Integrand size = 34, antiderivative size = 23 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=\frac {2}{5} e^{-3-x} \left (-e^{-3+e}+\log ^2(x)\right ) \]
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Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 2326} \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=-\frac {2 e^{-x-3} \left (e^{e-3} x-x \log ^2(x)\right )}{5 x} \]
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Rule 12
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{x} \, dx \\ & = -\frac {2 e^{-3-x} \left (e^{-3+e} x-x \log ^2(x)\right )}{5 x} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=\frac {1}{5} \left (-2 e^{-6+e-x}+2 e^{-3-x} \log ^2(x)\right ) \]
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Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {2 \ln \left (x \right )^{2} {\mathrm e}^{-3-x}}{5}-\frac {2 \,{\mathrm e}^{{\mathrm e}-6-x}}{5}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{-3} \left (2 \ln \left (x \right )^{2}-2 \,{\mathrm e}^{{\mathrm e}-3}\right ) {\mathrm e}^{-x}}{5}\) | \(25\) |
norman | \(\left (\frac {2 \,{\mathrm e}^{-3} \ln \left (x \right )^{2}}{5}-\frac {2 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{{\mathrm e}}}{5}\right ) {\mathrm e}^{-x}\) | \(28\) |
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=\frac {2}{5} \, e^{\left (-x - 3\right )} \log \left (x\right )^{2} - \frac {2}{5} \, e^{\left (-x + e - 6\right )} \]
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Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=\frac {\left (2 e^{3} \log {\left (x \right )}^{2} - 2 e^{e}\right ) e^{- x}}{5 e^{6}} \]
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Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=\frac {2}{5} \, e^{\left (-x - 3\right )} \log \left (x\right )^{2} - \frac {2}{5} \, e^{\left (-x + e - 6\right )} \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=\frac {2}{5} \, {\left (e^{\left (-x\right )} \log \left (x\right )^{2} - e^{\left (-x + e - 3\right )}\right )} e^{\left (-3\right )} \]
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Time = 12.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-3-x} \left (2 e^{-3+e} x+4 \log (x)-2 x \log ^2(x)\right )}{5 x} \, dx=-{\mathrm {e}}^{-x-6}\,\left (\frac {2\,{\mathrm {e}}^{\mathrm {e}}}{5}-\frac {2\,{\mathrm {e}}^3\,{\ln \left (x\right )}^2}{5}\right ) \]
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