Integrand size = 51, antiderivative size = 30 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=e^{(-1-x+\log (3))^2}+x^2-\left (\left (\frac {1}{5}-x\right )^2+x\right )^2 \]
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Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 2259, 2240} \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=-x^4-\frac {6 x^3}{5}+\frac {14 x^2}{25}-\frac {6 x}{125}+e^{(x+1-\log (3))^2} \]
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Rule 12
Rule 2240
Rule 2259
Rubi steps \begin{align*} \text {integral}& = \frac {1}{125} \int \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx \\ & = -\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4+\frac {1}{125} \int e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3)) \, dx \\ & = -\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4+\frac {1}{125} \int e^{(1+x-\log (3))^2} (250+250 x-250 \log (3)) \, dx \\ & = e^{(1+x-\log (3))^2}-\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4 \\ \end{align*}
Time = 0.50 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=3^{-2-2 x} e^{1+2 x+x^2+\log ^2(3)}-\frac {6 x}{125}+\frac {14 x^2}{25}-\frac {6 x^3}{5}-x^4 \]
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Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37
method | result | size |
norman | \(-\frac {6 x}{125}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}+{\mathrm e}^{\ln \left (3\right )^{2}+\left (-2-2 x \right ) \ln \left (3\right )+x^{2}+2 x +1}\) | \(41\) |
risch | \(-\frac {6 x}{125}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}+3^{-2-2 x} {\mathrm e}^{\ln \left (3\right )^{2}+1+x^{2}+2 x}\) | \(41\) |
parallelrisch | \(-\frac {6 x}{125}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}+{\mathrm e}^{\ln \left (3\right )^{2}+\left (-2-2 x \right ) \ln \left (3\right )+x^{2}+2 x +1}\) | \(41\) |
default | \(-\frac {6 x}{125}+{\mathrm e}^{x^{2}+\left (-2 \ln \left (3\right )+2\right ) x +\ln \left (3\right )^{2}-2 \ln \left (3\right )+1}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}\) | \(42\) |
parts | \(-\frac {6 x}{125}+{\mathrm e}^{x^{2}+\left (-2 \ln \left (3\right )+2\right ) x +\ln \left (3\right )^{2}-2 \ln \left (3\right )+1}+\frac {14 x^{2}}{25}-\frac {6 x^{3}}{5}-x^{4}\) | \(42\) |
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=-x^{4} - \frac {6}{5} \, x^{3} + \frac {14}{25} \, x^{2} - \frac {6}{125} \, x + e^{\left (x^{2} - 2 \, {\left (x + 1\right )} \log \left (3\right ) + \log \left (3\right )^{2} + 2 \, x + 1\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=- x^{4} - \frac {6 x^{3}}{5} + \frac {14 x^{2}}{25} - \frac {6 x}{125} + e^{x^{2} + 2 x + \left (- 2 x - 2\right ) \log {\left (3 \right )} + 1 + \log {\left (3 \right )}^{2}} \]
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Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=-x^{4} - \frac {6}{5} \, x^{3} + \frac {14}{25} \, x^{2} - \frac {6}{125} \, x + e^{\left (x^{2} - 2 \, {\left (x + 1\right )} \log \left (3\right ) + \log \left (3\right )^{2} + 2 \, x + 1\right )} \]
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Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=-x^{4} - \frac {6}{5} \, x^{3} + \frac {14}{25} \, x^{2} - \frac {6}{125} \, x + e^{\left (x^{2} - 2 \, x \log \left (3\right ) + \log \left (3\right )^{2} + 2 \, x - 2 \, \log \left (3\right ) + 1\right )} \]
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Time = 11.74 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {1}{125} \left (-6+140 x-450 x^2-500 x^3+e^{1+2 x+x^2+(-2-2 x) \log (3)+\log ^2(3)} (250+250 x-250 \log (3))\right ) \, dx=\frac {{\mathrm {e}}^{x^2+2\,x+{\ln \left (3\right )}^2+1}}{9\,3^{2\,x}}-\frac {6\,x}{125}+\frac {14\,x^2}{25}-\frac {6\,x^3}{5}-x^4 \]
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