Integrand size = 46, antiderivative size = 41 \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=\frac {2}{3 \left (1-\frac {x}{-\frac {2 e^{\frac {1}{5} \left (-e^{x/4}+\frac {x}{2}\right )}}{x}+x}\right )} \]
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Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.24, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 2306, 2326} \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=-\frac {e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (2 x-e^{x/4} x\right )}{3 \left (2-e^{x/4}\right )} \]
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Rule 12
Rule 2306
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{30} \int e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx \\ & = \frac {1}{30} \int \frac {1}{2} e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (-40+2 x-e^{x/4} x\right ) \, dx \\ & = \frac {1}{60} \int e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (-40+2 x-e^{x/4} x\right ) \, dx \\ & = -\frac {e^{\frac {1}{10} \left (2 e^{x/4}-x\right )} x \left (2 x-e^{x/4} x\right )}{3 \left (2-e^{x/4}\right )} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=-\frac {1}{3} e^{\frac {e^{x/4}}{5}-\frac {x}{10}} x^2 \]
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Time = 0.33 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.41
method | result | size |
risch | \(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\) | \(17\) |
default | \(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\) | \(23\) |
norman | \(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\) | \(23\) |
parallelrisch | \(-\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {x}{4}}}{5}-\frac {x}{10}}}{3}\) | \(23\) |
meijerg | \(-\frac {4 \left (1-{\mathrm e}^{-x \ln \left (2\right )}\right )}{3 \ln \left (2\right )}-\frac {8 \left (1-\frac {\left (2-\frac {x \left (-4 \ln \left (2\right )+1\right )}{2}\right ) {\mathrm e}^{\frac {x \left (-4 \ln \left (2\right )+1\right )}{4}}}{2}\right )}{15 \left (-4 \ln \left (2\right )+1\right )^{2}}+\frac {1-\frac {\left (2+2 x \ln \left (2\right )\right ) {\mathrm e}^{-x \ln \left (2\right )}}{2}}{15 \ln \left (2\right )^{2}}\) | \(76\) |
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.54 \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=-\frac {2}{3} \, x e^{\left (-\frac {1}{10} \, x + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )} - \log \left (\frac {2}{x}\right )\right )} \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.41 \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=- \frac {x^{2} e^{- \frac {x}{10} + \frac {e^{\frac {x}{4}}}{5}}}{3} \]
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\[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=\int { -\frac {1}{30} \, {\left (x e^{\left (\frac {1}{4} \, x\right )} - 2 \, x + 40\right )} e^{\left (-\frac {1}{10} \, x + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )} - \log \left (\frac {2}{x}\right )\right )} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.39 \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=-\frac {1}{3} \, x^{2} e^{\left (-\frac {1}{10} \, x + \frac {1}{5} \, e^{\left (\frac {1}{4} \, x\right )}\right )} \]
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Time = 13.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.39 \[ \int \frac {1}{30} e^{\frac {1}{10} \left (2 e^{x/4}-x-10 \log \left (\frac {2}{x}\right )\right )} \left (-40+2 x-e^{x/4} x\right ) \, dx=-\frac {x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{x/4}}{5}-\frac {x}{10}}}{3} \]
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