Integrand size = 196, antiderivative size = 32 \[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=e+25 \log ^2\left (2 x \left (e^5+\frac {x}{-\frac {e^x}{3-x}+x}\right )\right ) \]
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\[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=\int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {50 \left (e^{5+2 x}+2 e^{5+x} (-3+x) x+\left (1+e^5\right ) (-3+x)^2 x^2-e^x x \left (6-6 x+x^2\right )\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x \left (e^x+(-3+x) x\right ) \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )} \, dx \\ & = 50 \int \frac {\left (e^{5+2 x}+2 e^{5+x} (-3+x) x+\left (1+e^5\right ) (-3+x)^2 x^2-e^x x \left (6-6 x+x^2\right )\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x \left (e^x+(-3+x) x\right ) \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )} \, dx \\ & = 50 \int \left (\frac {\log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x}+\frac {\left (3-5 x+x^2\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{e^x-3 x+x^2}+\frac {\left (1+e^5\right ) \left (-3+5 x-x^2\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{e^{5+x}-3 \left (1+e^5\right ) x+\left (1+e^5\right ) x^2}\right ) \, dx \\ & = 50 \int \frac {\log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x} \, dx+50 \int \frac {\left (3-5 x+x^2\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{e^x-3 x+x^2} \, dx+\left (50 \left (1+e^5\right )\right ) \int \frac {\left (-3+5 x-x^2\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{e^{5+x}-3 \left (1+e^5\right ) x+\left (1+e^5\right ) x^2} \, dx \\ & = 50 \int \frac {\log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x} \, dx-50 \int \frac {\left (e^{5+2 x}+2 e^{5+x} (-3+x) x+\left (1+e^5\right ) (-3+x)^2 x^2-e^x x \left (6-6 x+x^2\right )\right ) \left (3 \int \frac {1}{e^x+(-3+x) x} \, dx-5 \int \frac {x}{e^x+(-3+x) x} \, dx+\int \frac {x^2}{e^x+(-3+x) x} \, dx\right )}{x \left (e^x+(-3+x) x\right ) \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )} \, dx-\left (50 \left (1+e^5\right )\right ) \int \frac {\left (e^{5+2 x}+2 e^{5+x} (-3+x) x+\left (1+e^5\right ) (-3+x)^2 x^2-e^x x \left (6-6 x+x^2\right )\right ) \left (3 \int -\frac {1}{e^{5+x}+(-3+x) x+e^5 (-3+x) x} \, dx+\int -\frac {x^2}{e^{5+x}+(-3+x) x+e^5 (-3+x) x} \, dx+5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx\right )}{x \left (e^x+(-3+x) x\right ) \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )} \, dx+\left (50 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x^2}{e^x-3 x+x^2} \, dx+\left (150 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {1}{e^x-3 x+x^2} \, dx-\left (250 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x}{e^x-3 x+x^2} \, dx+\left (50 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x^2}{-e^{5+x}+3 \left (1+e^5\right ) x-\left (1+e^5\right ) x^2} \, dx+\left (150 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {1}{-e^{5+x}+3 \left (1+e^5\right ) x-\left (1+e^5\right ) x^2} \, dx+\left (250 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx \\ & = 50 \int \frac {\log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x} \, dx-50 \int \left (\frac {\left (1+e^5\right ) \left (3-5 x+x^2\right ) \left (-3 \int \frac {1}{e^x+(-3+x) x} \, dx+5 \int \frac {x}{e^x+(-3+x) x} \, dx-\int \frac {x^2}{e^x+(-3+x) x} \, dx\right )}{e^{5+x}-3 \left (1+e^5\right ) x+\left (1+e^5\right ) x^2}+\frac {3 \int \frac {1}{e^x+(-3+x) x} \, dx-5 \int \frac {x}{e^x+(-3+x) x} \, dx+\int \frac {x^2}{e^x+(-3+x) x} \, dx}{x}+\frac {\left (3-5 x+x^2\right ) \left (3 \int \frac {1}{e^x+(-3+x) x} \, dx-5 \int \frac {x}{e^x+(-3+x) x} \, dx+\int \frac {x^2}{e^x+(-3+x) x} \, dx\right )}{e^x-3 x+x^2}\right ) \, dx-\left (50 \left (1+e^5\right )\right ) \int \left (\frac {-3 \int \frac {1}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx+5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx-\int \frac {x^2}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx}{x}-\frac {\left (3-5 x+x^2\right ) \left (3 \int \frac {1}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx-5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx+\int \frac {x^2}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx\right )}{e^x-3 x+x^2}+\frac {\left (1+e^5\right ) \left (3-5 x+x^2\right ) \left (3 \int \frac {1}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx-5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx+\int \frac {x^2}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx\right )}{e^{5+x}-3 \left (1+e^5\right ) x+\left (1+e^5\right ) x^2}\right ) \, dx+\left (50 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x^2}{e^x-3 x+x^2} \, dx+\left (150 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {1}{e^x-3 x+x^2} \, dx-\left (250 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x}{e^x-3 x+x^2} \, dx+\left (50 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x^2}{-e^{5+x}+3 \left (1+e^5\right ) x-\left (1+e^5\right ) x^2} \, dx+\left (150 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {1}{-e^{5+x}+3 \left (1+e^5\right ) x-\left (1+e^5\right ) x^2} \, dx+\left (250 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx \\ & = 50 \int \frac {\log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )}{x} \, dx-50 \int \frac {3 \int \frac {1}{e^x+(-3+x) x} \, dx-5 \int \frac {x}{e^x+(-3+x) x} \, dx+\int \frac {x^2}{e^x+(-3+x) x} \, dx}{x} \, dx-50 \int \frac {\left (3-5 x+x^2\right ) \left (3 \int \frac {1}{e^x+(-3+x) x} \, dx-5 \int \frac {x}{e^x+(-3+x) x} \, dx+\int \frac {x^2}{e^x+(-3+x) x} \, dx\right )}{e^x-3 x+x^2} \, dx-\left (50 \left (1+e^5\right )\right ) \int \frac {\left (3-5 x+x^2\right ) \left (-3 \int \frac {1}{e^x+(-3+x) x} \, dx+5 \int \frac {x}{e^x+(-3+x) x} \, dx-\int \frac {x^2}{e^x+(-3+x) x} \, dx\right )}{e^{5+x}-3 \left (1+e^5\right ) x+\left (1+e^5\right ) x^2} \, dx-\left (50 \left (1+e^5\right )\right ) \int \frac {-3 \int \frac {1}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx+5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx-\int \frac {x^2}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx}{x} \, dx+\left (50 \left (1+e^5\right )\right ) \int \frac {\left (3-5 x+x^2\right ) \left (3 \int \frac {1}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx-5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx+\int \frac {x^2}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx\right )}{e^x-3 x+x^2} \, dx-\left (50 \left (1+e^5\right )^2\right ) \int \frac {\left (3-5 x+x^2\right ) \left (3 \int \frac {1}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx-5 \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx+\int \frac {x^2}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx\right )}{e^{5+x}-3 \left (1+e^5\right ) x+\left (1+e^5\right ) x^2} \, dx+\left (50 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x^2}{e^x-3 x+x^2} \, dx+\left (150 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {1}{e^x-3 x+x^2} \, dx-\left (250 \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x}{e^x-3 x+x^2} \, dx+\left (50 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x^2}{-e^{5+x}+3 \left (1+e^5\right ) x-\left (1+e^5\right ) x^2} \, dx+\left (150 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {1}{-e^{5+x}+3 \left (1+e^5\right ) x-\left (1+e^5\right ) x^2} \, dx+\left (250 \left (1+e^5\right ) \log \left (\frac {2 x \left (e^{5+x}+\left (1+e^5\right ) (-3+x) x\right )}{e^x+(-3+x) x}\right )\right ) \int \frac {x}{e^{5+x}+\left (1+e^5\right ) (-3+x) x} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=25 \log ^2\left (\frac {2 x \left (e^{5+x}+(-3+x) x+e^5 (-3+x) x\right )}{e^x+(-3+x) x}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.72 (sec) , antiderivative size = 1829, normalized size of antiderivative = 57.16
