Integrand size = 56, antiderivative size = 27 \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=\frac {1}{4} e^{-2 x} \left (3+x-\left (-4+e^{\frac {9}{\log ^2(x)}}\right ) x^2\right ) \]
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Leaf count is larger than twice the leaf count of optimal. \(71\) vs. \(2(27)=54\).
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.63, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 6874, 2225, 2207, 2326} \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=e^{-2 x} x^2+\frac {1}{4} e^{-2 x} x+\frac {3 e^{-2 x}}{4}-\frac {x e^{\frac {9}{\log ^2(x)}-2 x} \left (x \log ^3(x)+9\right )}{4 \left (\frac {9}{x \log ^3(x)}+1\right ) \log ^3(x)} \]
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Rule 12
Rule 2207
Rule 2225
Rule 2326
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{\log ^3(x)} \, dx \\ & = \frac {1}{4} \int \left (-5 e^{-2 x}+6 e^{-2 x} x-8 e^{-2 x} x^2+\frac {2 e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9-\log ^3(x)+x \log ^3(x)\right )}{\log ^3(x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9-\log ^3(x)+x \log ^3(x)\right )}{\log ^3(x)} \, dx-\frac {5}{4} \int e^{-2 x} \, dx+\frac {3}{2} \int e^{-2 x} x \, dx-2 \int e^{-2 x} x^2 \, dx \\ & = \frac {5 e^{-2 x}}{8}-\frac {3}{4} e^{-2 x} x+e^{-2 x} x^2-\frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9+x \log ^3(x)\right )}{4 \left (1+\frac {9}{x \log ^3(x)}\right ) \log ^3(x)}+\frac {3}{4} \int e^{-2 x} \, dx-2 \int e^{-2 x} x \, dx \\ & = \frac {e^{-2 x}}{4}+\frac {1}{4} e^{-2 x} x+e^{-2 x} x^2-\frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9+x \log ^3(x)\right )}{4 \left (1+\frac {9}{x \log ^3(x)}\right ) \log ^3(x)}-\int e^{-2 x} \, dx \\ & = \frac {3 e^{-2 x}}{4}+\frac {1}{4} e^{-2 x} x+e^{-2 x} x^2-\frac {e^{-2 x+\frac {9}{\log ^2(x)}} x \left (9+x \log ^3(x)\right )}{4 \left (1+\frac {9}{x \log ^3(x)}\right ) \log ^3(x)} \\ \end{align*}
Time = 0.76 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=\frac {1}{4} e^{-2 x} \left (3+x-\left (-4+e^{\frac {9}{\log ^2(x)}}\right ) x^2\right ) \]
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Time = 0.66 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {\left (3-x^{2} {\mathrm e}^{\frac {9}{\ln \left (x \right )^{2}}}+4 x^{2}+x \right ) {\mathrm e}^{-2 x}}{4}\) | \(27\) |
risch | \(\frac {\left (4 x^{2}+x +3\right ) {\mathrm e}^{-2 x}}{4}-\frac {x^{2} {\mathrm e}^{-\frac {2 x \ln \left (x \right )^{2}-9}{\ln \left (x \right )^{2}}}}{4}\) | \(37\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=-\frac {1}{4} \, x^{2} e^{\left (-2 \, x + \frac {9}{\log \left (x\right )^{2}}\right )} + \frac {1}{4} \, {\left (4 \, x^{2} + x + 3\right )} e^{\left (-2 \, x\right )} \]
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Time = 7.52 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=\frac {\left (- x^{2} e^{\frac {9}{\log {\left (x \right )}^{2}}} + 4 x^{2} + x + 3\right ) e^{- 2 x}}{4} \]
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Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=-\frac {1}{4} \, x^{2} e^{\left (-2 \, x + \frac {9}{\log \left (x\right )^{2}}\right )} + \frac {1}{2} \, {\left (2 \, x^{2} + 2 \, x + 1\right )} e^{\left (-2 \, x\right )} - \frac {3}{8} \, {\left (2 \, x + 1\right )} e^{\left (-2 \, x\right )} + \frac {5}{8} \, e^{\left (-2 \, x\right )} \]
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\[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=\int { -\frac {{\left ({\left (8 \, x^{2} - 6 \, x + 5\right )} \log \left (x\right )^{3} - 2 \, {\left ({\left (x^{2} - x\right )} \log \left (x\right )^{3} + 9 \, x\right )} e^{\left (\frac {9}{\log \left (x\right )^{2}}\right )}\right )} e^{\left (-2 \, x\right )}}{4 \, \log \left (x\right )^{3}} \,d x } \]
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Timed out. \[ \int \frac {e^{-2 x} \left (\left (-5+6 x-8 x^2\right ) \log ^3(x)+e^{\frac {9}{\log ^2(x)}} \left (18 x+\left (-2 x+2 x^2\right ) \log ^3(x)\right )\right )}{4 \log ^3(x)} \, dx=\int -\frac {{\mathrm {e}}^{-2\,x}\,\left (\frac {{\ln \left (x\right )}^3\,\left (8\,x^2-6\,x+5\right )}{4}-\frac {{\mathrm {e}}^{\frac {9}{{\ln \left (x\right )}^2}}\,\left (18\,x-{\ln \left (x\right )}^3\,\left (2\,x-2\,x^2\right )\right )}{4}\right )}{{\ln \left (x\right )}^3} \,d x \]
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