Integrand size = 210, antiderivative size = 26 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x-\frac {2 e^3}{3-5 e^{-3+\frac {1}{x}}-e^x+x} \]
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\[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+\left (9+2 e^3\right ) x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx \\ & = \int \frac {10 e^{6+\frac {1}{x}}+2 e^9 x^2+25 e^{2/x} x^2-2 e^{9+x} x^2+10 e^{3+\frac {1}{x}+x} x^2+e^{6+2 x} x^2-10 e^{3+\frac {1}{x}} x^2 (3+x)-2 e^{6+x} x^2 (3+x)+e^6 x^2 (3+x)^2}{x^2 \left (5 e^{\frac {1}{x}}+e^{3+x}-e^3 (3+x)\right )^2} \, dx \\ & = \int \left (1-\frac {2 e^6}{-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x}-\frac {2 e^6 \left (-5 e^{\frac {1}{x}}+2 e^3 x^2-5 e^{\frac {1}{x}} x^2+e^3 x^3\right )}{x^2 \left (3 e^3-5 e^{\frac {1}{x}}-e^{3+x}+e^3 x\right )^2}\right ) \, dx \\ & = x-\left (2 e^6\right ) \int \frac {1}{-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x} \, dx-\left (2 e^6\right ) \int \frac {-5 e^{\frac {1}{x}}+2 e^3 x^2-5 e^{\frac {1}{x}} x^2+e^3 x^3}{x^2 \left (3 e^3-5 e^{\frac {1}{x}}-e^{3+x}+e^3 x\right )^2} \, dx \\ & = x-\left (2 e^6\right ) \int \frac {1}{-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x} \, dx-\left (2 e^6\right ) \int \left (\frac {2 e^3}{\left (-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x\right )^2}-\frac {5 e^{\frac {1}{x}}}{\left (-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x\right )^2}-\frac {5 e^{\frac {1}{x}}}{x^2 \left (-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x\right )^2}+\frac {e^3 x}{\left (3 e^3-5 e^{\frac {1}{x}}-e^{3+x}+e^3 x\right )^2}\right ) \, dx \\ & = x-\left (2 e^6\right ) \int \frac {1}{-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x} \, dx+\left (10 e^6\right ) \int \frac {e^{\frac {1}{x}}}{\left (-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x\right )^2} \, dx+\left (10 e^6\right ) \int \frac {e^{\frac {1}{x}}}{x^2 \left (-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x\right )^2} \, dx-\left (2 e^9\right ) \int \frac {x}{\left (3 e^3-5 e^{\frac {1}{x}}-e^{3+x}+e^3 x\right )^2} \, dx-\left (4 e^9\right ) \int \frac {1}{\left (-3 e^3+5 e^{\frac {1}{x}}+e^{3+x}-e^3 x\right )^2} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x+\frac {2 e^6}{5 e^{\frac {1}{x}}+e^{3+x}-e^3 (3+x)} \]
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Time = 0.69 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x -\frac {2 \,{\mathrm e}^{3}}{x -{\mathrm e}^{x}-5 \,{\mathrm e}^{-\frac {-1+3 x}{x}}+3}\) | \(29\) |
parallelrisch | \(-\frac {-x^{2}+{\mathrm e}^{x} x +x \,{\mathrm e}^{\frac {x \ln \left (5\right )-3 x +1}{x}}+2 \,{\mathrm e}^{3}-3 x}{x -{\mathrm e}^{x}-{\mathrm e}^{\frac {x \ln \left (5\right )-3 x +1}{x}}+3}\) | \(61\) |
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Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.27 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {x^{2} - x e^{x} - x e^{\left (\frac {x \log \left (5\right ) - 3 \, x + 1}{x}\right )} + 3 \, x - 2 \, e^{3}}{x - e^{x} - e^{\left (\frac {x \log \left (5\right ) - 3 \, x + 1}{x}\right )} + 3} \]
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Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x + \frac {2 e^{3}}{- x + e^{x} + e^{\frac {- 3 x + x \log {\left (5 \right )} + 1}{x}} - 3} \]
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Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (25) = 50\).
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\frac {x^{2} e^{3} + 3 \, x e^{3} - x e^{\left (x + 3\right )} - 5 \, x e^{\frac {1}{x}} - 2 \, e^{6}}{x e^{3} + 3 \, e^{3} - e^{\left (x + 3\right )} - 5 \, e^{\frac {1}{x}}} \]
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Timed out. \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=\text {Timed out} \]
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Time = 8.52 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {9 x^2+2 e^3 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 e^3 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (2 e^3-6 x^2+2 e^x x^2-2 x^3\right )}{9 x^2+e^{2 x} x^2+e^{\frac {2 (1-3 x+x \log (5))}{x}} x^2+6 x^3+x^4+e^x \left (-6 x^2-2 x^3\right )+e^{\frac {1-3 x+x \log (5)}{x}} \left (-6 x^2+2 e^x x^2-2 x^3\right )} \, dx=x-\frac {2\,{\mathrm {e}}^3}{x-{\mathrm {e}}^x-5\,{\mathrm {e}}^{1/x}\,{\mathrm {e}}^{-3}+3} \]
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