Integrand size = 43, antiderivative size = 26 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 e^{-x} \left (4+x+\frac {1}{4} e^5 (\log (3)+\log (4))-\log (x)\right ) \]
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Time = 0.33 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62, number of steps used = 15, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {6, 12, 6873, 6874, 2230, 2209, 2207, 2225, 2634} \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 e^{-x} x+3 e^{-x}-3 e^{-x} \log (x)+\frac {3}{4} e^{-x} \left (12+e^5 \log (12)\right ) \]
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Rule 6
Rule 12
Rule 2207
Rule 2209
Rule 2225
Rule 2230
Rule 2634
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (-12-12 x^2+x \left (-36-3 e^5 \log (3)\right )-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx \\ & = \int \frac {e^{-x} \left (-12-12 x^2+x \left (-36-3 e^5 \log (3)-3 e^5 \log (4)\right )+12 x \log (x)\right )}{4 x} \, dx \\ & = \frac {1}{4} \int \frac {e^{-x} \left (-12-12 x^2+x \left (-36-3 e^5 \log (3)-3 e^5 \log (4)\right )+12 x \log (x)\right )}{x} \, dx \\ & = \frac {1}{4} \int \frac {3 e^{-x} \left (-4-4 x^2-12 x \left (1+\frac {1}{12} e^5 \log (12)\right )+4 x \log (x)\right )}{x} \, dx \\ & = \frac {3}{4} \int \frac {e^{-x} \left (-4-4 x^2-12 x \left (1+\frac {1}{12} e^5 \log (12)\right )+4 x \log (x)\right )}{x} \, dx \\ & = \frac {3}{4} \int \left (\frac {e^{-x} \left (-4-4 x^2-x \left (12+e^5 \log (12)\right )\right )}{x}+4 e^{-x} \log (x)\right ) \, dx \\ & = \frac {3}{4} \int \frac {e^{-x} \left (-4-4 x^2-x \left (12+e^5 \log (12)\right )\right )}{x} \, dx+3 \int e^{-x} \log (x) \, dx \\ & = -3 e^{-x} \log (x)+\frac {3}{4} \int \left (-\frac {4 e^{-x}}{x}-4 e^{-x} x+e^{-x} \left (-12-e^5 \log (12)\right )\right ) \, dx+3 \int \frac {e^{-x}}{x} \, dx \\ & = 3 \operatorname {ExpIntegralEi}(-x)-3 e^{-x} \log (x)-3 \int \frac {e^{-x}}{x} \, dx-3 \int e^{-x} x \, dx-\frac {1}{4} \left (3 \left (12+e^5 \log (12)\right )\right ) \int e^{-x} \, dx \\ & = 3 e^{-x} x+\frac {3}{4} e^{-x} \left (12+e^5 \log (12)\right )-3 e^{-x} \log (x)-3 \int e^{-x} \, dx \\ & = 3 e^{-x}+3 e^{-x} x+\frac {3}{4} e^{-x} \left (12+e^5 \log (12)\right )-3 e^{-x} \log (x) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {3}{4} e^{-x} \left (16+4 x+e^5 \log (12)-4 \log (x)\right ) \]
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Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\left (3 x +12-3 \ln \left (x \right )+\frac {3 \,{\mathrm e}^{5} \ln \left (2\right )}{2}+\frac {3 \,{\mathrm e}^{5} \ln \left (3\right )}{4}\right ) {\mathrm e}^{-x}\) | \(27\) |
parallelrisch | \(\frac {\left (48+6 \,{\mathrm e}^{5} \ln \left (2\right )+3 \,{\mathrm e}^{5} \ln \left (3\right )+12 x -12 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{4}\) | \(28\) |
risch | \(-3 \ln \left (x \right ) {\mathrm e}^{-x}+\frac {3 \left (2 \,{\mathrm e}^{5} \ln \left (2\right )+{\mathrm e}^{5} \ln \left (3\right )+4 x +16\right ) {\mathrm e}^{-x}}{4}\) | \(32\) |
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Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {3}{4} \, {\left (e^{5} \log \left (3\right ) + 2 \, e^{5} \log \left (2\right ) + 4 \, x + 16\right )} e^{\left (-x\right )} - 3 \, e^{\left (-x\right )} \log \left (x\right ) \]
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Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {\left (12 x - 12 \log {\left (x \right )} + 48 + 3 e^{5} \log {\left (3 \right )} + 6 e^{5} \log {\left (2 \right )}\right ) e^{- x}}{4} \]
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Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {3}{4} \, e^{\left (-x + 5\right )} \log \left (3\right ) + \frac {3}{2} \, e^{\left (-x + 5\right )} \log \left (2\right ) - 3 \, e^{\left (-x\right )} \log \left (x\right ) + 9 \, e^{\left (-x\right )} \]
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Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=3 \, x e^{\left (-x\right )} + \frac {3}{4} \, e^{\left (-x + 5\right )} \log \left (3\right ) + \frac {3}{2} \, e^{\left (-x + 5\right )} \log \left (2\right ) - 3 \, e^{\left (-x\right )} \log \left (x\right ) + 12 \, e^{\left (-x\right )} \]
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Time = 12.93 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-12-36 x-12 x^2-3 e^5 x \log (3)-3 e^5 x \log (4)+12 x \log (x)\right )}{4 x} \, dx=\frac {3\,{\mathrm {e}}^{-x}\,\left (4\,x-4\,\ln \left (x\right )+2\,{\mathrm {e}}^5\,\ln \left (2\right )+{\mathrm {e}}^5\,\ln \left (3\right )+16\right )}{4} \]
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