Integrand size = 70, antiderivative size = 24 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{8-\left (-\frac {e^x}{x}+x\right )^2-\log ^2(2)} \]
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Time = 0.43 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6838} \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=\exp \left (-\frac {x^4-2 e^x x^2-8 x^2+x^2 \log ^2(2)+e^{2 x}}{x^2}\right ) \]
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Rule 6838
Rubi steps \begin{align*} \text {integral}& = \exp \left (-\frac {e^{2 x}-8 x^2-2 e^x x^2+x^4+x^2 \log ^2(2)}{x^2}\right ) \\ \end{align*}
Time = 0.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{8+2 e^x-\frac {e^{2 x}}{x^2}-x^2-\log ^2(2)} \]
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Time = 0.85 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46
method | result | size |
parallelrisch | \({\mathrm e}^{-\frac {x^{2} \ln \left (2\right )^{2}+x^{4}-2 \,{\mathrm e}^{x} x^{2}+{\mathrm e}^{2 x}-8 x^{2}}{x^{2}}}\) | \(35\) |
norman | \({\mathrm e}^{\frac {-{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x^{2}-x^{2} \ln \left (2\right )^{2}-x^{4}+8 x^{2}}{x^{2}}}\) | \(39\) |
risch | \({\mathrm e}^{\frac {-{\mathrm e}^{2 x}+2 \,{\mathrm e}^{x} x^{2}-x^{2} \ln \left (2\right )^{2}-x^{4}+8 x^{2}}{x^{2}}}\) | \(39\) |
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none
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\left (-\frac {x^{4} + x^{2} \log \left (2\right )^{2} - 2 \, x^{2} e^{x} - 8 \, x^{2} + e^{\left (2 \, x\right )}}{x^{2}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\frac {- x^{4} + 2 x^{2} e^{x} - x^{2} \log {\left (2 \right )}^{2} + 8 x^{2} - e^{2 x}}{x^{2}}} \]
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none
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\left (-x^{2} - \log \left (2\right )^{2} - \frac {e^{\left (2 \, x\right )}}{x^{2}} + 2 \, e^{x} + 8\right )} \]
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none
Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx=e^{\left (-x^{2} - \log \left (2\right )^{2} - \frac {e^{\left (2 \, x\right )}}{x^{2}} + 2 \, e^{x} + 8\right )} \]
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Time = 10.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {-e^{2 x}+8 x^2+2 e^x x^2-x^4-x^2 \log ^2(2)}{x^2}} \left (e^{2 x} (2-2 x)+2 e^x x^3-2 x^4\right )}{x^3} \, dx={\mathrm {e}}^8\,{\mathrm {e}}^{-{\ln \left (2\right )}^2}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}}{x^2}} \]
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