Integrand size = 32, antiderivative size = 20 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=e^{\frac {4 \log ^2(4)}{5}} x^2 \log \left (\frac {x}{e^4}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {2341} \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=x^2 e^{\frac {4 \log ^2(4)}{5}} \log \left (\frac {x}{e^4}\right ) \]
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Rule 2341
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} e^{\frac {4 \log ^2(4)}{5}} x^2+\left (2 e^{\frac {4 \log ^2(4)}{5}}\right ) \int x \log \left (\frac {x}{e^4}\right ) \, dx \\ & = e^{\frac {4 \log ^2(4)}{5}} x^2 \log \left (\frac {x}{e^4}\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=e^{\frac {4 \log ^2(4)}{5}} \left (-4 x^2+x^2 \log (x)\right ) \]
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Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85
method | result | size |
risch | \({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(17\) |
default | \({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
norman | \({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
parallelrisch | \({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
parts | \({\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2} \ln \left ({\mathrm e}^{-4} x \right )\) | \(19\) |
derivativedivides | \({\mathrm e}^{4} \left (\frac {{\mathrm e}^{-4} {\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} x^{2}}{2}+2 \,{\mathrm e}^{4} {\mathrm e}^{\frac {16 \ln \left (2\right )^{2}}{5}} \left (\frac {x^{2} {\mathrm e}^{-8} \ln \left ({\mathrm e}^{-4} x \right )}{2}-\frac {x^{2} {\mathrm e}^{-8}}{4}\right )\right )\) | \(58\) |
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Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=x^{2} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} \log \left (x e^{\left (-4\right )}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=x^{2} e^{\frac {16 \log {\left (2 \right )}^{2}}{5}} \log {\left (\frac {x}{e^{4}} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).
Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=\frac {1}{2} \, x^{2} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (x e^{\left (-4\right )}\right ) - x^{2}\right )} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=\frac {1}{2} \, x^{2} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} \log \left (x e^{\left (-4\right )}\right ) - x^{2}\right )} e^{\left (\frac {16}{5} \, \log \left (2\right )^{2}\right )} \]
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Time = 9.41 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \left (e^{\frac {4 \log ^2(4)}{5}} x+2 e^{\frac {4 \log ^2(4)}{5}} x \log \left (\frac {x}{e^4}\right )\right ) \, dx=x^2\,{\mathrm {e}}^{\frac {16\,{\ln \left (2\right )}^2}{5}}\,\left (\ln \left (x\right )-4\right ) \]
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