Integrand size = 294, antiderivative size = 31 \[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=x \left (2-e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}}+x\right ) \]
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\[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=\int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+2 e^8 (1+x)+2 e^{2 e^x} x^2 (1+x)-4 e^{e^x} x (1+x) \left (-3-e^4+x+x^2\right )-4 e^4 \left (-3-2 x+2 x^2+x^3\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+2 e^4 \left (-3+x+x^2\right )+e^{e^x} x \left (-5-2 e^4+2 x+e^x x+2 x^2\right )\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx \\ & = \int \left (\frac {18}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {6 x}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}-\frac {22 x^2}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}-\frac {6 x^3}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {6 x^4}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {2 x^5}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {2 e^8 (1+x)}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {2 e^{2 e^x} x^2 (1+x)}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {4 e^4 (1+x) \left (3-x-x^2\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {4 e^{e^x} x (1+x) \left (3+e^4-x-x^2\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}+\frac {e^{\frac {1}{3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2}} \left (-9 \left (1+\frac {1}{9} e^4 \left (6+e^4\right )\right )+5 \left (1+\frac {2 e^4}{5}\right ) x-5 e^{e^x} \left (1+\frac {2 e^4}{5}\right ) x+2 e^{e^x} x^2-e^{2 e^x} x^2+e^{e^x+x} x^2+3 \left (1+\frac {2 e^4}{3}\right ) x^2-2 x^3+2 e^{e^x} x^3-x^4\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2}\right ) \, dx \\ & = 2 \int \frac {x^5}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+2 \int \frac {e^{2 e^x} x^2 (1+x)}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+4 \int \frac {e^{e^x} x (1+x) \left (3+e^4-x-x^2\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+6 \int \frac {x}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx-6 \int \frac {x^3}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+6 \int \frac {x^4}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+18 \int \frac {1}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx-22 \int \frac {x^2}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+\left (4 e^4\right ) \int \frac {(1+x) \left (3-x-x^2\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+\left (2 e^8\right ) \int \frac {1+x}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx+\int \frac {e^{\frac {1}{3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2}} \left (-9 \left (1+\frac {1}{9} e^4 \left (6+e^4\right )\right )+5 \left (1+\frac {2 e^4}{5}\right ) x-5 e^{e^x} \left (1+\frac {2 e^4}{5}\right ) x+2 e^{e^x} x^2-e^{2 e^x} x^2+e^{e^x+x} x^2+3 \left (1+\frac {2 e^4}{3}\right ) x^2-2 x^3+2 e^{e^x} x^3-x^4\right )}{\left (3 \left (1+\frac {e^4}{3}\right )-x+e^{e^x} x-x^2\right )^2} \, dx \\ & = \text {Too large to display} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=x \left (2-e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}}+x\right ) \]
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Time = 95.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x^{2}-{\mathrm e}^{\frac {1}{x \,{\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{4}-x^{2}-x +3}} x +2 x\) | \(31\) |
parallelrisch | \(x^{2}-{\mathrm e}^{\frac {1}{x \,{\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{4}-x^{2}-x +3}} x +2 x\) | \(31\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=x^{2} - x e^{\left (-\frac {1}{x^{2} - x e^{\left (e^{x}\right )} + x - e^{4} - 3}\right )} + 2 \, x \]
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Time = 9.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=x^{2} - x e^{\frac {1}{- x^{2} + x e^{e^{x}} - x + 3 + e^{4}}} + 2 x \]
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\[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=\int { \frac {2 \, x^{5} + 6 \, x^{4} - 6 \, x^{3} - 22 \, x^{2} + 2 \, {\left (x + 1\right )} e^{8} - 4 \, {\left (x^{3} + 2 \, x^{2} - 2 \, x - 3\right )} e^{4} + 2 \, {\left (x^{3} + x^{2}\right )} e^{\left (2 \, e^{x}\right )} - {\left (x^{4} + 2 \, x^{3} + x^{2} e^{\left (2 \, e^{x}\right )} - 3 \, x^{2} - 2 \, {\left (x^{2} + x - 3\right )} e^{4} - {\left (2 \, x^{3} + x^{2} e^{x} + 2 \, x^{2} - 2 \, x e^{4} - 5 \, x\right )} e^{\left (e^{x}\right )} - 5 \, x + e^{8} + 9\right )} e^{\left (-\frac {1}{x^{2} - x e^{\left (e^{x}\right )} + x - e^{4} - 3}\right )} - 4 \, {\left (x^{4} + 2 \, x^{3} - 2 \, x^{2} - {\left (x^{2} + x\right )} e^{4} - 3 \, x\right )} e^{\left (e^{x}\right )} + 6 \, x + 18}{x^{4} + 2 \, x^{3} + x^{2} e^{\left (2 \, e^{x}\right )} - 5 \, x^{2} - 2 \, {\left (x^{2} + x - 3\right )} e^{4} - 2 \, {\left (x^{3} + x^{2} - x e^{4} - 3 \, x\right )} e^{\left (e^{x}\right )} - 6 \, x + e^{8} + 9} \,d x } \]
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Timed out. \[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=\text {Timed out} \]
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Time = 11.59 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {18+6 x-22 x^2-6 x^3+6 x^4+2 x^5+e^8 (2+2 x)+e^4 \left (12+8 x-8 x^2-4 x^3\right )+e^{2 e^x} \left (2 x^2+2 x^3\right )+e^{e^x} \left (12 x+8 x^2-8 x^3-4 x^4+e^4 \left (4 x+4 x^2\right )\right )+e^{\frac {1}{3+e^4-x+e^{e^x} x-x^2}} \left (-9-e^8+5 x+3 x^2-e^{2 e^x} x^2-2 x^3-x^4+e^4 \left (-6+2 x+2 x^2\right )+e^{e^x} \left (-5 x-2 e^4 x+2 x^2+e^x x^2+2 x^3\right )\right )}{9+e^8-6 x-5 x^2+e^{2 e^x} x^2+2 x^3+x^4+e^4 \left (6-2 x-2 x^2\right )+e^{e^x} \left (6 x+2 e^4 x-2 x^2-2 x^3\right )} \, dx=x\,\left (x-{\mathrm {e}}^{\frac {1}{{\mathrm {e}}^4-x+x\,{\mathrm {e}}^{{\mathrm {e}}^x}-x^2+3}}+2\right ) \]
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