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Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=25 \, \log \left (\frac {2 \, {\left ({\left (x^{3} - 3 \, x^{2}\right )} e^{10} + {\left (x^{3} - 3 \, x^{2}\right )} e^{5} + x e^{\left (x + 10\right )}\right )}}{{\left (x^{2} - 3 \, x\right )} e^{5} + e^{\left (x + 5\right )}}\right )^{2} \]
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Time = 0.50 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=25 \log {\left (\frac {2 x^{3} - 6 x^{2} + 2 x e^{5} e^{x} + \left (2 x^{3} - 6 x^{2}\right ) e^{5}}{x^{2} - 3 x + e^{x}} \right )}^{2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 5.91 \[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=50 \, {\left (\log \left (x^{2} - 3 \, x + e^{x}\right ) - \log \left (x\right ) + 5\right )} \log \left (x^{2} {\left (e^{5} + 1\right )} - 3 \, x {\left (e^{5} + 1\right )} + e^{\left (x + 5\right )}\right ) - 25 \, \log \left (x^{2} {\left (e^{5} + 1\right )} - 3 \, x {\left (e^{5} + 1\right )} + e^{\left (x + 5\right )}\right )^{2} + 50 \, {\left (\log \left (x\right ) - 5\right )} \log \left (x^{2} - 3 \, x + e^{x}\right ) - 25 \, \log \left (x^{2} - 3 \, x + e^{x}\right )^{2} - 25 \, \log \left (x\right )^{2} - 50 \, {\left (\log \left (x^{2} - 3 \, x + e^{x}\right ) - \log \left ({\left (x^{2} {\left (e^{5} + 1\right )} - 3 \, x {\left (e^{5} + 1\right )} + e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}\right ) - \log \left (x\right )\right )} \log \left (\frac {2 \, {\left (x^{3} - 3 \, x^{2} + {\left (x^{3} - 3 \, x^{2}\right )} e^{5} + x e^{\left (x + 5\right )}\right )}}{x^{2} - 3 \, x + e^{x}}\right ) + 250 \, \log \left (x\right ) \]
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\[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=\int { \frac {50 \, {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2} + {\left (x^{4} - 6 \, x^{3} + 9 \, x^{2}\right )} e^{5} - {\left (x^{3} - 6 \, x^{2} - 2 \, {\left (x^{2} - 3 \, x\right )} e^{5} + 6 \, x\right )} e^{x} + e^{\left (2 \, x + 5\right )}\right )} \log \left (\frac {2 \, {\left (x^{3} - 3 \, x^{2} + {\left (x^{3} - 3 \, x^{2}\right )} e^{5} + x e^{\left (x + 5\right )}\right )}}{x^{2} - 3 \, x + e^{x}}\right )}{x^{5} - 6 \, x^{4} + 9 \, x^{3} + {\left (x^{5} - 6 \, x^{4} + 9 \, x^{3}\right )} e^{5} + x e^{\left (2 \, x + 5\right )} + {\left (x^{3} - 3 \, x^{2} + 2 \, {\left (x^{3} - 3 \, x^{2}\right )} e^{5}\right )} e^{x}} \,d x } \]
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Time = 13.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56 \[ \int \frac {\left (50 e^{5+2 x}+450 x^2-300 x^3+50 x^4+e^5 \left (450 x^2-300 x^3+50 x^4\right )+e^x \left (-300 x+300 x^2-50 x^3+e^5 \left (-300 x+100 x^2\right )\right )\right ) \log \left (\frac {2 e^{5+x} x-6 x^2+2 x^3+e^5 \left (-6 x^2+2 x^3\right )}{e^x-3 x+x^2}\right )}{e^{5+2 x} x+9 x^3-6 x^4+x^5+e^5 \left (9 x^3-6 x^4+x^5\right )+e^x \left (-3 x^2+x^3+e^5 \left (-6 x^2+2 x^3\right )\right )} \, dx=25\,{\ln \left (-\frac {{\mathrm {e}}^5\,\left (6\,x^2-2\,x^3\right )+6\,x^2-2\,x^3-2\,x\,{\mathrm {e}}^5\,{\mathrm {e}}^x}{{\mathrm {e}}^x-3\,x+x^2}\right )}^2 \]
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