Integrand size = 220, antiderivative size = 34 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=4-e^{-x} x \left (-5+\frac {x}{5-e^{\frac {e^4 x}{(-2+2 x)^2}}}\right ) \]
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\[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=\int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (20 e^{\frac {e^4 x}{2 (-1+x)^2}} (-1+x)^4-e^{4+\frac {e^4 x}{4 (-1+x)^2}} x^2 (1+x)+4 e^{\frac {e^4 x}{4 (-1+x)^2}} (-1+x)^3 \left (50-52 x+x^2\right )-20 (-1+x)^3 \left (25-27 x+x^2\right )\right )}{4 \left (5-e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (1-x)^3} \, dx \\ & = \frac {1}{4} \int \frac {e^{-x} \left (20 e^{\frac {e^4 x}{2 (-1+x)^2}} (-1+x)^4-e^{4+\frac {e^4 x}{4 (-1+x)^2}} x^2 (1+x)+4 e^{\frac {e^4 x}{4 (-1+x)^2}} (-1+x)^3 \left (50-52 x+x^2\right )-20 (-1+x)^3 \left (25-27 x+x^2\right )\right )}{\left (5-e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (1-x)^3} \, dx \\ & = \frac {1}{4} \int \left (-20 e^{-x} (-1+x)+\frac {5 e^{4-x} x^2 (1+x)}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)^3}+\frac {e^{-x} x \left (-8+\left (28+e^4\right ) x-\left (36-e^4\right ) x^2+20 x^3-4 x^4\right )}{\left (5-e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (1-x)^3}\right ) \, dx \\ & = \frac {1}{4} \int \frac {e^{-x} x \left (-8+\left (28+e^4\right ) x-\left (36-e^4\right ) x^2+20 x^3-4 x^4\right )}{\left (5-e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (1-x)^3} \, dx+\frac {5}{4} \int \frac {e^{4-x} x^2 (1+x)}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)^3} \, dx-5 \int e^{-x} (-1+x) \, dx \\ & = -5 e^{-x} (1-x)+\frac {1}{4} \int \left (\frac {e^{4-x}}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}}+\frac {2 e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (-1+x)^3}+\frac {5 e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (-1+x)^2}+\frac {4 e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (-1+x)}+\frac {8 e^{-x} x}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}}-\frac {4 e^{-x} x^2}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}}\right ) \, dx+\frac {5}{4} \int \left (\frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2}+\frac {2 e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)^3}+\frac {5 e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)^2}+\frac {4 e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)}\right ) \, dx-5 \int e^{-x} \, dx \\ & = 5 e^{-x}-5 e^{-x} (1-x)+\frac {1}{4} \int \frac {e^{4-x}}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}} \, dx+\frac {1}{2} \int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (-1+x)^3} \, dx+\frac {5}{4} \int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2} \, dx+\frac {5}{4} \int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (-1+x)^2} \, dx+2 \int \frac {e^{-x} x}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}} \, dx+\frac {5}{2} \int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)^3} \, dx+5 \int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)} \, dx+\frac {25}{4} \int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right )^2 (-1+x)^2} \, dx+\int \frac {e^{4-x}}{\left (-5+e^{\frac {e^4 x}{4 (-1+x)^2}}\right ) (-1+x)} \, dx-\int \frac {e^{-x} x^2}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}} \, dx \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=\frac {e^{-x} x \left (-25+5 e^{\frac {e^4 x}{4 (-1+x)^2}}+x\right )}{-5+e^{\frac {e^4 x}{4 (-1+x)^2}}} \]
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Time = 9.88 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94
method | result | size |
risch | \(5 x \,{\mathrm e}^{-x}+\frac {x^{2} {\mathrm e}^{-x}}{{\mathrm e}^{\frac {x \,{\mathrm e}^{4}}{4 \left (-1+x \right )^{2}}}-5}\) | \(32\) |
parallelrisch | \(\frac {\left (4 x^{2}+20 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{4}}{4 x^{2}-8 x +4}} x -100 x \right ) {\mathrm e}^{-x}}{4 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{4}}{4 x^{2}-8 x +4}}-20}\) | \(55\) |
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Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (27) = 54\).
Time = 0.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 5.97 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=\frac {{\left (x^{2} - 25 \, x\right )} e^{\left (\frac {4 \, x^{3} - 8 \, x^{2} + x e^{4} + 4 \, x}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {2 \, x^{3} - 4 \, x^{2} + x e^{4} + 2 \, x}{2 \, {\left (x^{2} - 2 \, x + 1\right )}}\right )} + 5 \, x e^{\left (\frac {2 \, x^{3} - 4 \, x^{2} + x e^{4} + 2 \, x}{x^{2} - 2 \, x + 1}\right )}}{e^{\left (\frac {4 \, x^{3} - 8 \, x^{2} + x e^{4} + 4 \, x}{2 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {2 \, x^{3} - 4 \, x^{2} + x e^{4} + 2 \, x}{2 \, {\left (x^{2} - 2 \, x + 1\right )}}\right )} - 5 \, e^{\left (\frac {3 \, {\left (4 \, x^{3} - 8 \, x^{2} + x e^{4} + 4 \, x\right )}}{4 \, {\left (x^{2} - 2 \, x + 1\right )}}\right )}} \]
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Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=\frac {x^{2}}{e^{x} e^{\frac {x e^{4}}{4 x^{2} - 8 x + 4}} - 5 e^{x}} + 5 x e^{- x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (27) = 54\).
Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.03 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=\frac {x^{2} + 5 \, x e^{\left (\frac {e^{4}}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {e^{4}}{4 \, {\left (x - 1\right )}}\right )} - 25 \, x}{e^{\left (x + \frac {e^{4}}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} + \frac {e^{4}}{4 \, {\left (x - 1\right )}}\right )} - 5 \, e^{x}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1417 vs. \(2 (27) = 54\).
Time = 0.36 (sec) , antiderivative size = 1417, normalized size of antiderivative = 41.68 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=\text {Too large to display} \]
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Time = 14.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {-500+2040 x-3140 x^2+2180 x^3-600 x^4+20 x^5+e^{\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-20+80 x-120 x^2+80 x^3-20 x^4\right )+e^{\frac {e^4 x}{4-8 x+4 x^2}} \left (200-808 x+1228 x^2-836 x^3+220 x^4-4 x^5+e^4 \left (x^2+x^3\right )\right )}{e^{x+\frac {e^4 x}{4-8 x+4 x^2}} \left (40-120 x+120 x^2-40 x^3\right )+e^{x+\frac {2 e^4 x}{4-8 x+4 x^2}} \left (-4+12 x-12 x^2+4 x^3\right )+e^x \left (-100+300 x-300 x^2+100 x^3\right )} \, dx=5\,x\,{\mathrm {e}}^{-x}+\frac {x^2\,{\mathrm {e}}^{-x}}{{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^4}{4\,x^2-8\,x+4}}-5} \]
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