Integrand size = 119, antiderivative size = 25 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=1+\log \left (-e^4+e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \]
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Time = 0.92 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6873, 6816} \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x x-10 x^2-\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{\left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx \\ & = \log \left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (-e^4+e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \]
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Time = 26.68 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\ln \left (-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{x}-5 x^{2}\right )}}\right )\) | \(21\) |
parallelrisch | \(\ln \left (-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{x}-5 x^{2}\right )}}\right )\) | \(21\) |
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Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (-e^{4} + e^{\left (\frac {x}{\log \left (-{\left (5 \, x^{2} e^{4} - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )}\right )}\right )}\right ) \]
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Timed out. \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\text {Timed out} \]
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (-e^{4} + e^{\left (\frac {x}{\log \left (-5 \, x^{2} + e^{x}\right )}\right )}\right ) \]
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Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (e^{4} - e^{\left (\frac {x}{\log \left (-5 \, x^{2} + e^{x}\right )}\right )}\right ) \]
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Time = 8.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\ln \left ({\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^x-5\,x^2\right )}}-{\mathrm {e}}^4\right ) \]
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