Integrand size = 57, antiderivative size = 22 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=\log \left (9+\frac {9}{x}+x+\left (2+4 e^{x^2}\right ) x^2\right ) \]
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\[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=\int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 \left (1+x^2\right )}{x}-\frac {27+18 x+19 x^2+18 x^3+2 x^4+4 x^5}{x \left (9+9 x+x^2+2 x^3+4 e^{x^2} x^3\right )}\right ) \, dx \\ & = 2 \int \frac {1+x^2}{x} \, dx-\int \frac {27+18 x+19 x^2+18 x^3+2 x^4+4 x^5}{x \left (9+9 x+x^2+2 x^3+4 e^{x^2} x^3\right )} \, dx \\ & = 2 \int \left (\frac {1}{x}+x\right ) \, dx-\int \left (\frac {18}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3}+\frac {27}{x \left (9+9 x+x^2+2 x^3+4 e^{x^2} x^3\right )}+\frac {19 x}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3}+\frac {18 x^2}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3}+\frac {2 x^3}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3}+\frac {4 x^4}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3}\right ) \, dx \\ & = x^2+2 \log (x)-2 \int \frac {x^3}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3} \, dx-4 \int \frac {x^4}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3} \, dx-18 \int \frac {1}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3} \, dx-18 \int \frac {x^2}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3} \, dx-19 \int \frac {x}{9+9 x+x^2+2 x^3+4 e^{x^2} x^3} \, dx-27 \int \frac {1}{x \left (9+9 x+x^2+2 x^3+4 e^{x^2} x^3\right )} \, dx \\ \end{align*}
Time = 5.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=-\log (x)+\log \left (9+9 x+x^2+2 x^3+4 e^{x^2} x^3\right ) \]
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Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
norman | \(-\ln \left (x \right )+\ln \left (4 x^{3} {\mathrm e}^{x^{2}}+2 x^{3}+x^{2}+9 x +9\right )\) | \(29\) |
risch | \(2 \ln \left (x \right )+\ln \left ({\mathrm e}^{x^{2}}+\frac {2 x^{3}+x^{2}+9 x +9}{4 x^{3}}\right )\) | \(30\) |
parallelrisch | \(-\ln \left (x \right )+\ln \left (x^{3} {\mathrm e}^{x^{2}}+\frac {x^{3}}{2}+\frac {x^{2}}{4}+\frac {9 x}{4}+\frac {9}{4}\right )\) | \(30\) |
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Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=2 \, \log \left (x\right ) + \log \left (\frac {4 \, x^{3} e^{\left (x^{2}\right )} + 2 \, x^{3} + x^{2} + 9 \, x + 9}{x^{3}}\right ) \]
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Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=2 \log {\left (x \right )} + \log {\left (e^{x^{2}} + \frac {2 x^{3} + x^{2} + 9 x + 9}{4 x^{3}} \right )} \]
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Time = 0.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=2 \, \log \left (x\right ) + \log \left (\frac {4 \, x^{3} e^{\left (x^{2}\right )} + 2 \, x^{3} + x^{2} + 9 \, x + 9}{4 \, x^{3}}\right ) \]
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Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=\log \left (4 \, x^{3} e^{\left (x^{2}\right )} + 2 \, x^{3} + x^{2} + 9 \, x + 9\right ) - \log \left (x\right ) \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-9+x^2+4 x^3+e^{x^2} \left (8 x^3+8 x^5\right )}{9 x+9 x^2+x^3+2 x^4+4 e^{x^2} x^4} \, dx=\ln \left (9\,x+4\,x^3\,{\mathrm {e}}^{x^2}+x^2+2\,x^3+9\right )-\ln \left (x\right ) \]
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