Integrand size = 124, antiderivative size = 32 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=-e^{5 x}+4 \log \left (-3 e^{-x}+x\right ) \log \left (-x+\frac {x^2}{3}\right ) \]
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\[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=\int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-5 e^{5 x}+\frac {4 \left (3+e^x\right ) \log \left (\frac {1}{3} (-3+x) x\right )}{-3+e^x x}+\frac {4 (-3+2 x) \log \left (-3 e^{-x}+x\right )}{(-3+x) x}\right ) \, dx \\ & = 4 \int \frac {\left (3+e^x\right ) \log \left (\frac {1}{3} (-3+x) x\right )}{-3+e^x x} \, dx+4 \int \frac {(-3+2 x) \log \left (-3 e^{-x}+x\right )}{(-3+x) x} \, dx-5 \int e^{5 x} \, dx \\ & = -e^{5 x}+4 \int \left (\frac {\log \left (\frac {1}{3} (-3+x) x\right )}{x}+\frac {3 (1+x) \log \left (\frac {1}{3} (-3+x) x\right )}{x \left (-3+e^x x\right )}\right ) \, dx+4 \int \left (\frac {\log \left (-3 e^{-x}+x\right )}{-3+x}+\frac {\log \left (-3 e^{-x}+x\right )}{x}\right ) \, dx \\ & = -e^{5 x}+4 \int \frac {\log \left (\frac {1}{3} (-3+x) x\right )}{x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx+12 \int \frac {(1+x) \log \left (\frac {1}{3} (-3+x) x\right )}{x \left (-3+e^x x\right )} \, dx \\ & = -e^{5 x}+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )-4 \int \frac {\log (x)}{-3+x} \, dx-4 \int \frac {\log (x)}{x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \frac {(3-2 x) \left (\int \frac {1}{-3+e^x x} \, dx+\int \frac {1}{x \left (-3+e^x x\right )} \, dx\right )}{(3-x) x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx \\ & = -e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )-4 \int \frac {\log \left (\frac {x}{3}\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \left (\frac {(-3+2 x) \int \frac {1}{-3+e^x x} \, dx}{(-3+x) x}+\frac {(-3+2 x) \int \frac {1}{x \left (-3+e^x x\right )} \, dx}{(-3+x) x}\right ) \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx \\ & = -e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )+4 \operatorname {PolyLog}\left (2,1-\frac {x}{3}\right )+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \frac {(-3+2 x) \int \frac {1}{-3+e^x x} \, dx}{(-3+x) x} \, dx-12 \int \frac {(-3+2 x) \int \frac {1}{x \left (-3+e^x x\right )} \, dx}{(-3+x) x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx \\ & = -e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )+4 \operatorname {PolyLog}\left (2,1-\frac {x}{3}\right )+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \left (\frac {\int \frac {1}{-3+e^x x} \, dx}{-3+x}+\frac {\int \frac {1}{-3+e^x x} \, dx}{x}\right ) \, dx-12 \int \left (\frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{-3+x}+\frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{x}\right ) \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx \\ & = -e^{5 x}-4 \log (3) \log (-3+x)-2 \log ^2(x)+4 \log (x) \log \left (-\frac {1}{3} (3-x) x\right )+4 \operatorname {PolyLog}\left (2,1-\frac {x}{3}\right )+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{-3+x} \, dx+4 \int \frac {\log \left (-3 e^{-x}+x\right )}{x} \, dx-12 \int \frac {\int \frac {1}{-3+e^x x} \, dx}{-3+x} \, dx-12 \int \frac {\int \frac {1}{-3+e^x x} \, dx}{x} \, dx-12 \int \frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{-3+x} \, dx-12 \int \frac {\int \frac {1}{x \left (-3+e^x x\right )} \, dx}{x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{-3+e^x x} \, dx+\left (12 \log \left (\frac {1}{3} (-3+x) x\right )\right ) \int \frac {1}{x \left (-3+e^x x\right )} \, dx \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(95\) vs. \(2(32)=64\).
Time = 0.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.97 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=-e^{5 x}-4 x \log \left (\frac {1}{3} (-3+x) x\right )+4 \log (3-x) \left (x+\log \left (-3 e^{-x}+x\right )-\log \left (-3+e^x x\right )\right )+4 \log (x) \left (x+\log \left (-3 e^{-x}+x\right )-\log \left (-3+e^x x\right )\right )+4 \log \left (\frac {1}{3} (-3+x) x\right ) \log \left (-3+e^x x\right ) \]
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Time = 84.62 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(4 \ln \left (\left ({\mathrm e}^{x} x -3\right ) {\mathrm e}^{-x}\right ) \ln \left (\frac {1}{3} x^{2}-x \right )-{\mathrm e}^{5 x}\) | \(32\) |
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=4 \, \log \left (\frac {1}{3} \, x^{2} - x\right ) \log \left ({\left (x e^{x} - 3\right )} e^{\left (-x\right )}\right ) - e^{\left (5 \, x\right )} \]
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Time = 0.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=- e^{5 x} + 4 \log {\left (\left (x e^{x} - 3\right ) e^{- x} \right )} \log {\left (\frac {x^{2}}{3} - x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (28) = 56\).
Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=4 \, x \log \left (3\right ) + 4 \, {\left (\log \left (x - 3\right ) + \log \left (x\right )\right )} \log \left (x e^{x} - 3\right ) - 4 \, x \log \left (x - 3\right ) - 4 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right ) - 4 \, \log \left (3\right ) \log \left (\frac {x e^{x} - 3}{x}\right ) - e^{\left (5 \, x\right )} \]
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\[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=\int { -\frac {5 \, {\left (3 \, x^{2} - {\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 9 \, x\right )} e^{\left (5 \, x\right )} + 4 \, {\left (3 \, x^{2} + {\left (x^{2} - 3 \, x\right )} e^{x} - 9 \, x\right )} \log \left (\frac {1}{3} \, x^{2} - x\right ) + 4 \, {\left ({\left (2 \, x^{2} - 3 \, x\right )} e^{x} - 6 \, x + 9\right )} \log \left ({\left (x e^{x} - 3\right )} e^{\left (-x\right )}\right )}{3 \, x^{2} - {\left (x^{3} - 3 \, x^{2}\right )} e^{x} - 9 \, x} \,d x } \]
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Time = 15.87 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{5 x} \left (-45 x+15 x^2+e^x \left (15 x^2-5 x^3\right )\right )+\left (36-24 x+e^x \left (-12 x+8 x^2\right )\right ) \log \left (e^{-x} \left (-3+e^x x\right )\right )+\left (-36 x+12 x^2+e^x \left (-12 x+4 x^2\right )\right ) \log \left (\frac {1}{3} \left (-3 x+x^2\right )\right )}{9 x-3 x^2+e^x \left (-3 x^2+x^3\right )} \, dx=4\,\ln \left (\frac {x^2}{3}-x\right )\,\ln \left (x-3\,{\mathrm {e}}^{-x}\right )-{\mathrm {e}}^{5\,x} \]
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