Integrand size = 93, antiderivative size = 27 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x+\frac {-\frac {2}{5}-e^{5 x}}{1+e^{-4/x} x} \]
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\[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{8/x} \left (5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )\right )}{5 x \left (e^{4/x}+x\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {e^{8/x} \left (5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )\right )}{x \left (e^{4/x}+x\right )^2} \, dx \\ & = \frac {1}{5} \int \left (-\frac {5 e^{\frac {4}{x}+5 x} \left (-4-x+5 e^{4/x} x+5 x^2\right )}{x \left (e^{4/x}+x\right )^2}+\frac {8 e^{4/x}+2 e^{4/x} x+5 e^{8/x} x+10 e^{4/x} x^2+5 x^3}{x \left (e^{4/x}+x\right )^2}\right ) \, dx \\ & = \frac {1}{5} \int \frac {8 e^{4/x}+2 e^{4/x} x+5 e^{8/x} x+10 e^{4/x} x^2+5 x^3}{x \left (e^{4/x}+x\right )^2} \, dx-\int \frac {e^{\frac {4}{x}+5 x} \left (-4-x+5 e^{4/x} x+5 x^2\right )}{x \left (e^{4/x}+x\right )^2} \, dx \\ & = \frac {1}{5} \int \frac {5 e^{8/x} x+5 x^3+2 e^{4/x} \left (4+x+5 x^2\right )}{x \left (e^{4/x}+x\right )^2} \, dx-\int \left (-\frac {e^{\frac {4}{x}+5 x} (4+x)}{x \left (e^{4/x}+x\right )^2}+\frac {5 e^{\frac {4}{x}+5 x}}{e^{4/x}+x}\right ) \, dx \\ & = \frac {1}{5} \int \left (5-\frac {2 (4+x)}{\left (e^{4/x}+x\right )^2}+\frac {2 (4+x)}{x \left (e^{4/x}+x\right )}\right ) \, dx-5 \int \frac {e^{\frac {4}{x}+5 x}}{e^{4/x}+x} \, dx+\int \frac {e^{\frac {4}{x}+5 x} (4+x)}{x \left (e^{4/x}+x\right )^2} \, dx \\ & = x-\frac {2}{5} \int \frac {4+x}{\left (e^{4/x}+x\right )^2} \, dx+\frac {2}{5} \int \frac {4+x}{x \left (e^{4/x}+x\right )} \, dx-5 \int \frac {e^{\frac {4}{x}+5 x}}{e^{4/x}+x} \, dx+\int \left (\frac {e^{\frac {4}{x}+5 x}}{\left (e^{4/x}+x\right )^2}+\frac {4 e^{\frac {4}{x}+5 x}}{x \left (e^{4/x}+x\right )^2}\right ) \, dx \\ & = x-\frac {2}{5} \int \left (\frac {4}{\left (e^{4/x}+x\right )^2}+\frac {x}{\left (e^{4/x}+x\right )^2}\right ) \, dx+\frac {2}{5} \int \left (\frac {1}{e^{4/x}+x}+\frac {4}{x \left (e^{4/x}+x\right )}\right ) \, dx+4 \int \frac {e^{\frac {4}{x}+5 x}}{x \left (e^{4/x}+x\right )^2} \, dx-5 \int \frac {e^{\frac {4}{x}+5 x}}{e^{4/x}+x} \, dx+\int \frac {e^{\frac {4}{x}+5 x}}{\left (e^{4/x}+x\right )^2} \, dx \\ & = x-\frac {2}{5} \int \frac {x}{\left (e^{4/x}+x\right )^2} \, dx+\frac {2}{5} \int \frac {1}{e^{4/x}+x} \, dx-\frac {8}{5} \int \frac {1}{\left (e^{4/x}+x\right )^2} \, dx+\frac {8}{5} \int \frac {1}{x \left (e^{4/x}+x\right )} \, dx+4 \int \frac {e^{\frac {4}{x}+5 x}}{x \left (e^{4/x}+x\right )^2} \, dx-5 \int \frac {e^{\frac {4}{x}+5 x}}{e^{4/x}+x} \, dx+\int \frac {e^{\frac {4}{x}+5 x}}{\left (e^{4/x}+x\right )^2} \, dx \\ \end{align*}
Timed out. \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\text {\$Aborted} \]
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Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(x -\frac {5 \,{\mathrm e}^{5 x}+2}{5 \left (x \,{\mathrm e}^{-\frac {4}{x}}+1\right )}\) | \(25\) |
norman | \(\frac {x +x^{2} {\mathrm e}^{-\frac {4}{x}}-{\mathrm e}^{5 x}-\frac {2}{5}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}\) | \(33\) |
parallelrisch | \(\frac {5 x^{2} {\mathrm e}^{-\frac {4}{x}}+2 x \,{\mathrm e}^{-\frac {4}{x}}+5 x -5 \,{\mathrm e}^{5 x}}{5 x \,{\mathrm e}^{-\frac {4}{x}}+5}\) | \(45\) |
parts | \(\frac {x +x^{2} {\mathrm e}^{-\frac {4}{x}}-\frac {2}{5}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}-\frac {{\mathrm e}^{5 x}}{x \,{\mathrm e}^{-\frac {4}{x}}+1}\) | \(46\) |
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} e^{\left (-\frac {4}{x}\right )} + 5 \, x - 5 \, e^{\left (5 \, x\right )} - 2}{5 \, {\left (x e^{\left (-\frac {4}{x}\right )} + 1\right )}} \]
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Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x - \frac {2}{5 x e^{- \frac {4}{x}} + 5} - \frac {e^{5 x}}{x e^{- \frac {4}{x}} + 1} \]
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Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} + 5 \, {\left (x - e^{\left (5 \, x\right )}\right )} e^{\frac {4}{x}} + 2 \, x}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=\frac {5 \, x^{2} + 5 \, x e^{\frac {4}{x}} + 2 \, x - 5 \, e^{\left (5 \, x + \frac {4}{x}\right )}}{5 \, {\left (x + e^{\frac {4}{x}}\right )}} \]
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Time = 12.76 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {5 x+5 e^{-8/x} x^3+e^{-4/x} \left (8+2 x+10 x^2\right )+e^{5 x} \left (-25 x+e^{-4/x} \left (20+5 x-25 x^2\right )\right )}{5 x+10 e^{-4/x} x^2+5 e^{-8/x} x^3} \, dx=x-\frac {\frac {2\,{\mathrm {e}}^{4/x}}{5}+{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{4/x}}{x+{\mathrm {e}}^{4/x}} \]
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