Integrand size = 125, antiderivative size = 33 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{x+\frac {\log \left (e^{4+x}+\frac {1}{4} \left (x-4 x \left (\frac {3}{2 x}+x\right )\right )\right )}{x}} \]
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Time = 4.36 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {6838} \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=4^{-1/x} e^x \left (-4 x^2+x+4 e^{x+4}-6\right )^{\frac {1}{x}} \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = 4^{-1/x} e^x \left (-6+4 e^{4+x}+x-4 x^2\right )^{\frac {1}{x}} \\ \end{align*}
\[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=\int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx \]
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Time = 5.94 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\left ({\mathrm e}^{4+x}-x^{2}+\frac {x}{4}-\frac {3}{2}\right )^{\frac {1}{x}} {\mathrm e}^{x}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {\ln \left ({\mathrm e}^{4+x}-x^{2}+\frac {x}{4}-\frac {3}{2}\right )+x^{2}}{x}}\) | \(25\) |
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Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (\frac {x^{2} + \log \left (-x^{2} + \frac {1}{4} \, x + e^{\left (x + 4\right )} - \frac {3}{2}\right )}{x}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=\text {Timed out} \]
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Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (x - \frac {2 \, \log \left (2\right )}{x} + \frac {\log \left (-4 \, x^{2} + x + 4 \, e^{\left (x + 4\right )} - 6\right )}{x}\right )} \]
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Time = 0.57 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx=e^{\left (x + \frac {\log \left (-x^{2} + \frac {1}{4} \, x + e^{\left (x + 4\right )} - \frac {3}{2}\right )}{x}\right )} \]
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Time = 15.66 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\frac {x^2+\log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )}{x}} \left (x-14 x^2+x^3-4 x^4+e^{4+x} \left (4 x+4 x^2\right )+\left (6-4 e^{4+x}-x+4 x^2\right ) \log \left (\frac {1}{4} \left (-6+4 e^{4+x}+x-4 x^2\right )\right )\right )}{-6 x^2+4 e^{4+x} x^2+x^3-4 x^4} \, dx={\mathrm {e}}^x\,{\left (\frac {x}{4}+{\mathrm {e}}^{x+4}-x^2-\frac {3}{2}\right )}^{1/x} \]
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