Integrand size = 159, antiderivative size = 32 \[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=e^{e^{3+\frac {1}{-5+\frac {1}{e^x+x}}}-e^2 \left (1+\frac {x}{2}\right )^2} \]
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\[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=\int \frac {\exp \left (\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2 \left (1-5 e^x-5 x\right )^2} \, dx \\ & = \frac {1}{2} \int \frac {\exp \left (\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{\left (1-5 e^x-5 x\right )^2} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 \exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (1+e^x\right )}{\left (1-5 e^x-5 x\right )^2}-\exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (2+x)\right ) \, dx \\ & = -\left (\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (2+x) \, dx\right )+\int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \left (1+e^x\right )}{\left (1-5 e^x-5 x\right )^2} \, dx \\ & = -\left (\frac {1}{2} \int \left (2 \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )+\exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x\right ) \, dx\right )+\int \left (-\frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (-6+5 x)}{5 \left (-1+5 e^x+5 x\right )^2}+\frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{5 \left (-1+5 e^x+5 x\right )}\right ) \, dx \\ & = -\left (\frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) (-6+5 x)}{\left (-1+5 e^x+5 x\right )^2} \, dx\right )+\frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{-1+5 e^x+5 x} \, dx-\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x \, dx-\int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \, dx \\ & = \frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{-1+5 e^x+5 x} \, dx-\frac {1}{5} \int \left (-\frac {6 \exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{\left (-1+5 e^x+5 x\right )^2}+\frac {5 \exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x}{\left (-1+5 e^x+5 x\right )^2}\right ) \, dx-\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x \, dx-\int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \, dx \\ & = \frac {1}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{-1+5 e^x+5 x} \, dx-\frac {1}{2} \int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x \, dx+\frac {6}{5} \int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right )}{\left (-1+5 e^x+5 x\right )^2} \, dx-\int \exp \left (2+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) \, dx-\int \frac {\exp \left (\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}+\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )\right ) x}{\left (-1+5 e^x+5 x\right )^2} \, dx \\ \end{align*}
Time = 0.39 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=e^{\frac {1}{4} e^2 \left (4 e^{\frac {4}{5}+\frac {1}{5-25 e^x-25 x}}-(2+x)^2\right )} \]
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Time = 3.56 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.28
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {14 \,{\mathrm e}^{x}+14 x -3}{5 \,{\mathrm e}^{x}+5 x -1}}-\frac {x^{2} {\mathrm e}^{2}}{4}-{\mathrm e}^{2} x -{\mathrm e}^{2}}\) | \(41\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {14 \,{\mathrm e}^{x}+14 x -3}{5 \,{\mathrm e}^{x}+5 x -1}}+\frac {\left (-x^{2}-4 x -4\right ) {\mathrm e}^{2}}{4}}\) | \(41\) |
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Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=e^{\left (-\frac {1}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{2} + e^{\left (\frac {{\left (14 \, x - 3\right )} e^{2} + 14 \, e^{\left (x + 2\right )}}{{\left (5 \, x - 1\right )} e^{2} + 5 \, e^{\left (x + 2\right )}}\right )}\right )} \]
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Time = 1.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=e^{\left (- \frac {x^{2}}{4} - x - 1\right ) e^{2} + e^{\frac {14 x + 14 e^{x} - 3}{5 x + 5 e^{x} - 1}}} \]
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\[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=\int { -\frac {{\left ({\left (25 \, x^{3} + 40 \, x^{2} - 19 \, x + 2\right )} e^{2} + 25 \, {\left (x + 2\right )} e^{\left (2 \, x + 2\right )} + 10 \, {\left (5 \, x^{2} + 9 \, x - 2\right )} e^{\left (x + 2\right )} - 2 \, {\left (e^{x} + 1\right )} e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{2} + e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )}}{2 \, {\left (25 \, x^{2} + 10 \, {\left (5 \, x - 1\right )} e^{x} - 10 \, x + 25 \, e^{\left (2 \, x\right )} + 1\right )}} \,d x } \]
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\[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx=\int { -\frac {{\left ({\left (25 \, x^{3} + 40 \, x^{2} - 19 \, x + 2\right )} e^{2} + 25 \, {\left (x + 2\right )} e^{\left (2 \, x + 2\right )} + 10 \, {\left (5 \, x^{2} + 9 \, x - 2\right )} e^{\left (x + 2\right )} - 2 \, {\left (e^{x} + 1\right )} e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} + 4 \, x + 4\right )} e^{2} + e^{\left (\frac {14 \, x + 14 \, e^{x} - 3}{5 \, x + 5 \, e^{x} - 1}\right )}\right )}}{2 \, {\left (25 \, x^{2} + 10 \, {\left (5 \, x - 1\right )} e^{x} - 10 \, x + 25 \, e^{\left (2 \, x\right )} + 1\right )}} \,d x } \]
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Time = 14.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.09 \[ \int \frac {e^{\frac {1}{4} \left (4 e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}}+e^2 \left (-4-4 x-x^2\right )\right )} \left (e^{\frac {-3+14 e^x+14 x}{-1+5 e^x+5 x}} \left (2+2 e^x\right )+e^{2+2 x} (-50-25 x)+e^{2+x} \left (20-90 x-50 x^2\right )+e^2 \left (-2+19 x-40 x^2-25 x^3\right )\right )}{2+50 e^{2 x}-20 x+50 x^2+e^x (-20+100 x)} \, dx={\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^2}{4}}\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{-x\,{\mathrm {e}}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {14\,x}{5\,x+5\,{\mathrm {e}}^x-1}}\,{\mathrm {e}}^{\frac {14\,{\mathrm {e}}^x}{5\,x+5\,{\mathrm {e}}^x-1}}\,{\mathrm {e}}^{-\frac {3}{5\,x+5\,{\mathrm {e}}^x-1}}} \]
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