Integrand size = 111, antiderivative size = 31 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=-x+9 x^2+\frac {3 e^{3-x}}{x (4 x-2 \log (x))} \]
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Time = 0.98 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {6873, 12, 6874, 2326} \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {3 e^{3-x} \left (2 x^2-x \log (x)\right )}{2 x^2 (2 x-\log (x))^2}+\frac {1}{36} (1-18 x)^2 \]
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Rule 12
Rule 2326
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)\right )}{2 x^2 (2 x-\log (x))^2} \, dx \\ & = \frac {1}{2} \int \frac {e^{-x} \left (e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)\right )}{x^2 (2 x-\log (x))^2} \, dx \\ & = \frac {1}{2} \int \left (2 (-1+18 x)-\frac {3 e^{3-x} \left (-1+4 x+2 x^2-\log (x)-x \log (x)\right )}{x^2 (2 x-\log (x))^2}\right ) \, dx \\ & = \frac {1}{36} (1-18 x)^2-\frac {3}{2} \int \frac {e^{3-x} \left (-1+4 x+2 x^2-\log (x)-x \log (x)\right )}{x^2 (2 x-\log (x))^2} \, dx \\ & = \frac {1}{36} (1-18 x)^2+\frac {3 e^{3-x} \left (2 x^2-x \log (x)\right )}{2 x^2 (2 x-\log (x))^2} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {1}{2} \left (-2 x+18 x^2-\frac {3 e^{3-x}}{x (-2 x+\log (x))}\right ) \]
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Time = 0.58 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \(9 x^{2}-x +\frac {3 \,{\mathrm e}^{-x +3}}{2 \left (2 x -\ln \left (x \right )\right ) x}\) | \(31\) |
parallelrisch | \(\frac {\left (36 \,{\mathrm e}^{x} x^{4}-18 x^{3} {\mathrm e}^{x} \ln \left (x \right )-4 \,{\mathrm e}^{x} x^{3}+2 x^{2} {\mathrm e}^{x} \ln \left (x \right )+3 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-x}}{2 x \left (2 x -\ln \left (x \right )\right )}\) | \(57\) |
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Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=-\frac {2 \, {\left (9 \, x^{3} - x^{2}\right )} e^{x} \log \left (x\right ) - 4 \, {\left (9 \, x^{4} - x^{3}\right )} e^{x} - 3 \, e^{3}}{2 \, {\left (2 \, x^{2} e^{x} - x e^{x} \log \left (x\right )\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=9 x^{2} - x + \frac {3 e^{3} e^{- x}}{4 x^{2} - 2 x \log {\left (x \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {36 \, x^{4} - 4 \, x^{3} - 2 \, {\left (9 \, x^{3} - x^{2}\right )} \log \left (x\right ) + 3 \, e^{\left (-x + 3\right )}}{2 \, {\left (2 \, x^{2} - x \log \left (x\right )\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {36 \, x^{4} - 18 \, x^{3} \log \left (x\right ) - 4 \, x^{3} + 2 \, x^{2} \log \left (x\right ) + 3 \, e^{\left (-x + 3\right )}}{2 \, {\left (2 \, x^{2} - x \log \left (x\right )\right )}} \]
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Time = 14.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 3.39 \[ \int \frac {e^3 \left (3-12 x-6 x^2\right )+e^x \left (-8 x^4+144 x^5\right )+\left (e^3 (3+3 x)+e^x \left (8 x^3-144 x^4\right )\right ) \log (x)+e^x \left (-2 x^2+36 x^3\right ) \log ^2(x)}{8 e^x x^4-8 e^x x^3 \log (x)+2 e^x x^2 \log ^2(x)} \, dx=\frac {\frac {3\,{\mathrm {e}}^{-x}\,\left (2\,{\mathrm {e}}^3\,x^2+4\,{\mathrm {e}}^3\,x-{\mathrm {e}}^3\right )}{2\,x\,\left (2\,x-1\right )}-\frac {3\,{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left ({\mathrm {e}}^3+x\,{\mathrm {e}}^3\right )}{2\,x\,\left (2\,x-1\right )}}{2\,x-\ln \left (x\right )}-x+9\,x^2+\frac {{\mathrm {e}}^{-x}\,\left (\frac {3\,{\mathrm {e}}^3}{4}+\frac {3\,x\,{\mathrm {e}}^3}{4}\right )}{\frac {x}{2}-x^2} \]
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