Integrand size = 46, antiderivative size = 24 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=10 e^{4-e^4+e^{x/2}-x^2} x \]
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Leaf count is larger than twice the leaf count of optimal. \(56\) vs. \(2(24)=48\).
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2326} \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=\frac {10 e^{-x^2+e^{x/2}-e^4+4} \left (e^{x/2} x^2-4 x^3\right )}{\left (e^{x/2}-4 x\right ) x} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {10 e^{4-e^4+e^{x/2}-x^2} \left (e^{x/2} x^2-4 x^3\right )}{\left (e^{x/2}-4 x\right ) x} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=10 e^{4-e^4+e^{x/2}-x^2} x \]
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Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
risch | \(10 x \,{\mathrm e}^{4-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{2}}-x^{2}}\) | \(20\) |
parallelrisch | \(10 \,{\mathrm e}^{4-{\mathrm e}^{4}} x \,{\mathrm e}^{{\mathrm e}^{\frac {x}{2}}-x^{2}}\) | \(28\) |
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=10 \, x^{2} e^{\left (-x^{2} - e^{4} + e^{\left (\frac {1}{2} \, x\right )} - \log \left (x\right ) + 4\right )} \]
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Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=\frac {10 x e^{4} e^{- x^{2} + e^{\frac {x}{2}}}}{e^{e^{4}}} \]
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Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=10 \, x e^{\left (-x^{2} - e^{4} + e^{\left (\frac {1}{2} \, x\right )} + 4\right )} \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=10 \, x e^{\left (-x^{2} - e^{4} + e^{\left (\frac {1}{2} \, x\right )} + 4\right )} \]
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Time = 13.72 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{4-e^4+e^{x/2}-x^2} \left (10 x+5 e^{x/2} x^2-20 x^3\right )}{x} \, dx=10\,x\,{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^4\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{{\mathrm {e}}^{x/2}} \]
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