Integrand size = 170, antiderivative size = 33 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=5-\frac {\left (x-\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right )\right ) (x+2 \log (x+\log (9)))}{x} \]
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\[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=\int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (-2+e^x-x\right ) \log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) (-x+(x+\log (9)) \log (x+\log (9)))-x \left (x \left (4-2 e^x+x^2+\log (9)+x (3+\log (9))\right )+2 \left (-1+e^x\right ) (x+\log (9)) \log (x+\log (9))\right )}{x^2 \left (2-e^x+x\right ) (x+\log (9))} \, dx \\ & = \int \left (\frac {(1+x) (x+2 \log (x+\log (9)))}{\left (-2+e^x-x\right ) x}+\frac {2 \left (x+\log (5)-\log \left (-2+e^x-x\right )\right ) (-x+x \log (x+\log (9))+\log (9) \log (x+\log (9)))}{x^2 (x+\log (9))}\right ) \, dx \\ & = 2 \int \frac {\left (x+\log (5)-\log \left (-2+e^x-x\right )\right ) (-x+x \log (x+\log (9))+\log (9) \log (x+\log (9)))}{x^2 (x+\log (9))} \, dx+\int \frac {(1+x) (x+2 \log (x+\log (9)))}{\left (-2+e^x-x\right ) x} \, dx \\ & = 2 \int \frac {\left (x+\log (5)-\log \left (-2+e^x-x\right )\right ) (-x+(x+\log (9)) \log (x+\log (9)))}{x^2 (x+\log (9))} \, dx+\int \left (\frac {x+2 \log (x+\log (9))}{-2+e^x-x}+\frac {x+2 \log (x+\log (9))}{\left (-2+e^x-x\right ) x}\right ) \, dx \\ & = 2 \int \left (\frac {-x-\log (5)+\log \left (-2+e^x-x\right )}{x (x+\log (9))}+\frac {\left (x+\log (5)-\log \left (-2+e^x-x\right )\right ) \log (x+\log (9))}{x^2}\right ) \, dx+\int \frac {x+2 \log (x+\log (9))}{-2+e^x-x} \, dx+\int \frac {x+2 \log (x+\log (9))}{\left (-2+e^x-x\right ) x} \, dx \\ & = 2 \int \frac {-x-\log (5)+\log \left (-2+e^x-x\right )}{x (x+\log (9))} \, dx+2 \int \frac {\left (x+\log (5)-\log \left (-2+e^x-x\right )\right ) \log (x+\log (9))}{x^2} \, dx+\int \left (\frac {x}{-2+e^x-x}+\frac {2 \log (x+\log (9))}{-2+e^x-x}\right ) \, dx+\int \left (\frac {1}{-2+e^x-x}+\frac {2 \log (x+\log (9))}{\left (-2+e^x-x\right ) x}\right ) \, dx \\ & = 2 \int \left (\frac {-x-\log (5)}{x (x+\log (9))}+\frac {\log \left (-2+e^x-x\right )}{x (x+\log (9))}\right ) \, dx+2 \int \frac {\log (x+\log (9))}{-2+e^x-x} \, dx+2 \int \frac {\log (x+\log (9))}{\left (-2+e^x-x\right ) x} \, dx+2 \int \left (\frac {\log (x+\log (9))}{x}+\frac {\log (5) \log (x+\log (9))}{x^2}-\frac {\log \left (-2+e^x-x\right ) \log (x+\log (9))}{x^2}\right ) \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = 2 \int \frac {-x-\log (5)}{x (x+\log (9))} \, dx+2 \int \frac {\log \left (-2+e^x-x\right )}{x (x+\log (9))} \, dx+2 \int \frac {\log (x+\log (9))}{x} \, dx-2 \int \frac {\log \left (-2+e^x-x\right ) \log (x+\log (9))}{x^2} \, dx-2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)} \, dx-2 \int \frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+(2 \log (5)) \int \frac {\log (x+\log (9))}{x^2} \, dx+(2 \log (x+\log (9))) \int \frac {1}{-2+e^x-x} \, dx+(2 \log (x+\log (9))) \int \frac {1}{\left (-2+e^x-x\right ) x} \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = 2 \log (x) \log (\log (9))-\frac {2 \log (5) \log (x+\log (9))}{x}+\frac {2 \log \left (-2+e^x-x\right ) \log (x+\log (9))}{x}+2 \int \left (-\frac {\log (5)}{x \log (9)}-\frac {\log \left (\frac {9}{5}\right )}{\log (9) (x+\log (9))}\right ) \, dx-2 \int \frac {\log \left (-2+e^x-x\right )}{x (x+\log (9))} \, dx+2 \int \left (\frac {\log \left (-2+e^x-x\right )}{x \log (9)}-\frac {\log \left (-2+e^x-x\right )}{\log (9) (x+\log (9))}\right ) \, dx+2 \int \frac {\log \left (1+\frac {x}{\log (9)}\right )}{x} \, dx-2 \int \frac {\left (-1+e^x\right ) \log (x+\log (9))}{\left (-2+e^x-x\right ) x} \, dx-2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)} \, dx-2 \int \frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+(2 \log (5)) \int \frac {1}{x (x+\log (9))} \, dx+(2 \log (x+\log (9))) \int \frac {1}{-2+e^x-x} \, dx+(2 \log (x+\log (9))) \int \frac {1}{\left (-2+e^x-x\right ) x} \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = -\frac {2 \log (5) \log (x)}{\log (9)}+2 \log (x) \log (\log (9))-\frac {2 \log (5) \log (x+\log (9))}{x}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}+\frac {2 \log \left (-2+e^x-x\right ) \log (x+\log (9))}{x}-2 \operatorname {PolyLog}\left (2,-\frac {x}{\log (9)}\right )-2 \int \left (\frac {\log \left (-2+e^x-x\right )}{x \log (9)}-\frac {\log \left (-2+e^x-x\right )}{\log (9) (x+\log (9))}\right ) \, dx-2 \int \left (\frac {\log (x+\log (9))}{x}+\frac {(1+x) \log (x+\log (9))}{\left (-2+e^x-x\right ) x}\right ) \, dx-2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)} \, dx-2 \int \frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+\frac {2 \int \frac {\log \left (-2+e^x-x\right )}{x} \, dx}{\log (9)}-\frac {2 \int \frac {\log \left (-2+e^x-x\right )}{x+\log (9)} \, dx}{\log (9)}+\frac {(2 \log (5)) \int \frac {1}{x} \, dx}{\log (9)}-\frac {(2 \log (5)) \int \frac {1}{x+\log (9)} \, dx}{\log (9)}+(2 \log (x+\log (9))) \int \frac {1}{-2+e^x-x} \, dx+(2 \log (x+\log (9))) \int \frac {1}{\left (-2+e^x-x\right ) x} \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = 2 \log (x) \log (\log (9))-\frac {2 \log (5) \log (x+\log (9))}{x}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}-\frac {2 \log (5) \log (x+\log (9))}{\log (9)}+\frac {2 \log \left (-2+e^x-x\right ) \log (x+\log (9))}{x}-2 \operatorname {PolyLog}\left (2,-\frac {x}{\log (9)}\right )-2 \int \frac {\log (x+\log (9))}{x} \, dx-2 \int \frac {(1+x) \log (x+\log (9))}{\left (-2+e^x-x\right ) x} \, dx-2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)} \, dx-2 \int \frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+(2 \log (x+\log (9))) \int \frac {1}{-2+e^x-x} \, dx+(2 \log (x+\log (9))) \int \frac {1}{\left (-2+e^x-x\right ) x} \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = -\frac {2 \log (5) \log (x+\log (9))}{x}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}-\frac {2 \log (5) \log (x+\log (9))}{\log (9)}+\frac {2 \log \left (-2+e^x-x\right ) \log (x+\log (9))}{x}-2 \operatorname {PolyLog}\left (2,-\frac {x}{\log (9)}\right )-2 \int \frac {\log \left (1+\frac {x}{\log (9)}\right )}{x} \, dx-2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)} \, dx-2 \int \frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx+\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = -\frac {2 \log (5) \log (x+\log (9))}{x}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}-\frac {2 \log (5) \log (x+\log (9))}{\log (9)}+\frac {2 \log \left (-2+e^x-x\right ) \log (x+\log (9))}{x}-2 \int \frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)} \, dx-2 \int \frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)} \, dx+2 \int \left (\frac {\int \frac {1}{-2+e^x-x} \, dx}{x+\log (9)}+\frac {\int \frac {1}{\left (-2+e^x-x\right ) x} \, dx}{x+\log (9)}\right ) \, dx+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ & = -\frac {2 \log (5) \log (x+\log (9))}{x}-\frac {2 \log \left (\frac {9}{5}\right ) \log (x+\log (9))}{\log (9)}-\frac {2 \log (5) \log (x+\log (9))}{\log (9)}+\frac {2 \log \left (-2+e^x-x\right ) \log (x+\log (9))}{x}+\int \frac {1}{-2+e^x-x} \, dx+\int \frac {x}{-2+e^x-x} \, dx \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-x+\log \left (2-e^x+x\right )-2 \log (x+\log (9))+\frac {2 \log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \log (x+\log (9))}{x} \]
