Integrand size = 119, antiderivative size = 27 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\left (3-\frac {5}{3} e^{\frac {e^x}{x \left (x+x \log \left (\frac {1}{x}\right )\right )}}\right )^2 \]
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Time = 2.39 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {6820, 12, 6840} \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25}{9} e^{\frac {2 e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}}-10 e^{\frac {e^x}{x^2 \left (\log \left (\frac {1}{x}\right )+1\right )}} \]
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Rule 12
Rule 6820
Rule 6840
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e^{x+\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \left (9-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \left (1-x-(-2+x) \log \left (\frac {1}{x}\right )\right )}{9 x^3 \left (1+\log \left (\frac {1}{x}\right )\right )^2} \, dx \\ & = \frac {10}{9} \int \frac {e^{x+\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \left (9-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \left (1-x-(-2+x) \log \left (\frac {1}{x}\right )\right )}{x^3 \left (1+\log \left (\frac {1}{x}\right )\right )^2} \, dx \\ & = \frac {2}{9} \text {Subst}\left (\int (9+x) \, dx,x,-5 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \\ & = -10 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {25}{9} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}} \\ \end{align*}
Time = 5.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {10}{9} \left (-9 e^{\frac {e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}+\frac {5}{2} e^{\frac {2 e^x}{x^2 \left (1+\log \left (\frac {1}{x}\right )\right )}}\right ) \]
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Time = 7.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26
method | result | size |
risch | \(\frac {25 \,{\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}}{9}-10 \,{\mathrm e}^{-\frac {{\mathrm e}^{x}}{x^{2} \left (\ln \left (x \right )-1\right )}}\) | \(34\) |
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25}{9} \, e^{\left (\frac {2 \, e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} - 10 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (\frac {1}{x}\right ) + x^{2}}\right )} \]
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Time = 0.50 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {25 e^{\frac {2 e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}}}{9} - 10 e^{\frac {e^{x}}{x^{2} \log {\left (\frac {1}{x} \right )} + x^{2}}} \]
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Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=-\frac {5}{9} \, {\left (18 \, e^{\left (\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} - 5\right )} e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=-10 \, e^{\left (-\frac {e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} + \frac {25}{9} \, e^{\left (-\frac {2 \, e^{x}}{x^{2} \log \left (x\right ) - x^{2}}\right )} \]
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Time = 16.58 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (90-90 x)+e^x (180-90 x) \log \left (\frac {1}{x}\right )\right )+e^{\frac {2 e^x}{x^2+x^2 \log \left (\frac {1}{x}\right )}} \left (e^x (-50+50 x)+e^x (-100+50 x) \log \left (\frac {1}{x}\right )\right )}{9 x^3+18 x^3 \log \left (\frac {1}{x}\right )+9 x^3 \log ^2\left (\frac {1}{x}\right )} \, dx=\frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}\,\left (5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x^2\,\ln \left (\frac {1}{x}\right )+x^2}}-18\right )}{9} \]
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