Integrand size = 130, antiderivative size = 34 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=25 e^{-5+e^{2 e^{-x}}} \left (x+x \left (1-\frac {x}{e^3-\log (x)}\right )\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(34)=68\).
Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.32, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2326} \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {25 e^{x+e^{2 e^{-x}}-5} \left (-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 e^6 x+2 x \log ^2(x)\right )}{e^{x+6}+e^x \log ^2(x)-2 e^{x+3} \log (x)} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {25 e^{-5+e^{2 e^{-x}}+x} \left (2 e^6 x-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 x \log ^2(x)\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \\ \end{align*}
Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=25 e^{-5+e^{2 e^{-x}}} x \left (2+\frac {x}{-e^3+\log (x)}\right ) \]
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Time = 39.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {25 \left (2 \,{\mathrm e}^{3}-x -2 \ln \left (x \right )\right ) x \,{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5}}{{\mathrm e}^{3}-\ln \left (x \right )}\) | \(35\) |
parallelrisch | \(-\frac {-50 \,{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5} x +25 \,{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5} x^{2}+50 x \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5}}{{\mathrm e}^{3}-\ln \left (x \right )}\) | \(58\) |
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Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=-\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{3} - \log \left (x\right )} \]
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Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {\left (- 25 x^{2} - 50 x \log {\left (x \right )} + 50 x e^{3}\right ) e^{e^{2 e^{- x}} - 5}}{- \log {\left (x \right )} + e^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )}\right )}}{e^{5} \log \left (x\right ) - e^{8}} \]
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\[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\int { \frac {25 \, {\left (2 \, {\left (x - 2 \, e^{3}\right )} e^{x} \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )^{2} - {\left (2 \, x e^{3} + x - 2 \, e^{6}\right )} e^{x} + 2 \, {\left (x^{2} e^{3} - 2 \, x \log \left (x\right )^{2} - 2 \, x e^{6} - {\left (x^{2} - 4 \, x e^{3}\right )} \log \left (x\right )\right )} e^{\left (2 \, e^{\left (-x\right )}\right )}\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{x} \log \left (x\right )^{2} - 2 \, e^{\left (x + 3\right )} \log \left (x\right ) + e^{\left (x + 6\right )}} \,d x } \]
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Time = 12.81 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=50\,x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}-\frac {25\,x^2\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}}{{\mathrm {e}}^3-\ln \left (x\right )} \]
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