Integrand size = 76, antiderivative size = 29 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 e^{-\frac {\left (\frac {e^3}{5}+\frac {10}{x}-x^2\right )^2}{x}} \]
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Time = 0.83 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {12, 6820, 6838} \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 e^{-\frac {\left (-5 x^3+e^3 x+50\right )^2}{25 x^3}} \]
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Rule 12
Rule 6820
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \frac {\exp \left (-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}\right ) \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{x^4} \, dx \\ & = \frac {1}{25} \int \frac {2 e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \left (e^6 x^2+10 e^3 x \left (20+x^3\right )-75 \left (-100+x^6\right )\right )}{x^4} \, dx \\ & = \frac {2}{25} \int \frac {e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \left (e^6 x^2+10 e^3 x \left (20+x^3\right )-75 \left (-100+x^6\right )\right )}{x^4} \, dx \\ & = 2 e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 e^{-\frac {\left (50+e^3 x-5 x^3\right )^2}{25 x^3}} \]
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Time = 0.32 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38
method | result | size |
risch | \(2 \,{\mathrm e}^{\frac {-25 x^{6}+10 x^{4} {\mathrm e}^{3}-x^{2} {\mathrm e}^{6}+500 x^{3}-100 x \,{\mathrm e}^{3}-2500}{25 x^{3}}}\) | \(40\) |
gosper | \(2 \,{\mathrm e}^{-\frac {25 x^{6}-10 x^{4} {\mathrm e}^{3}+x^{2} {\mathrm e}^{6}-500 x^{3}+100 x \,{\mathrm e}^{3}+2500}{25 x^{3}}}\) | \(43\) |
norman | \(2 \,{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{6}+\left (-10 x^{4}+100 x \right ) {\mathrm e}^{3}+25 x^{6}-500 x^{3}+2500}{25 x^{3}}}\) | \(43\) |
parallelrisch | \(2 \,{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{6}+\left (-10 x^{4}+100 x \right ) {\mathrm e}^{3}+25 x^{6}-500 x^{3}+2500}{25 x^{3}}}\) | \(43\) |
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Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 \, e^{\left (-\frac {25 \, x^{6} - 500 \, x^{3} + x^{2} e^{6} - 10 \, {\left (x^{4} - 10 \, x\right )} e^{3} + 2500}{25 \, x^{3}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).
Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 e^{- \frac {x^{6} - 20 x^{3} + \frac {x^{2} e^{6}}{25} + \frac {\left (- 10 x^{4} + 100 x\right ) e^{3}}{25} + 100}{x^{3}}} \]
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Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 \, e^{\left (-x^{3} + \frac {2}{5} \, x e^{3} - \frac {e^{6}}{25 \, x} - \frac {4 \, e^{3}}{x^{2}} - \frac {100}{x^{3}} + 20\right )} \]
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Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2 \, e^{\left (-\frac {25 \, x^{6} - 10 \, x^{4} e^{3} - 500 \, x^{3} + x^{2} e^{6} + 100 \, x e^{3} + 2500}{25 \, x^{3}}\right )} \]
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Time = 8.54 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-\frac {2500+e^6 x^2-500 x^3+25 x^6+e^3 \left (100 x-10 x^4\right )}{25 x^3}} \left (15000+2 e^6 x^2-150 x^6+e^3 \left (400 x+20 x^4\right )\right )}{25 x^4} \, dx=2\,{\mathrm {e}}^{-\frac {4\,{\mathrm {e}}^3}{x^2}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^6}{25\,x}}\,{\mathrm {e}}^{20}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{-\frac {100}{x^3}}\,{\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^3}{5}} \]
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