Integrand size = 76, antiderivative size = 25 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )\right ) \]
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Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {12, 6816} \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{e^3 \log \left (\frac {2-x}{x}\right )-e^3+2}\right )\right ) \]
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Rule 12
Rule 6816
Rubi steps \begin{align*} \text {integral}& = -\left (\left (2 e^3\right ) \int \frac {1}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx\right ) \\ & = \log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )\right ) \\ \end{align*}
Time = 0.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{2-e^3+e^3 \log \left (-1+\frac {2}{x}\right )}\right )\right ) \]
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Time = 7.46 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\ln \left (\ln \left (\frac {1}{{\mathrm e}^{3} \ln \left (-\frac {-2+x}{x}\right )-{\mathrm e}^{3}+2}\right )\right )\) | \(23\) |
norman | \(\ln \left (\ln \left (\frac {1}{{\mathrm e}^{3} \ln \left (\frac {2-x}{x}\right )-{\mathrm e}^{3}+2}\right )\right )\) | \(24\) |
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Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (\frac {1}{e^{3} \log \left (-\frac {x - 2}{x}\right ) - e^{3} + 2}\right )\right ) \]
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Time = 0.18 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log {\left (\log {\left (\frac {1}{e^{3} \log {\left (\frac {2 - x}{x} \right )} - e^{3} + 2} \right )} \right )} \]
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Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\log \left (\log \left (-e^{3} \log \left (x\right ) + e^{3} \log \left (-x + 2\right ) - e^{3} + 2\right )\right ) \]
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\[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\int { -\frac {2 \, e^{3}}{{\left ({\left (x^{2} - 2 \, x\right )} e^{3} \log \left (-\frac {x - 2}{x}\right ) + 2 \, x^{2} - {\left (x^{2} - 2 \, x\right )} e^{3} - 4 \, x\right )} \log \left (\frac {1}{e^{3} \log \left (-\frac {x - 2}{x}\right ) - e^{3} + 2}\right )} \,d x } \]
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Time = 32.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int -\frac {2 e^3}{\left (-4 x+2 x^2+e^3 \left (2 x-x^2\right )+e^3 \left (-2 x+x^2\right ) \log \left (\frac {2-x}{x}\right )\right ) \log \left (\frac {1}{2-e^3+e^3 \log \left (\frac {2-x}{x}\right )}\right )} \, dx=\ln \left (\ln \left (\frac {1}{{\mathrm {e}}^3\,\ln \left (-\frac {x-2}{x}\right )-{\mathrm {e}}^3+2}\right )\right ) \]
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