Integrand size = 168, antiderivative size = 24 \[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=-\frac {e^{x+\frac {100}{\left (\frac {e^x}{3}+x\right )^2}}}{x}+x \]
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\[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=\int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {27 \left (\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5-\frac {1}{27} e^{x+\frac {900}{\left (e^x+3 x\right )^2}} \left (e^{3 x} (-1+x)+9 e^{2 x} (-1+x) x+9 e^x x \left (-200-3 x+3 x^2\right )+27 x \left (-200-x^2+x^3\right )\right )\right )}{x^2 \left (e^x+3 x\right )^3} \, dx \\ & = 27 \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5-\frac {1}{27} e^{x+\frac {900}{\left (e^x+3 x\right )^2}} \left (e^{3 x} (-1+x)+9 e^{2 x} (-1+x) x+9 e^x x \left (-200-3 x+3 x^2\right )+27 x \left (-200-x^2+x^3\right )\right )}{x^2 \left (e^x+3 x\right )^3} \, dx \\ & = 27 \int \left (\frac {e^{3 x}}{27 \left (e^x+3 x\right )^3}+\frac {e^{2 x} x}{3 \left (e^x+3 x\right )^3}+\frac {e^x x^2}{\left (e^x+3 x\right )^3}+\frac {x^3}{\left (e^x+3 x\right )^3}-\frac {e^{x+\frac {900}{\left (e^x+3 x\right )^2}} \left (-e^{3 x}-5400 x-1800 e^x x-9 e^{2 x} x+e^{3 x} x-27 e^x x^2+9 e^{2 x} x^2-27 x^3+27 e^x x^3+27 x^4\right )}{27 x^2 \left (e^x+3 x\right )^3}\right ) \, dx \\ & = 9 \int \frac {e^{2 x} x}{\left (e^x+3 x\right )^3} \, dx+27 \int \frac {e^x x^2}{\left (e^x+3 x\right )^3} \, dx+27 \int \frac {x^3}{\left (e^x+3 x\right )^3} \, dx+\int \frac {e^{3 x}}{\left (e^x+3 x\right )^3} \, dx-\int \frac {e^{x+\frac {900}{\left (e^x+3 x\right )^2}} \left (-e^{3 x}-5400 x-1800 e^x x-9 e^{2 x} x+e^{3 x} x-27 e^x x^2+9 e^{2 x} x^2-27 x^3+27 e^x x^3+27 x^4\right )}{x^2 \left (e^x+3 x\right )^3} \, dx \\ & = -\frac {27 x^2}{2 \left (e^x+3 x\right )^2}+\frac {e^{x+\frac {900}{\left (e^x+3 x\right )^2}} \left (5400 x+1800 e^x x-e^{3 x} x-9 e^{2 x} x^2-27 e^x x^3-27 x^4\right )}{x^2 \left (e^x+3 x\right )^3 \left (1-\frac {1800 \left (3+e^x\right )}{\left (e^x+3 x\right )^3}\right )}+9 \int \frac {e^{2 x} x}{\left (e^x+3 x\right )^3} \, dx+27 \int \frac {x^3}{\left (e^x+3 x\right )^3} \, dx+27 \int \frac {x}{\left (e^x+3 x\right )^2} \, dx-81 \int \frac {x^2}{\left (e^x+3 x\right )^3} \, dx+\int \frac {e^{3 x}}{\left (e^x+3 x\right )^3} \, dx \\ \end{align*}
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=27 \left (-\frac {e^{x+\frac {900}{\left (e^x+3 x\right )^2}}}{27 x}+\frac {x}{27}\right ) \]
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Time = 3.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83
method | result | size |
parallelrisch | \(\frac {15 x^{2}-15 \,{\mathrm e}^{x} {\mathrm e}^{\frac {100}{\frac {{\mathrm e}^{2 x}}{9}+2 x \,{\mathrm e}^{-\ln \left (3\right )+x}+x^{2}}}}{15 x}\) | \(44\) |
risch | \(x -\frac {{\mathrm e}^{\frac {6 \,{\mathrm e}^{x} x^{2}+9 x^{3}+x \,{\mathrm e}^{2 x}+900}{6 \,{\mathrm e}^{x} x +9 x^{2}+{\mathrm e}^{2 x}}}}{x}\) | \(47\) |
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Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (23) = 46\).
