Integrand size = 105, antiderivative size = 25 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{\frac {2 x-\frac {9}{\log \left (3+e^x+4 x^2\right )}}{x}} \]
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\[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=\int \frac {\exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {9 \exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (3-8 x+4 x^2\right )}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}+\frac {9 \exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (x+\log \left (3+e^x+4 x^2\right )\right )}{x^2 \log ^2\left (3+e^x+4 x^2\right )}\right ) \, dx \\ & = -\left (9 \int \frac {\exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (3-8 x+4 x^2\right )}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx\right )+9 \int \frac {\exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (x+\log \left (3+e^x+4 x^2\right )\right )}{x^2 \log ^2\left (3+e^x+4 x^2\right )} \, dx \\ & = -\left (9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \left (3-8 x+4 x^2\right )}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx\right )+9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \left (x+\log \left (3+e^x+4 x^2\right )\right )}{x^2 \log ^2\left (3+e^x+4 x^2\right )} \, dx \\ & = -\left (9 \int \left (-\frac {8 e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}+\frac {3 e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}+\frac {4 e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} x}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}\right ) \, dx\right )+9 \int \left (\frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \log ^2\left (3+e^x+4 x^2\right )}+\frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x^2 \log \left (3+e^x+4 x^2\right )}\right ) \, dx \\ & = 9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \log ^2\left (3+e^x+4 x^2\right )} \, dx+9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x^2 \log \left (3+e^x+4 x^2\right )} \, dx-27 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx-36 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} x}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx+72 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \]
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Time = 34.56 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
risch | \({\mathrm e}^{\frac {2 x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )-9}{x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )}}\) | \(33\) |
parallelrisch | \({\mathrm e}^{\frac {2 x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )-9}{x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )}}\) | \(33\) |
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{\left (\frac {2 \, x \log \left (4 \, x^{2} + e^{x} + 3\right ) - 9}{x \log \left (4 \, x^{2} + e^{x} + 3\right )}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx=e^{\left (-\frac {9}{x \log \left (4 \, x^{2} + e^{x} + 3\right )} + 2\right )} \]
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Time = 12.82 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}} \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx={\mathrm {e}}^2\,{\mathrm {e}}^{-\frac {9}{x\,\ln \left ({\mathrm {e}}^x+4\,x^2+3\right )}} \]
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