Integrand size = 130, antiderivative size = 32 \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=e^{4+\frac {5 x^2 \left (1+e^x+2 x\right ) \log \left (e^{-x} x\right )}{3-x}} \]
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\[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=\int \frac {\exp \left (\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}\right ) \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}\right ) \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{(-3+x)^2} \, dx \\ & = \int \frac {5 e^4 x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \left (\left (1+e^x+2 x\right ) \left (3-4 x+x^2\right )+\left (6+17 x-4 x^2+e^x \left (6+2 x-x^2\right )\right ) \log \left (e^{-x} x\right )\right )}{(3-x)^2} \, dx \\ & = \left (5 e^4\right ) \int \frac {x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \left (\left (1+e^x+2 x\right ) \left (3-4 x+x^2\right )+\left (6+17 x-4 x^2+e^x \left (6+2 x-x^2\right )\right ) \log \left (e^{-x} x\right )\right )}{(3-x)^2} \, dx \\ & = \left (5 e^4\right ) \int \left (\frac {3 x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{(-3+x)^2}-\frac {4 x^2 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{(-3+x)^2}+\frac {2 (-1+x) x^2 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{-3+x}+\frac {x^3 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{(-3+x)^2}+\frac {6 x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \log \left (e^{-x} x\right )}{(-3+x)^2}+\frac {17 x^2 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \log \left (e^{-x} x\right )}{(-3+x)^2}-\frac {4 x^3 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \log \left (e^{-x} x\right )}{(-3+x)^2}-\frac {e^x x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \left (-3+4 x-x^2-6 \log \left (e^{-x} x\right )-2 x \log \left (e^{-x} x\right )+x^2 \log \left (e^{-x} x\right )\right )}{(-3+x)^2}\right ) \, dx \\ & = \left (5 e^4\right ) \int \frac {x^3 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{(-3+x)^2} \, dx-\left (5 e^4\right ) \int \frac {e^x x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \left (-3+4 x-x^2-6 \log \left (e^{-x} x\right )-2 x \log \left (e^{-x} x\right )+x^2 \log \left (e^{-x} x\right )\right )}{(-3+x)^2} \, dx+\left (10 e^4\right ) \int \frac {(-1+x) x^2 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{-3+x} \, dx+\left (15 e^4\right ) \int \frac {x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{(-3+x)^2} \, dx-\left (20 e^4\right ) \int \frac {x^2 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}}}{(-3+x)^2} \, dx-\left (20 e^4\right ) \int \frac {x^3 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \log \left (e^{-x} x\right )}{(-3+x)^2} \, dx+\left (30 e^4\right ) \int \frac {x \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \log \left (e^{-x} x\right )}{(-3+x)^2} \, dx+\left (85 e^4\right ) \int \frac {x^2 \left (e^{-x} x\right )^{-\frac {5 x^2 \left (1+e^x+2 x\right )}{-3+x}} \log \left (e^{-x} x\right )}{(-3+x)^2} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(32)=64\).
Time = 0.41 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.06 \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=e^{4+\frac {5 x \left (21 (-3+x)+\left (-21+x+e^x x+2 x^2\right ) \log (x)-\left (-21+x+e^x x+2 x^2\right ) \log \left (e^{-x} x\right )\right )}{-3+x}} x^{-\frac {5 \left (-63+\left (1+e^x\right ) x^2+2 x^3\right )}{-3+x}} \left (e^{-x} x\right )^{-\frac {315}{-3+x}} \]
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Time = 7.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (-5 \,{\mathrm e}^{x} x^{2}-10 x^{3}-5 x^{2}\right ) \ln \left (x \,{\mathrm e}^{-x}\right )+4 x -12}{-3+x}}\) | \(39\) |
risch | \(x^{-\frac {5 \,{\mathrm e}^{x} x^{2}}{-3+x}} x^{-\frac {10 x^{3}}{-3+x}} \left ({\mathrm e}^{x}\right )^{\frac {5 \,{\mathrm e}^{x} x^{2}}{-3+x}} \left ({\mathrm e}^{x}\right )^{\frac {10 x^{3}}{-3+x}} x^{-\frac {5 x^{2}}{-3+x}} \left ({\mathrm e}^{x}\right )^{\frac {5 x^{2}}{-3+x}} {\mathrm e}^{\frac {5 i \pi \,{\mathrm e}^{x} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} x^{2}-5 i \pi \,{\mathrm e}^{x} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{2}-5 i \pi \,{\mathrm e}^{x} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) x^{2}+5 i \pi \,{\mathrm e}^{x} \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) x^{2}+10 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} x^{3}-10 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{3}-10 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) x^{3}+10 i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) x^{3}+5 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3} x^{2}-5 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{2}-5 i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) x^{2}+5 i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) x^{2}+8 x -24}{2 x -6}}\) | \(390\) |
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Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=e^{\left (-\frac {5 \, {\left (2 \, x^{3} + x^{2} e^{x} + x^{2}\right )} \log \left (x e^{\left (-x\right )}\right ) - 4 \, x + 12}{x - 3}\right )} \]
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Timed out. \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (27) = 54\).
Time = 0.93 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.19 \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=e^{\left (-\frac {10 \, x^{3} \log \left (x e^{\left (-x\right )}\right )}{x - 3} - \frac {5 \, x^{2} e^{x} \log \left (x e^{\left (-x\right )}\right )}{x - 3} - \frac {5 \, x^{2} \log \left (x e^{\left (-x\right )}\right )}{x - 3} + \frac {4 \, x}{x - 3} - \frac {12}{x - 3}\right )} \]
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Time = 9.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.47 \[ \int \frac {e^{\frac {-12+4 x+\left (-5 x^2-5 e^x x^2-10 x^3\right ) \log \left (e^{-x} x\right )}{-3+x}} \left (15 x+10 x^2-35 x^3+10 x^4+e^x \left (15 x-20 x^2+5 x^3\right )+\left (30 x+85 x^2-20 x^3+e^x \left (30 x+10 x^2-5 x^3\right )\right ) \log \left (e^{-x} x\right )\right )}{9-6 x+x^2} \, dx=\frac {{\mathrm {e}}^{\frac {4\,x}{x-3}}\,{\mathrm {e}}^{\frac {5\,x^3\,{\mathrm {e}}^x}{x-3}}\,{\mathrm {e}}^{\frac {5\,x^3}{x-3}}\,{\mathrm {e}}^{\frac {10\,x^4}{x-3}}\,{\mathrm {e}}^{-\frac {12}{x-3}}}{x^{\frac {5\,\left (x^2\,{\mathrm {e}}^x+x^2+2\,x^3\right )}{x-3}}} \]
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