Integrand size = 218, antiderivative size = 23 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \log (x) \left (5+\log \left (-5 e^{-\frac {2-x}{e^9}}+x\right )\right ) \]
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\[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=\int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-5 \left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)+5 e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right )}{e^9 x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )} \, dx \\ & = \frac {\int \frac {-5 \left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)+5 e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right )}{x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )} \, dx}{e^9} \\ & = \frac {\int \left (\frac {5 e^{\frac {2}{e^9}} \left (e^9-x\right ) \log (x)}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x}+\frac {5 \left (5 e^9+x \log (x)+e^9 \log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right )}{x}\right ) \, dx}{e^9} \\ & = \frac {5 \int \frac {5 e^9+x \log (x)+e^9 \log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx}{e^9}+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\left (e^9-x\right ) \log (x)}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx \\ & = \frac {5 \int \left (\frac {5 e^9+x \log (x)}{x}+\frac {e^9 \log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x}\right ) \, dx}{e^9}-\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {-e^9 \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx-\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx \\ & = 5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\frac {5 \int \frac {5 e^9+x \log (x)}{x} \, dx}{e^9}-\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \left (-\frac {e^9 \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x}-\frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x}\right ) \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx \\ & = 5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\frac {5 \int \left (\frac {5 e^9}{x}+\log (x)\right ) \, dx}{e^9}+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx+\left (5 e^{\frac {2}{e^9}}\right ) \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx \\ & = 25 \log (x)+5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\frac {5 \int \log (x) \, dx}{e^9}+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx+\left (5 e^{\frac {2}{e^9}}\right ) \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx \\ & = -\frac {5 x}{e^9}+25 \log (x)+\frac {5 x \log (x)}{e^9}+5 \int \frac {\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )}{x} \, dx+\left (5 e^{-9+\frac {2}{e^9}}\right ) \int \frac {\int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx}{x} \, dx+\left (5 e^{\frac {2}{e^9}}\right ) \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx}{x} \, dx-\left (5 e^{-9+\frac {2}{e^9}} \log (x)\right ) \int \frac {x}{-5 e^{\frac {x}{e^9}}+e^{\frac {2}{e^9}} x} \, dx-\left (5 e^{\frac {2}{e^9}} \log (x)\right ) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x} \, dx \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \log (x) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.41 (sec) , antiderivative size = 212, normalized size of antiderivative = 9.22
\[-5 \ln \left (x \right ) \ln \left ({\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}\right )+5 \ln \left (x \right ) \ln \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )-\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )}{2}+\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )^{2}}{2}+\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )^{2}}{2}-\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )^{3}}{2}+25 \ln \left (x \right )\]
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Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \, \log \left ({\left (x e^{\left (-{\left (x - 9 \, e^{9} - 2\right )} e^{\left (-9\right )}\right )} - 5 \, e^{9}\right )} e^{\left ({\left (x - 9 \, e^{9} - 2\right )} e^{\left (-9\right )}\right )}\right ) \log \left (x\right ) + 25 \, \log \left (x\right ) \]
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Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \log {\left (x \right )} \log {\left (\left (x e^{\frac {2 - x}{e^{9}}} - 5\right ) e^{- \frac {2 - x}{e^{9}}} \right )} + 25 \log {\left (x \right )} \]
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Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \, {\left (e^{9} \log \left (x e^{\left (2 \, e^{\left (-9\right )}\right )} - 5 \, e^{\left (x e^{\left (-9\right )}\right )}\right ) \log \left (x\right ) + {\left (5 \, e^{9} - 2\right )} \log \left (x\right )\right )} e^{\left (-9\right )} \]
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Time = 0.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \, {\left (e^{9} \log \left (x e^{\left (2 \, e^{\left (-9\right )}\right )} - 5 \, e^{\left (x e^{\left (-9\right )}\right )}\right ) \log \left (x\right ) + 5 \, e^{9} \log \left (x\right ) - 2 \, \log \left (x\right )\right )} e^{\left (-9\right )} \]
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Time = 12.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5\,\ln \left (x\right )\,\left (\ln \left (x-5\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-9}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-9}}\right )+5\right ) \]
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