Integrand size = 141, antiderivative size = 26 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)} \]
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\[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {4 e^{-\frac {e^4}{x}+x} \left (-e^x x^2-5 x^3-5 e^4 (1+x)+e^{4+x} (1+x)\right )}{\left (5-e^x\right )^2 x^2 (1+x)^2} \, dx \\ & = 4 \int \frac {e^{-\frac {e^4}{x}+x} \left (-e^x x^2-5 x^3-5 e^4 (1+x)+e^{4+x} (1+x)\right )}{\left (5-e^x\right )^2 x^2 (1+x)^2} \, dx \\ & = 4 \int \left (-\frac {5 e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right )^2 (1+x)}-\frac {e^{-\frac {e^4}{x}+x} \left (-e^4-e^4 x+x^2\right )}{\left (-5+e^x\right ) x^2 (1+x)^2}\right ) \, dx \\ & = -\left (4 \int \frac {e^{-\frac {e^4}{x}+x} \left (-e^4-e^4 x+x^2\right )}{\left (-5+e^x\right ) x^2 (1+x)^2} \, dx\right )-20 \int \frac {e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right )^2 (1+x)} \, dx \\ & = -\left (4 \int \left (-\frac {e^{4-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) x^2}+\frac {e^{4-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) x}+\frac {e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)^2}-\frac {e^{4-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)}\right ) \, dx\right )-20 \int \frac {e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right )^2 (1+x)} \, dx \\ & = 4 \int \frac {e^{4-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) x^2} \, dx-4 \int \frac {e^{4-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) x} \, dx-4 \int \frac {e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)^2} \, dx+4 \int \frac {e^{4-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)} \, dx-20 \int \frac {e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right )^2 (1+x)} \, dx \\ \end{align*}
Time = 3.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 e^{-\frac {e^4}{x}+x}}{\left (-5+e^x\right ) (1+x)} \]
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Time = 2.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{-\frac {{\mathrm e}^{4}-x^{2}}{x}}}{\left (1+x \right ) \left ({\mathrm e}^{x}-5\right )}\) | \(28\) |
norman | \(\frac {{\mathrm e}^{x} {\mathrm e}^{-\frac {-2 x \ln \left (2\right )+{\mathrm e}^{4}}{x}}}{\left ({\mathrm e}^{x}-5\right ) \left (1+x \right )}\) | \(30\) |
parallelrisch | \(\frac {{\mathrm e}^{x} {\mathrm e}^{\frac {2 x \ln \left (2\right )-{\mathrm e}^{4}}{x}}}{\left ({\mathrm e}^{x}-5\right ) \left (1+x \right )}\) | \(33\) |
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Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {e^{\left (3 \, x\right )}}{{\left (x + 1\right )} e^{\left (x + \frac {2 \, x^{2} - 2 \, x \log \left (2\right ) + e^{4}}{x}\right )} - 5 \, {\left (x + 1\right )} e^{\left (\frac {2 \, x^{2} - 2 \, x \log \left (2\right ) + e^{4}}{x}\right )}} \]
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Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {e^{x} e^{- \frac {- 2 x \log {\left (2 \right )} + e^{4}}{x}}}{x e^{x} - 5 x + e^{x} - 5} \]
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Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 \, e^{\left (x - \frac {e^{4}}{x}\right )}}{{\left (x + 1\right )} e^{x} - 5 \, x - 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (26) = 52\).
Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 3.85 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=\frac {4 \, e^{\left (\frac {2 \, {\left (2 \, x^{2} + e^{4}\right )}}{x}\right )}}{x e^{\left (\frac {2 \, x^{2} + e^{4}}{x} + \frac {2 \, {\left (x^{2} + e^{4}\right )}}{x}\right )} - 5 \, x e^{\left (\frac {3 \, {\left (x^{2} + e^{4}\right )}}{x}\right )} + e^{\left (\frac {2 \, x^{2} + e^{4}}{x} + \frac {2 \, {\left (x^{2} + e^{4}\right )}}{x}\right )} - 5 \, e^{\left (\frac {3 \, {\left (x^{2} + e^{4}\right )}}{x}\right )}} \]
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Time = 12.75 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (e^4 (-5-5 x)-5 x^3\right )+e^{2 x} \left (-x^2+e^4 (1+x)\right )}{e^{x+\frac {e^4-x \log (4)}{x}} \left (-10 x^2-20 x^3-10 x^4\right )+e^{2 x+\frac {e^4-x \log (4)}{x}} \left (x^2+2 x^3+x^4\right )+e^{\frac {e^4-x \log (4)}{x}} \left (25 x^2+50 x^3+25 x^4\right )} \, dx=-\frac {4\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^4}{x}}\,{\mathrm {e}}^x}{5\,x-{\mathrm {e}}^x-x\,{\mathrm {e}}^x+5} \]
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