ln \left (\left (-50 x +50\right ) {\mathrm e}^{{\mathrm e}^{x}+x}+1\right )}\) | \(17\) |
parallelrisch | \(\frac {1}{\ln \left (\left (-50 x +50\right ) {\mathrm e}^{{\mathrm e}^{x}+x}+1\right )}\) | \(17\) |
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Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\log \left (-50 \, {\left (x - 1\right )} e^{\left (x + e^{x}\right )} + 1\right )} \]
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Time = 0.32 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\log {\left (\left (50 - 50 x\right ) e^{x + e^{x}} + 1 \right )}} \]
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Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\log \left (-50 \, {\left (x - 1\right )} e^{\left (x + e^{x}\right )} + 1\right )} \]
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Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\log \left (-50 \, x e^{\left (x + e^{x}\right )} + 50 \, e^{\left (x + e^{x}\right )} + 1\right )} \]
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Time = 0.34 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{e^x+x} \left (e^x (50-50 x)-50 x\right )}{\left (-1+e^{e^x+x} (-50+50 x)\right ) \log ^2\left (1+e^{e^x+x} (50-50 x)\right )} \, dx=\frac {1}{\ln \left (1-{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\left (50\,x-50\right )\right )} \]
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