Integrand size = 126, antiderivative size = 32 \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\frac {1}{2 \left (-e^x+e^{-3+\frac {1}{10} e^{(2-x) x} x}+\log (2)\right )} \]
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Time = 0.53 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6820, 12, 6818} \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\frac {e^3}{2 \left (e^{\frac {1}{10} e^{(2-x) x} x}-e^{x+3}+e^3 \log (2)\right )} \]
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Rule 12
Rule 6818
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e^{6+x}+\exp \left (3+\left (2+\frac {1}{10} e^{-((-2+x) x)}\right ) x-x^2\right ) \left (-1-2 x+2 x^2\right )}{20 \left (e^{\frac {1}{10} e^{-((-2+x) x)} x}-e^{3+x}+e^3 \log (2)\right )^2} \, dx \\ & = \frac {1}{20} \int \frac {10 e^{6+x}+\exp \left (3+\left (2+\frac {1}{10} e^{-((-2+x) x)}\right ) x-x^2\right ) \left (-1-2 x+2 x^2\right )}{\left (e^{\frac {1}{10} e^{-((-2+x) x)} x}-e^{3+x}+e^3 \log (2)\right )^2} \, dx \\ & = \frac {e^3}{2 \left (e^{\frac {1}{10} e^{(2-x) x} x}-e^{3+x}+e^3 \log (2)\right )} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\frac {e^3}{2 \left (e^{\frac {1}{10} e^{-((-2+x) x)} x}-e^{3+x}+e^3 \log (2)\right )} \]
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Time = 1.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {1}{2 \ln \left (2\right )-2 \,{\mathrm e}^{x}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-\left (-2+x \right ) x}}{10}-3}}\) | \(25\) |
parallelrisch | \(\frac {1}{2 \ln \left (2\right )-2 \,{\mathrm e}^{x}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-x^{2}+2 x}}{10}-3}}\) | \(28\) |
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\frac {1}{2 \, {\left (e^{\left (\frac {1}{10} \, x e^{\left (-x^{2} + 2 \, x\right )} - 3\right )} - e^{x} + \log \left (2\right )\right )}} \]
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Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\frac {1}{- 2 e^{x} + 2 e^{\frac {x e^{- x^{2} + 2 x}}{10} - 3} + 2 \log {\left (2 \right )}} \]
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\[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\int { -\frac {{\left (2 \, x^{2} - 2 \, x - 1\right )} e^{\left (-x^{2} + \frac {1}{10} \, x e^{\left (-x^{2} + 2 \, x\right )} + 2 \, x - 3\right )} + 10 \, e^{x}}{20 \, {\left (2 \, {\left (e^{x} - \log \left (2\right )\right )} e^{\left (\frac {1}{10} \, x e^{\left (-x^{2} + 2 \, x\right )} - 3\right )} + 2 \, e^{x} \log \left (2\right ) - \log \left (2\right )^{2} - e^{\left (\frac {1}{5} \, x e^{\left (-x^{2} + 2 \, x\right )} - 6\right )} - e^{\left (2 \, x\right )}\right )}} \,d x } \]
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Exception generated. \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {10 e^x+e^{2 x-x^2+\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-1-2 x+2 x^2\right )}{20 e^{2 x}+20 e^{\frac {1}{5} \left (-30+e^{2 x-x^2} x\right )}-40 e^x \log (2)+20 \log ^2(2)+e^{\frac {1}{10} \left (-30+e^{2 x-x^2} x\right )} \left (-40 e^x+40 \log (2)\right )} \, dx=\int \frac {10\,{\mathrm {e}}^x-{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x-x^2}}{10}-3}\,{\mathrm {e}}^{2\,x-x^2}\,\left (-2\,x^2+2\,x+1\right )}{20\,{\mathrm {e}}^{2\,x}+20\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x-x^2}}{5}-6}+{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{2\,x-x^2}}{10}-3}\,\left (40\,\ln \left (2\right )-40\,{\mathrm {e}}^x\right )-40\,{\mathrm {e}}^x\,\ln \left (2\right )+20\,{\ln \left (2\right )}^2} \,d x \]
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