Integrand size = 43, antiderivative size = 24 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{e \left (\frac {e^4}{4}-x\right )^2}}{5 x} \]
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Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2326} \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{\frac {1}{16} \left (16 e x^2-8 e^5 x+e^9\right )-1} \left (e^5 x-4 e x^2\right )}{5 \left (e^4-4 x\right ) x^2} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{x^2} \, dx \\ & = \frac {2 e^{-1+\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (e^5 x-4 e x^2\right )}{5 \left (e^4-4 x\right ) x^2} \\ \end{align*}
Time = 1.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{\frac {1}{16} e \left (e^4-4 x\right )^2}}{5 x} \]
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Time = 0.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e}^{9}}{16}-\frac {x \,{\mathrm e}^{5}}{2}+x^{2} {\mathrm e}}}{5 x}\) | \(23\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e} \left ({\mathrm e}^{8}-8 x \,{\mathrm e}^{4}+16 x^{2}\right )}{16}}}{5 x}\) | \(28\) |
gosper | \(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e} \,{\mathrm e}^{8}}{16}-\frac {x \,{\mathrm e} \,{\mathrm e}^{4}}{2}+x^{2} {\mathrm e}}}{5 x}\) | \(31\) |
norman | \(\frac {2 \,{\mathrm e}^{\frac {{\mathrm e} \,{\mathrm e}^{8}}{16}-\frac {x \,{\mathrm e} \,{\mathrm e}^{4}}{2}+x^{2} {\mathrm e}}}{5 x}\) | \(31\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (x^{2} e - \frac {1}{2} \, x e^{5} + \frac {1}{16} \, e^{9}\right )}}{5 \, x} \]
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Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 e^{e x^{2} - \frac {x e^{5}}{2} + \frac {e^{9}}{16}}}{5 x} \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (x^{2} e - \frac {1}{2} \, x e^{5} + \frac {1}{16} \, e^{9}\right )}}{5 \, x} \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2 \, e^{\left (x^{2} e - \frac {1}{2} \, x e^{5} + \frac {1}{16} \, e^{9}\right )}}{5 \, x} \]
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Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{16} \left (e^9-8 e^5 x+16 e x^2\right )} \left (-2-e^5 x+4 e x^2\right )}{5 x^2} \, dx=\frac {2\,{\mathrm {e}}^{x^2\,\mathrm {e}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^9}{16}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{2}}}{5\,x} \]
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