Integrand size = 73, antiderivative size = 23 \[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=e^{3-\frac {e^{-1-x}}{(-3+x) x}} x \]
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\[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=\int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{x \left (9-6 x+x^2\right )} \, dx \\ & = \int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{(-3+x)^2 x} \, dx \\ & = \int \left (\frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x}+\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (-3-x+x^2\right )}{(-3+x)^2 x^2}\right ) \, dx \\ & = \int \frac {\exp \left (\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x} \, dx+\int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right ) \left (-3-x+x^2\right )}{(-3+x)^2 x^2} \, dx \\ & = \int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx+\int \left (\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 (-3+x)^2}+\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 (-3+x)}-\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 x^2}-\frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{3 x}\right ) \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx+\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{-3+x} \, dx-\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x^2} \, dx-\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx \\ & = -\left (\frac {1}{3} \int e^{2-\frac {e^{-1-x}}{(-3+x) x}-x} \, dx\right )+\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx-\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{x} \, dx+\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x} x}{-3+x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx \\ & = -\left (\frac {1}{3} \int e^{2-\frac {e^{-1-x}}{(-3+x) x}-x} \, dx\right )+\frac {1}{3} \int \left (e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}+\frac {3 e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{-3+x}\right ) \, dx+\frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx-\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx \\ & = \frac {1}{3} \int \frac {\exp \left (-1-x+\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}\right )}{(-3+x)^2} \, dx-\frac {1}{3} \int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{x} \, dx+\int e^{3-\frac {e^{-1-x}}{(-3+x) x}} \, dx+\int \frac {e^{2-\frac {e^{-1-x}}{(-3+x) x}-x}}{-3+x} \, dx \\ \end{align*}
Time = 2.36 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=e^{3-\frac {e^{-1-x}}{(-3+x) x}} x \]
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Time = 4.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {-{\mathrm e}^{-\ln \left (x \right )-x -1}+\left (-3+x \right ) \ln \left (x \,{\mathrm e}^{3}\right )}{-3+x}}\) | \(30\) |
risch | \({\mathrm e}^{\frac {x^{2} \ln \left (x \right )-3 x \ln \left (x \right )+3 x^{2}-{\mathrm e}^{-1-x}-9 x}{x \left (-3+x \right )}}\) | \(39\) |
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Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=e^{\left (\frac {{\left (x - 3\right )} \log \left (x\right ) + 3 \, x - e^{\left (-x - \log \left (x\right ) - 1\right )} - 9}{x - 3}\right )} \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=e^{\frac {\left (x - 3\right ) \log {\left (x e^{3} \right )} - \frac {e^{- x - 1}}{x}}{x - 3}} \]
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\[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=\int { \frac {{\left (x^{2} + {\left (x^{2} - x - 3\right )} e^{\left (-x - \log \left (x\right ) - 1\right )} - 6 \, x + 9\right )} e^{\left (\frac {{\left (x - 3\right )} \log \left (x e^{3}\right ) - e^{\left (-x - \log \left (x\right ) - 1\right )}}{x - 3}\right )}}{x^{3} - 6 \, x^{2} + 9 \, x} \,d x } \]
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Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=e^{\left (\frac {x \log \left (x e^{3}\right )}{x - 3} - \frac {e^{\left (-x - \log \left (x\right ) - 1\right )}}{x - 3} - \frac {3 \, \log \left (x e^{3}\right )}{x - 3}\right )} \]
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Time = 12.92 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {-\frac {e^{-1-x}}{x}+(-3+x) \log \left (e^3 x\right )}{-3+x}} \left (9-6 x+x^2+\frac {e^{-1-x} \left (-3-x+x^2\right )}{x}\right )}{9 x-6 x^2+x^3} \, dx=x\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-1}}{3\,x-x^2}}\,{\mathrm {e}}^{\frac {3\,x}{x-3}}\,{\mathrm {e}}^{-\frac {9}{x-3}} \]
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