Integrand size = 38, antiderivative size = 20 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {1}{128} e^{2 e^2-x} x^2 \log (x) \]
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Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {12, 6820, 6874, 2207, 2225, 2227, 2634} \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {1}{128} e^{2 e^2-x} x^2 \log (x) \]
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Rule 12
Rule 2207
Rule 2225
Rule 2227
Rule 2634
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{128} \int e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx \\ & = \frac {1}{128} \int e^{2 e^2-x} x (1-(-2+x) \log (x)) \, dx \\ & = \frac {1}{128} \int \left (e^{2 e^2-x} x-e^{2 e^2-x} (-2+x) x \log (x)\right ) \, dx \\ & = \frac {1}{128} \int e^{2 e^2-x} x \, dx-\frac {1}{128} \int e^{2 e^2-x} (-2+x) x \log (x) \, dx \\ & = -\frac {1}{128} e^{2 e^2-x} x+\frac {1}{128} e^{2 e^2-x} x^2 \log (x)+\frac {1}{128} \int e^{2 e^2-x} \, dx-\frac {1}{128} \int e^{2 e^2-x} x \, dx \\ & = -\frac {1}{128} e^{2 e^2-x}+\frac {1}{128} e^{2 e^2-x} x^2 \log (x)-\frac {1}{128} \int e^{2 e^2-x} \, dx \\ & = \frac {1}{128} e^{2 e^2-x} x^2 \log (x) \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {1}{128} e^{2 e^2-x} x^2 \log (x) \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85
method | result | size |
norman | \(\frac {{\mathrm e}^{2 \,{\mathrm e}^{2}} {\mathrm e}^{-x} x^{2} \ln \left (x \right )}{128}\) | \(17\) |
risch | \(\frac {x^{2} \ln \left (x \right ) {\mathrm e}^{2 \,{\mathrm e}^{2}-x}}{128}\) | \(17\) |
parallelrisch | \(\frac {{\mathrm e}^{2 \,{\mathrm e}^{2}} {\mathrm e}^{-x} x^{2} \ln \left (x \right )}{128}\) | \(17\) |
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {1}{128} \, x^{2} e^{\left (-x + 2 \, e^{2}\right )} \log \left (x\right ) \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {x^{2} e^{- x} e^{2 e^{2}} \log {\left (x \right )}}{128} \]
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Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (16) = 32\).
Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.50 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {1}{128} \, {\left (x^{2} e^{\left (2 \, e^{2}\right )} \log \left (x\right ) + x e^{\left (2 \, e^{2}\right )} + e^{\left (2 \, e^{2}\right )}\right )} e^{\left (-x\right )} - \frac {1}{128} \, {\left (x e^{\left (2 \, e^{2}\right )} + e^{\left (2 \, e^{2}\right )}\right )} e^{\left (-x\right )} \]
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Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {1}{128} \, x^{2} e^{\left (-x + 2 \, e^{2}\right )} \log \left (x\right ) \]
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Time = 9.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{128} e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx=\frac {x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}\,\ln \left (x\right )}{128} \]
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