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Time = 3.84 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33
method | result | size |
risch | \(\frac {2 \ln \left (2 \ln \left (3\right )+x \right ) \ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right )}{x}-2 \ln \left (2 \ln \left (3\right )+x \right )-x +\ln \left ({\mathrm e}^{x}-2-x \right )\) | \(44\) |
parallelrisch | \(\frac {4 x \ln \left (3\right )-x^{2}-2 \ln \left (2 \ln \left (3\right )+x \right ) x +\ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right ) x +2 \ln \left (\frac {{\mathrm e}^{x}}{5}-\frac {2}{5}-\frac {x}{5}\right ) \ln \left (2 \ln \left (3\right )+x \right )}{x}\) | \(57\) |
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} + 2 \, x \log \left (x + 2 \, \log \left (3\right )\right ) - {\left (x + 2 \, \log \left (x + 2 \, \log \left (3\right )\right )\right )} \log \left (-\frac {1}{5} \, x + \frac {1}{5} \, e^{x} - \frac {2}{5}\right )}{x} \]
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Time = 0.55 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=- x - 2 \log {\left (x + 2 \log {\left (3 \right )} \right )} + \log {\left (- x + e^{x} - 2 \right )} + \frac {2 \log {\left (x + 2 \log {\left (3 \right )} \right )} \log {\left (- \frac {x}{5} + \frac {e^{x}}{5} - \frac {2}{5} \right )}}{x} \]
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Time = 0.35 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} + 2 \, {\left (x + \log \left (5\right )\right )} \log \left (x + 2 \, \log \left (3\right )\right ) - {\left (x + 2 \, \log \left (x + 2 \, \log \left (3\right )\right )\right )} \log \left (-x + e^{x} - 2\right )}{x} \]
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Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.76 \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=-\frac {x^{2} - x \log \left (x - e^{x} + 2\right ) + 2 \, x \log \left (x + 2 \, \log \left (3\right )\right ) + 2 \, \log \left (5\right ) \log \left (x + 2 \, \log \left (3\right )\right ) - 2 \, \log \left (x + 2 \, \log \left (3\right )\right ) \log \left (-x + e^{x} - 2\right )}{x} \]
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Timed out. \[ \int \frac {4 x^2-2 e^x x^2+3 x^3+x^4+\left (x^2+x^3\right ) \log (9)+\left (-2 x^2-2 x \log (9)+e^x \left (2 x^2+2 x \log (9)\right )\right ) \log (x+\log (9))+\log \left (\frac {1}{5} \left (-2+e^x-x\right )\right ) \left (-4 x+2 e^x x-2 x^2+\left (4 x+2 x^2+e^x (-2 x-2 \log (9))+(4+2 x) \log (9)\right ) \log (x+\log (9))\right )}{-2 x^3-x^4+\left (-2 x^2-x^3\right ) \log (9)+e^x \left (x^3+x^2 \log (9)\right )} \, dx=\int -\frac {4\,x^2-\ln \left (x+2\,\ln \left (3\right )\right )\,\left (4\,x\,\ln \left (3\right )-{\mathrm {e}}^x\,\left (2\,x^2+4\,\ln \left (3\right )\,x\right )+2\,x^2\right )-\ln \left (\frac {{\mathrm {e}}^x}{5}-\frac {x}{5}-\frac {2}{5}\right )\,\left (4\,x-\ln \left (x+2\,\ln \left (3\right )\right )\,\left (4\,x+2\,\ln \left (3\right )\,\left (2\,x+4\right )-{\mathrm {e}}^x\,\left (2\,x+4\,\ln \left (3\right )\right )+2\,x^2\right )-2\,x\,{\mathrm {e}}^x+2\,x^2\right )-2\,x^2\,{\mathrm {e}}^x+3\,x^3+x^4+2\,\ln \left (3\right )\,\left (x^3+x^2\right )}{2\,\ln \left (3\right )\,\left (x^3+2\,x^2\right )-{\mathrm {e}}^x\,\left (x^3+2\,\ln \left (3\right )\,x^2\right )+2\,x^3+x^4} \,d x \]
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