Integrand size = 92, antiderivative size = 32 \[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=x-e^{-e^{\frac {-4-x+x^2+(5-5 x) x^2}{x}}} x \]
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\[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=\int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-e^{-1-\frac {4}{x}+6 x-5 x^2}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx \\ & = \int \left (1-e^{-e^{-1-\frac {4}{x}+6 x-5 x^2}}-\frac {2 \exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right ) \left (-2-3 x^2+5 x^3\right )}{x}\right ) \, dx \\ & = x-2 \int \frac {\exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right ) \left (-2-3 x^2+5 x^3\right )}{x} \, dx-\int e^{-e^{-1-\frac {4}{x}+6 x-5 x^2}} \, dx \\ & = x-2 \int \left (-\frac {2 \exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right )}{x}-3 \exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right ) x+5 \exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right ) x^2\right ) \, dx-\int e^{-e^{-1-\frac {4}{x}+6 x-5 x^2}} \, dx \\ & = x+4 \int \frac {\exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right )}{x} \, dx+6 \int \exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right ) x \, dx-10 \int \exp \left (-1-e^{-1-\frac {4}{x}+6 x-5 x^2}-\frac {4}{x}+6 x-5 x^2\right ) x^2 \, dx-\int e^{-e^{-1-\frac {4}{x}+6 x-5 x^2}} \, dx \\ \end{align*}
\[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=\int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx \]
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Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
risch | \(x -x \,{\mathrm e}^{-{\mathrm e}^{-\frac {5 x^{3}-6 x^{2}+x +4}{x}}}\) | \(28\) |
parallelrisch | \(-\left (-x \,{\mathrm e}^{{\mathrm e}^{-\frac {5 x^{3}-6 x^{2}+x +4}{x}}}+x \right ) {\mathrm e}^{-{\mathrm e}^{-\frac {5 x^{3}-6 x^{2}+x +4}{x}}}\) | \(50\) |
norman | \(\left (x \,{\mathrm e}^{{\mathrm e}^{\frac {-5 x^{3}+6 x^{2}-x -4}{x}}}-x \right ) {\mathrm e}^{-{\mathrm e}^{\frac {-5 x^{3}+6 x^{2}-x -4}{x}}}\) | \(52\) |
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Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx={\left (x e^{\left (e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )}\right )} - x\right )} e^{\left (-e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )}\right )} \]
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Timed out. \[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=\text {Timed out} \]
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\[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=\int { -\frac {{\left (2 \, {\left (5 \, x^{3} - 3 \, x^{2} - 2\right )} e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )} - x e^{\left (e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )}\right )} + x\right )} e^{\left (-e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )}\right )}}{x} \,d x } \]
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\[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=\int { -\frac {{\left (2 \, {\left (5 \, x^{3} - 3 \, x^{2} - 2\right )} e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )} - x e^{\left (e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )}\right )} + x\right )} e^{\left (-e^{\left (-\frac {5 \, x^{3} - 6 \, x^{2} + x + 4}{x}\right )}\right )}}{x} \,d x } \]
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Time = 13.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-e^{\frac {-4-x+6 x^2-5 x^3}{x}}} \left (-x+e^{e^{\frac {-4-x+6 x^2-5 x^3}{x}}} x+e^{\frac {-4-x+6 x^2-5 x^3}{x}} \left (4+6 x^2-10 x^3\right )\right )}{x} \, dx=-x\,\left ({\mathrm {e}}^{-{\mathrm {e}}^{6\,x}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-\frac {4}{x}}\,{\mathrm {e}}^{-5\,x^2}}-1\right ) \]
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