Integrand size = 149, antiderivative size = 25 \[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\frac {\log \left (\log \left (7-e^{-\frac {1}{x}+\frac {6+x}{9}}\right )\right )}{x} \]
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\[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {e^{\frac {2}{3}+\frac {x}{9}} \left (9+x^2\right )}{9 \left (-7 e^{\frac {1}{x}}+e^{\frac {6+x}{9}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )}-\frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2}\right ) \, dx \\ & = \frac {1}{9} \int \frac {e^{\frac {2}{3}+\frac {x}{9}} \left (9+x^2\right )}{\left (-7 e^{\frac {1}{x}}+e^{\frac {6+x}{9}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )} \, dx-\int \frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2} \, dx \\ & = \frac {1}{9} \int \left (\frac {9 e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )}+\frac {e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )}\right ) \, dx-\int \frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2} \, dx \\ & = \frac {1}{9} \int \frac {e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )} \, dx+\int \frac {e^{\frac {2}{3}+\frac {x}{9}}}{\left (e^{\frac {2}{3}+\frac {x}{9}}-7 e^{\frac {1}{x}}\right ) x^3 \log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )} \, dx-\int \frac {\log \left (\log \left (7-e^{\frac {1}{9} \left (6-\frac {9}{x}+x\right )}\right )\right )}{x^2} \, dx \\ \end{align*}
Time = 1.73 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\frac {\log \left (\log \left (7-e^{\frac {2}{3}-\frac {1}{x}+\frac {x}{9}}\right )\right )}{x} \]
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Time = 1.88 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left (\ln \left (-{\mathrm e}^{\frac {x^{2}+6 x -9}{9 x}}+7\right )\right )}{x}\) | \(25\) |
parallelrisch | \(\frac {\ln \left (\ln \left (-{\mathrm e}^{\frac {x^{2}+6 x -9}{9 x}}+7\right )\right )}{x}\) | \(25\) |
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Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\frac {\log \left (\log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right )\right )}{x} \]
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Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\frac {\log {\left (\log {\left (7 - e^{\frac {\frac {x^{2}}{9} + \frac {2 x}{3} - 1}{x}} \right )} \right )}}{x} \]
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Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\frac {\log \left (x \log \left (-e^{\left (\frac {1}{9} \, x + \frac {2}{3}\right )} + 7 \, e^{\frac {1}{x}}\right ) - 1\right ) - \log \left (x\right )}{x} \]
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\[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\int { -\frac {9 \, {\left (x e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} - 7 \, x\right )} \log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right ) \log \left (\log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right )\right ) - {\left (x^{2} + 9\right )} e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )}}{9 \, {\left (x^{3} e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} - 7 \, x^{3}\right )} \log \left (-e^{\left (\frac {x^{2} + 6 \, x - 9}{9 \, x}\right )} + 7\right )} \,d x } \]
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Time = 14.58 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-9+6 x+x^2}{9 x}} \left (9+x^2\right )+\left (63 x-9 e^{\frac {-9+6 x+x^2}{9 x}} x\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right ) \log \left (\log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )\right )}{\left (-63 x^3+9 e^{\frac {-9+6 x+x^2}{9 x}} x^3\right ) \log \left (7-e^{\frac {-9+6 x+x^2}{9 x}}\right )} \, dx=\frac {\ln \left (\ln \left (7-{\mathrm {e}}^{x/9}\,{\mathrm {e}}^{2/3}\,{\mathrm {e}}^{-\frac {1}{x}}\right )\right )}{x} \]
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