Integrand size = 98, antiderivative size = 32 \[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=x^2+\log \left (e^{-e^{-e^{e^x}+\frac {e^{5+x}}{x+x^2}}}\right ) \]
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\[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=\int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2 \left (1+2 x+x^2\right )} \, dx \\ & = \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2 (1+x)^2} \, dx \\ & = \int \left (2 x+\frac {e^{-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}} \left (e^5+e^5 x-e^5 x^2+e^{e^x} x^2+2 e^{e^x} x^3+e^{e^x} x^4\right )}{x^2 (1+x)^2}\right ) \, dx \\ & = x^2+\int \frac {e^{-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}} \left (e^5+e^5 x-e^5 x^2+e^{e^x} x^2+2 e^{e^x} x^3+e^{e^x} x^4\right )}{x^2 (1+x)^2} \, dx \\ & = x^2+\int \frac {e^{-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}} \left (e^{e^x} x^2 (1+x)^2+e^5 \left (1+x-x^2\right )\right )}{x^2 (1+x)^2} \, dx \\ & = x^2+\int \left (e^{-e^{e^x}+e^x+x+\frac {e^{5+x}}{x+x^2}}-\frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}} \left (-1-x+x^2\right )}{x^2 (1+x)^2}\right ) \, dx \\ & = x^2+\int e^{-e^{e^x}+e^x+x+\frac {e^{5+x}}{x+x^2}} \, dx-\int \frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}} \left (-1-x+x^2\right )}{x^2 (1+x)^2} \, dx \\ & = x^2+\int e^{-e^{e^x}+e^x+x+\frac {e^{5+x}}{x+x^2}} \, dx-\int \left (\frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{-1-x}-\frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{x^2}+\frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{x}+\frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{(1+x)^2}\right ) \, dx \\ & = x^2+\int e^{-e^{e^x}+e^x+x+\frac {e^{5+x}}{x+x^2}} \, dx-\int \frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{-1-x} \, dx+\int \frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{x^2} \, dx-\int \frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{x} \, dx-\int \frac {e^{5-e^{e^x}+x+\frac {e^{5+x}}{x+x^2}}}{(1+x)^2} \, dx \\ \end{align*}
Time = 5.36 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=-e^{-e^{e^x}+\frac {e^{5+x}}{x (1+x)}}+x^2 \]
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Time = 95.69 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16
method | result | size |
risch | \(x^{2}-{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}} x^{2}+x \,{\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{5+x}}{\left (1+x \right ) x}}\) | \(37\) |
parallelrisch | \(x^{2}-{\mathrm e}^{\frac {\left (-x^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{5} {\mathrm e}^{x}}{\left (1+x \right ) x}}-6\) | \(37\) |
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Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=x^{2} - e^{\left (-\frac {{\left ({\left (x^{2} + x\right )} e^{\left ({\left (x e^{5} + e^{\left (x + 5\right )}\right )} e^{\left (-5\right )} + 5\right )} - e^{\left (2 \, x + 10\right )}\right )} e^{\left (-x - 5\right )}}{x^{2} + x}\right )} \]
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Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=x^{2} - e^{\frac {\left (- x^{2} - x\right ) e^{e^{x}} + e^{5} e^{x}}{x^{2} + x}} \]
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Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=x^{2} - e^{\left (-\frac {e^{\left (x + 5\right )}}{x + 1} + \frac {e^{\left (x + 5\right )}}{x} - e^{\left (e^{x}\right )}\right )} \]
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\[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=\int { \frac {2 \, x^{5} + 4 \, x^{4} + 2 \, x^{3} + {\left ({\left (x^{4} + 2 \, x^{3} + x^{2}\right )} e^{\left (x + e^{x}\right )} - {\left (x^{2} - x - 1\right )} e^{\left (x + 5\right )}\right )} e^{\left (-\frac {{\left (x^{2} + x\right )} e^{\left (e^{x}\right )} - e^{\left (x + 5\right )}}{x^{2} + x}\right )}}{x^{4} + 2 \, x^{3} + x^{2}} \,d x } \]
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Time = 13.96 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {2 x^3+4 x^4+2 x^5+e^{\frac {e^{5+x}+e^{e^x} \left (-x-x^2\right )}{x+x^2}} \left (e^{5+x} \left (1+x-x^2\right )+e^{e^x+x} \left (x^2+2 x^3+x^4\right )\right )}{x^2+2 x^3+x^4} \, dx=x^2-{\mathrm {e}}^{\frac {{\mathrm {e}}^5\,{\mathrm {e}}^x}{x^2+x}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{x^2+x}}\,{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{x^2+x}} \]
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