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.71 \[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=\frac {x^{2} - e^{\left (\frac {x^{3} + 2 \, x^{2} e^{\left (x - \log \left (3\right )\right )} + x e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + 100}{x^{2} + 2 \, x e^{\left (x - \log \left (3\right )\right )} + e^{\left (2 \, x - 2 \, \log \left (3\right )\right )}}\right )}}{x} \]
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Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=x - \frac {e^{x} e^{\frac {100}{x^{2} + \frac {2 x e^{x}}{3} + \frac {e^{2 x}}{9}}}}{x} \]
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\[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=\int { \frac {x^{5} + 3 \, x^{4} e^{\left (x - \log \left (3\right )\right )} + 3 \, x^{3} e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + x^{2} e^{\left (3 \, x - 3 \, \log \left (3\right )\right )} - {\left (x^{4} - x^{3} + {\left (x - 1\right )} e^{\left (3 \, x - 3 \, \log \left (3\right )\right )} + 3 \, {\left (x^{2} - x\right )} e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + {\left (3 \, x^{3} - 3 \, x^{2} - 200 \, x\right )} e^{\left (x - \log \left (3\right )\right )} - 200 \, x\right )} e^{\left (x + \frac {100}{x^{2} + 2 \, x e^{\left (x - \log \left (3\right )\right )} + e^{\left (2 \, x - 2 \, \log \left (3\right )\right )}}\right )}}{x^{5} + 3 \, x^{4} e^{\left (x - \log \left (3\right )\right )} + 3 \, x^{3} e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + x^{2} e^{\left (3 \, x - 3 \, \log \left (3\right )\right )}} \,d x } \]
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\[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=\int { \frac {x^{5} + 3 \, x^{4} e^{\left (x - \log \left (3\right )\right )} + 3 \, x^{3} e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + x^{2} e^{\left (3 \, x - 3 \, \log \left (3\right )\right )} - {\left (x^{4} - x^{3} + {\left (x - 1\right )} e^{\left (3 \, x - 3 \, \log \left (3\right )\right )} + 3 \, {\left (x^{2} - x\right )} e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + {\left (3 \, x^{3} - 3 \, x^{2} - 200 \, x\right )} e^{\left (x - \log \left (3\right )\right )} - 200 \, x\right )} e^{\left (x + \frac {100}{x^{2} + 2 \, x e^{\left (x - \log \left (3\right )\right )} + e^{\left (2 \, x - 2 \, \log \left (3\right )\right )}}\right )}}{x^{5} + 3 \, x^{4} e^{\left (x - \log \left (3\right )\right )} + 3 \, x^{3} e^{\left (2 \, x - 2 \, \log \left (3\right )\right )} + x^{2} e^{\left (3 \, x - 3 \, \log \left (3\right )\right )}} \,d x } \]
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Time = 10.73 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5+e^{x+\frac {100}{\frac {e^{2 x}}{9}+\frac {2 e^x x}{3}+x^2}} \left (\frac {1}{27} e^{3 x} (1-x)+200 x+x^3-x^4+\frac {1}{9} e^{2 x} \left (3 x-3 x^2\right )+\frac {1}{3} e^x \left (200 x+3 x^2-3 x^3\right )\right )}{\frac {1}{27} e^{3 x} x^2+\frac {1}{3} e^{2 x} x^3+e^x x^4+x^5} \, dx=x-\frac {{\mathrm {e}}^{\frac {100}{\frac {{\mathrm {e}}^{2\,x}}{9}+\frac {2\,x\,{\mathrm {e}}^x}{3}+x^2}}\,{\mathrm {e}}^x}{x} \]
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