Integrand size = 342, antiderivative size = 33 \[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=-e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}+\log \left (-\frac {9 x}{5}\right ) \]
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\[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=\int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2+2 x^3+x^4+e^8 (1+x)^2+2 e^4 x (1+x)^2+e^{\frac {4}{e^4+x}} \left (e^4+x\right )^2-2 e^{\frac {2}{e^4+x}} (1+x) \left (e^4+x\right )^2+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x^2 \left (2 x^2+x^3+e^8 (2+x)+2 e^4 x (2+x)-2 e^{\frac {2}{e^4+x}} \left (e^8+x+2 e^4 x+x^2\right )\right )}{x \left (e^4+x\right )^2 \left (1-e^{\frac {2}{e^4+x}}+x\right )^2} \, dx \\ & = \int \left (\frac {1}{x}+\frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x^2 \left (-2-e^8-2 \left (1+e^4\right ) x-x^2\right )}{\left (e^4+x\right )^2 \left (1-e^{\frac {2}{e^4+x}}+x\right )^2}+\frac {2 e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x \left (e^8+\left (1+2 e^4\right ) x+x^2\right )}{\left (e^4+x\right )^2 \left (1-e^{\frac {2}{e^4+x}}+x\right )}\right ) \, dx \\ & = \log (x)+2 \int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x \left (e^8+\left (1+2 e^4\right ) x+x^2\right )}{\left (e^4+x\right )^2 \left (1-e^{\frac {2}{e^4+x}}+x\right )} \, dx+\int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x^2 \left (-2-e^8-2 \left (1+e^4\right ) x-x^2\right )}{\left (e^4+x\right )^2 \left (1-e^{\frac {2}{e^4+x}}+x\right )^2} \, dx \\ & = \log (x)+2 \int \left (-\frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{-1+e^{\frac {2}{e^4+x}}-x}-\frac {e^{8+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right ) \left (e^4+x\right )^2}+\frac {2 e^{4+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right ) \left (e^4+x\right )}+\frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x}{1-e^{\frac {2}{e^4+x}}+x}\right ) \, dx+\int \left (\frac {2 e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (-1+2 e^4\right )}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2}-\frac {2 e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2}-\frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x^2}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2}+\frac {2 e^{8+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (-1+e^4\right )}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2 \left (e^4+x\right )^2}-\frac {2 e^{4+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (-2+3 e^4\right )}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2 \left (e^4+x\right )}\right ) \, dx \\ & = \log (x)-2 \int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{-1+e^{\frac {2}{e^4+x}}-x} \, dx-2 \int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2} \, dx-2 \int \frac {e^{8+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right ) \left (e^4+x\right )^2} \, dx+2 \int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x}{1-e^{\frac {2}{e^4+x}}+x} \, dx+4 \int \frac {e^{4+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right ) \left (e^4+x\right )} \, dx+\left (2 \left (2-3 e^4\right )\right ) \int \frac {e^{4+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2 \left (e^4+x\right )} \, dx-\left (2 \left (1-2 e^4\right )\right ) \int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2} \, dx-\left (2 \left (1-e^4\right )\right ) \int \frac {e^{8+\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2 \left (e^4+x\right )^2} \, dx-\int \frac {e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} x^2}{\left (-1+e^{\frac {2}{e^4+x}}-x\right )^2} \, dx \\ \end{align*}
Time = 0.85 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=-e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}}+\log (x) \]
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Time = 19.69 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \(\ln \left (x \right )-{\mathrm e}^{\frac {x^{2}}{{\mathrm e}^{\frac {2}{x +{\mathrm e}^{4}}}-x -1}}\) | \(27\) |
risch | \(\ln \left (x \right )-{\mathrm e}^{-\frac {x^{2}}{-{\mathrm e}^{\frac {2}{x +{\mathrm e}^{4}}}+x +1}}\) | \(28\) |
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Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=-e^{\left (-\frac {x^{2}}{x - e^{\left (\frac {2}{x + e^{4}}\right )} + 1}\right )} + \log \left (x\right ) \]
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Time = 0.52 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.58 \[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=- e^{\frac {x^{2}}{- x + e^{\frac {2}{x + e^{4}}} - 1}} + \log {\left (x \right )} \]
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\[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=\int { \frac {x^{4} + 2 \, x^{3} + x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{8} + 2 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{4} + {\left (x^{5} + 2 \, x^{4} + {\left (x^{3} + 2 \, x^{2}\right )} e^{8} + 2 \, {\left (x^{4} + 2 \, x^{3}\right )} e^{4} - 2 \, {\left (x^{4} + 2 \, x^{3} e^{4} + x^{3} + x^{2} e^{8}\right )} e^{\left (\frac {2}{x + e^{4}}\right )}\right )} e^{\left (-\frac {x^{2}}{x - e^{\left (\frac {2}{x + e^{4}}\right )} + 1}\right )} + {\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} e^{\left (\frac {4}{x + e^{4}}\right )} - 2 \, {\left (x^{3} + x^{2} + {\left (x + 1\right )} e^{8} + 2 \, {\left (x^{2} + x\right )} e^{4}\right )} e^{\left (\frac {2}{x + e^{4}}\right )}}{x^{5} + 2 \, x^{4} + x^{3} + {\left (x^{3} + 2 \, x^{2} + x\right )} e^{8} + 2 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{4} + {\left (x^{3} + 2 \, x^{2} e^{4} + x e^{8}\right )} e^{\left (\frac {4}{x + e^{4}}\right )} - 2 \, {\left (x^{4} + x^{3} + {\left (x^{2} + x\right )} e^{8} + 2 \, {\left (x^{3} + x^{2}\right )} e^{4}\right )} e^{\left (\frac {2}{x + e^{4}}\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=\text {Timed out} \]
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Time = 13.76 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {x^2+2 x^3+x^4+e^8 \left (1+2 x+x^2\right )+e^{\frac {4}{e^4+x}} \left (e^8+2 e^4 x+x^2\right )+e^4 \left (2 x+4 x^2+2 x^3\right )+e^{\frac {2}{e^4+x}} \left (e^8 (-2-2 x)-2 x^2-2 x^3+e^4 \left (-4 x-4 x^2\right )\right )+e^{\frac {x^2}{-1+e^{\frac {2}{e^4+x}}-x}} \left (2 x^4+x^5+e^8 \left (2 x^2+x^3\right )+e^{\frac {2}{e^4+x}} \left (-2 e^8 x^2-2 x^3-4 e^4 x^3-2 x^4\right )+e^4 \left (4 x^3+2 x^4\right )\right )}{x^3+2 x^4+x^5+e^8 \left (x+2 x^2+x^3\right )+e^{\frac {4}{e^4+x}} \left (e^8 x+2 e^4 x^2+x^3\right )+e^4 \left (2 x^2+4 x^3+2 x^4\right )+e^{\frac {2}{e^4+x}} \left (-2 x^3-2 x^4+e^8 \left (-2 x-2 x^2\right )+e^4 \left (-4 x^2-4 x^3\right )\right )} \, dx=\ln \left (x\right )-{\mathrm {e}}^{-\frac {x^2}{x-{\mathrm {e}}^{\frac {2}{x+{\mathrm {e}}^4}}+1}} \]
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