Integrand size = 41, antiderivative size = 29 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{2-x+\left (4+\frac {e^{-2+x}}{5}\right ) x+5 \left (2+4 x^2\right )} \]
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Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6838} \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\frac {1}{5} \left (100 x^2+e^{x-2} x+15 x+60\right )} \]
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Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx \\ & = e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \\ \end{align*}
Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{12+3 x+\frac {1}{5} e^{-2+x} x+20 x^2} \]
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Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66
method | result | size |
norman | \({\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{5}+20 x^{2}+3 x +12}\) | \(19\) |
risch | \({\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{5}+20 x^{2}+3 x +12}\) | \(19\) |
parallelrisch | \({\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{5}+20 x^{2}+3 x +12}\) | \(19\) |
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Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\left (20 \, x^{2} + \frac {1}{5} \, x e^{\left (x - 2\right )} + 3 \, x + 12\right )} \]
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Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{20 x^{2} + \frac {x e^{x - 2}}{5} + 3 x + 12} \]
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none
Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\left (20 \, x^{2} + \frac {1}{5} \, x e^{\left (x - 2\right )} + 3 \, x + 12\right )} \]
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Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\left (20 \, x^{2} + \frac {1}{5} \, x e^{\left (x - 2\right )} + 3 \, x + 12\right )} \]
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Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx={\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^{20\,x^2} \]
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