Integrand size = 132, antiderivative size = 29 \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (x-x^2\right )} \]
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\[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=\int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} \left (2 (2+x)^2 (-2+3 x)+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{2 (1-x)^2 x^3 \left (2+x+\log \left (x^2\right )\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} \left (2 (2+x)^2 (-2+3 x)+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{(1-x)^2 x^3 \left (2+x+\log \left (x^2\right )\right )^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (-2+3 x)}{(-1+x)^2 x^3}+\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (2+x)}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )^2}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (2+x)}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )^2} \, dx-\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )} \, dx+\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (-2+3 x)}{(-1+x)^2 x^3} \, dx \\ & = \frac {1}{2} \int \left (\frac {3 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )^2}-\frac {2 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )^2}-\frac {3 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )^2}\right ) \, dx-\frac {1}{2} \int \left (\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )}\right ) \, dx+\int \left (\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x)^2}-\frac {2 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^3}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )} \, dx\right )+\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )} \, dx+\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )} \, dx+\frac {3}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )^2} \, dx-2 \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^3} \, dx+\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x)^2} \, dx-\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2} \, dx-\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )^2} \, dx \\ \end{align*}
Time = 2.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) x^2} \]
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Time = 1.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {{\mathrm e}^{\frac {x}{2 \ln \left (x^{2}\right )+2 x +4}}}{x^{2} \left (-1+x \right )}\) | \(24\) |
parallelrisch | \(-\frac {{\mathrm e}^{\frac {x}{2 \ln \left (x^{2}\right )+2 x +4}}}{x^{2} \left (-1+x \right )}\) | \(24\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=-\frac {e^{\left (\frac {x}{2 \, {\left (x + \log \left (x^{2}\right ) + 2\right )}}\right )}}{x^{3} - x^{2}} \]
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Exception generated. \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=\text {Exception raised: TypeError} \]
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Exception generated. \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=-\frac {e^{\left (\frac {x}{2 \, {\left (x + \log \left (x^{2}\right ) + 2\right )}}\right )}}{x^{3} - x^{2}} \]
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Time = 10.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{\frac {x}{2\,x+\ln \left (x^4\right )+4}}}{x^2-x^3} \]
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