Integrand size = 50, antiderivative size = 33 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-e^{3+x}+x-x \left (2 x-\frac {e^{-x^2} x \log (\log (4))}{\log (5)}\right ) \]
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Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 6874, 2225, 2258, 2240, 2243} \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)}+x-e^{x+3} \]
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Rule 12
Rule 2225
Rule 2240
Rule 2243
Rule 2258
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right ) \, dx}{\log (5)} \\ & = \frac {\int \left (-\left (\left (-1+e^{3+x}+4 x\right ) \log (5)\right )-2 e^{-x^2} x \left (-1+x^2\right ) \log (\log (4))\right ) \, dx}{\log (5)} \\ & = -\frac {(2 \log (\log (4))) \int e^{-x^2} x \left (-1+x^2\right ) \, dx}{\log (5)}-\int \left (-1+e^{3+x}+4 x\right ) \, dx \\ & = x-2 x^2-\frac {(2 \log (\log (4))) \int \left (-e^{-x^2} x+e^{-x^2} x^3\right ) \, dx}{\log (5)}-\int e^{3+x} \, dx \\ & = -e^{3+x}+x-2 x^2+\frac {(2 \log (\log (4))) \int e^{-x^2} x \, dx}{\log (5)}-\frac {(2 \log (\log (4))) \int e^{-x^2} x^3 \, dx}{\log (5)} \\ & = -e^{3+x}+x-2 x^2-\frac {e^{-x^2} \log (\log (4))}{\log (5)}+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)}-\frac {(2 \log (\log (4))) \int e^{-x^2} x \, dx}{\log (5)} \\ & = -e^{3+x}+x-2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-e^{3+x}+x-2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)} \]
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Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00
method | result | size |
parts | \(-2 x^{2}+x +\frac {\ln \left (2 \ln \left (2\right )\right ) {\mathrm e}^{-x^{2}} x^{2}}{\ln \left (5\right )}-{\mathrm e}^{3+x}\) | \(33\) |
risch | \(-2 x^{2}+x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x^{2} {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}-{\mathrm e}^{3+x}\) | \(34\) |
norman | \(\left ({\mathrm e}^{x^{2}} x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x^{2}}{\ln \left (5\right )}-2 x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}} {\mathrm e}^{3+x}\right ) {\mathrm e}^{-x^{2}}\) | \(48\) |
parallelrisch | \(\frac {\left (-2 \ln \left (5\right ) x^{2} {\mathrm e}^{x^{2}}+x^{2} \ln \left (2 \ln \left (2\right )\right )+{\mathrm e}^{x^{2}} \ln \left (5\right ) x -{\mathrm e}^{x^{2}} {\mathrm e}^{3+x} \ln \left (5\right )\right ) {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}\) | \(53\) |
default | \(\frac {-2 x^{2} \ln \left (5\right )-{\mathrm e}^{x} \ln \left (5\right ) {\mathrm e}^{3}-\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{-x^{2}}+x^{2} {\mathrm e}^{-x^{2}} \ln \left (2\right )-2 \ln \left (\ln \left (2\right )\right ) \left (-\frac {x^{2} {\mathrm e}^{-x^{2}}}{2}-\frac {{\mathrm e}^{-x^{2}}}{2}\right )+x \ln \left (5\right )}{\ln \left (5\right )}\) | \(74\) |
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Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=\frac {{\left (x^{2} \log \left (2 \, \log \left (2\right )\right ) - {\left ({\left (2 \, x^{2} - x\right )} \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right )\right )} e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{\log \left (5\right )} \]
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Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=- 2 x^{2} + x + \frac {\left (x^{2} \log {\left (\log {\left (2 \right )} \right )} + x^{2} \log {\left (2 \right )}\right ) e^{- x^{2}}}{\log {\left (5 \right )}} - e^{x + 3} \]
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Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - {\left (x^{2} + 1\right )} e^{\left (-x^{2}\right )} \log \left (2 \, \log \left (2\right )\right ) - x \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right ) + e^{\left (-x^{2}\right )} \log \left (2 \, \log \left (2\right )\right )}{\log \left (5\right )} \]
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Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - {\left (x^{2} \log \left (2\right ) + x^{2} \log \left (\log \left (2\right )\right )\right )} e^{\left (-x^{2}\right )} - x \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right )}{\log \left (5\right )} \]
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Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=x-{\mathrm {e}}^{x+3}-2\,x^2+\frac {x^2\,{\mathrm {e}}^{-x^2}\,\ln \left (2\right )}{\ln \left (5\right )}+\frac {x^2\,{\mathrm {e}}^{-x^2}\,\ln \left (\ln \left (2\right )\right )}{\ln \left (5\right )} \]
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