Integrand size = 103, antiderivative size = 31 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\frac {5+e^{\frac {5 \left (3-x+x^2\right )}{2+x}}+x-x^2}{x}} \]
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\[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=\int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{x^2 \left (4+4 x+x^2\right )} \, dx \\ & = \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{x^2 (2+x)^2} \, dx \\ & = \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{x^2 (2+x)^2} \, dx \\ & = \int \left (-\frac {9 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2}-\frac {20 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x^2 (2+x)^2}-\frac {20 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x (2+x)^2}-\frac {4 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x}{(2+x)^2}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x^2}{(2+x)^2}+\frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right ) \left (-4-29 x+19 x^2+5 x^3\right )}{x^2 (2+x)^2}\right ) \, dx \\ & = -\left (4 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x}{(2+x)^2} \, dx\right )-9 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-20 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x^2 (2+x)^2} \, dx-20 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x (2+x)^2} \, dx-\int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} x^2}{(2+x)^2} \, dx+\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right ) \left (-4-29 x+19 x^2+5 x^3\right )}{x^2 (2+x)^2} \, dx \\ & = -\left (4 \int \left (-\frac {2 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{2+x}\right ) \, dx\right )-9 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-20 \int \left (\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 x}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{2 (2+x)^2}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 (2+x)}\right ) \, dx-20 \int \left (\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 x^2}-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 x}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 (2+x)^2}+\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{4 (2+x)}\right ) \, dx-\int \left (e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}+\frac {4 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2}-\frac {4 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{2+x}\right ) \, dx+\int \left (-\frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{x^2}-\frac {25 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{4 x}+\frac {45 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{2 (2+x)^2}+\frac {45 e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}}}{4 (2+x)}\right ) \, dx \\ & = -\left (4 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx\right )-5 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{x^2} \, dx-5 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-\frac {25}{4} \int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{x} \, dx+8 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx-9 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx+10 \int \frac {e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x}}{(2+x)^2} \, dx+\frac {45}{4} \int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{2+x} \, dx+\frac {45}{2} \int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{(2+x)^2} \, dx-\int e^{1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x} \, dx-\int \frac {\exp \left (1+\frac {5}{x}+\frac {e^{\frac {15-5 x+5 x^2}{2+x}}}{x}-x+\frac {5 \left (3-x+x^2\right )}{2+x}\right )}{x^2} \, dx \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{1+\frac {5}{x}+\frac {e^{-15+5 x+\frac {45}{2+x}}}{x}-x} \]
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Time = 0.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}-x^{2}+x +5}{x}}\) | \(30\) |
risch | \({\mathrm e}^{-\frac {-x -{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}+x^{2}-5}{x}}\) | \(33\) |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}-x^{2}+x +5}{x}}+2 x \,{\mathrm e}^{\frac {{\mathrm e}^{\frac {5 x^{2}-5 x +15}{2+x}}-x^{2}+x +5}{x}}}{x \left (2+x \right )}\) | \(78\) |
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\left (-\frac {x^{2} - x - e^{\left (\frac {5 \, {\left (x^{2} - x + 3\right )}}{x + 2}\right )} - 5}{x}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\frac {- x^{2} + x + e^{\frac {5 x^{2} - 5 x + 15}{x + 2}} + 5}{x}} \]
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Time = 0.48 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\left (-x + \frac {e^{\left (5 \, x + \frac {45}{x + 2} - 15\right )}}{x} + \frac {5}{x} + 1\right )} \]
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Time = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx=e^{\left (-x + \frac {e^{\left (\frac {5 \, x^{2}}{x + 2} - \frac {5 \, x}{x + 2} + \frac {15}{x + 2}\right )}}{x} + \frac {5}{x} + 1\right )} \]
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Time = 7.83 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {5+e^{\frac {15-5 x+5 x^2}{2+x}}+x-x^2}{x}} \left (-20-20 x-9 x^2-4 x^3-x^4+e^{\frac {15-5 x+5 x^2}{2+x}} \left (-4-29 x+19 x^2+5 x^3\right )\right )}{4 x^2+4 x^3+x^4} \, dx={\mathrm {e}}^{-x}\,\mathrm {e}\,{\mathrm {e}}^{5/x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {5\,x}{x+2}}\,{\mathrm {e}}^{\frac {5\,x^2}{x+2}}\,{\mathrm {e}}^{\frac {15}{x+2}}}{x}} \]
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