Integrand size = 139, antiderivative size = 27 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=-5+\frac {\log \left (10 \left (e^{3 e^x}+\frac {4}{x}\right )+x\right )}{\log (-2+x)} \]
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\[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\frac {\left (-40+\left (1+30 e^{3 e^x+x}\right ) x^2\right ) \log (-2+x)}{x \left (40+10 e^{3 e^x} x+x^2\right )}-\frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{-2+x}}{\log ^2(-2+x)} \, dx \\ & = \int \left (\frac {30 e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}+\frac {80 \log (-2+x)-40 x \log (-2+x)-2 x^2 \log (-2+x)+x^3 \log (-2+x)-40 x \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-10 e^{3 e^x} x^2 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-x^3 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) x \left (40+10 e^{3 e^x} x+x^2\right ) \log ^2(-2+x)}\right ) \, dx \\ & = 30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {80 \log (-2+x)-40 x \log (-2+x)-2 x^2 \log (-2+x)+x^3 \log (-2+x)-40 x \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-10 e^{3 e^x} x^2 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )-x^3 \log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) x \left (40+10 e^{3 e^x} x+x^2\right ) \log ^2(-2+x)} \, dx \\ & = 30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {-\frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)}{x \left (40+10 e^{3 e^x} x+x^2\right )}+\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(2-x) \log ^2(-2+x)} \, dx \\ & = 30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \left (\frac {-40+x^2}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}-\frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)}\right ) \, dx \\ & = 30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {-40+x^2}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx-\int \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)} \, dx \\ & = 30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \left (-\frac {40}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}+\frac {x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)}\right ) \, dx-\int \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)} \, dx \\ & = 30 \int \frac {e^{3 e^x+x} x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx-40 \int \frac {1}{x \left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx+\int \frac {x}{\left (40+10 e^{3 e^x} x+x^2\right ) \log (-2+x)} \, dx-\int \frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{(-2+x) \log ^2(-2+x)} \, dx \\ \end{align*}
Time = 0.95 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (10 e^{3 e^x}+\frac {40}{x}+x\right )}{\log (-2+x)} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 186, normalized size of antiderivative = 6.89
\[\frac {\ln \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{\ln \left (-2+x \right )}-\frac {-i \pi \,\operatorname {csgn}\left (i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )\right ) {\operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )\right ) \operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+i \pi {\operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )}^{3}-i \pi {\operatorname {csgn}\left (\frac {i \left (10 x \,{\mathrm e}^{3 \,{\mathrm e}^{x}}+x^{2}+40\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 \ln \left (x \right )}{2 \ln \left (-2+x \right )}\]
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (\frac {{\left (10 \, x e^{\left (x + 3 \, e^{x}\right )} + {\left (x^{2} + 40\right )} e^{x}\right )} e^{\left (-x\right )}}{x}\right )}{\log \left (x - 2\right )} \]
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Exception generated. \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (x^{2} + 10 \, x e^{\left (3 \, e^{x}\right )} + 40\right ) - \log \left (x\right )}{\log \left (x - 2\right )} \]
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Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\log \left (x^{2} + 10 \, x e^{\left (3 \, e^{x}\right )} + 40\right ) - \log \left (x\right )}{\log \left (x - 2\right )} \]
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Time = 13.37 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (80-40 x-2 x^2+x^3\right ) \log (-2+x)+e^{3 e^x+x} \left (-60 x^2+30 x^3\right ) \log (-2+x)+\left (-40 x-10 e^{3 e^x} x^2-x^3\right ) \log \left (\frac {40+10 e^{3 e^x} x+x^2}{x}\right )}{e^{3 e^x} \left (-20 x^2+10 x^3\right ) \log ^2(-2+x)+\left (-80 x+40 x^2-2 x^3+x^4\right ) \log ^2(-2+x)} \, dx=\frac {\ln \left (\frac {10\,x\,{\mathrm {e}}^{3\,{\mathrm {e}}^x}+x^2+40}{x}\right )}{\ln \left (x-2\right )} \]
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