Integrand size = 85, antiderivative size = 21 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=x-\frac {5}{e^{-8 x}+\frac {x^2}{5+x}} \]
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\[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {(5+x)^2+e^{16 x} x \left (50+5 x+x^3\right )+2 e^{8 x} \left (-500-200 x-15 x^2+x^3\right )}{\left (5+x+e^{8 x} x^2\right )^2} \, dx \\ & = \int \left (\frac {5 (5+x)^2 \left (10+41 x+8 x^2\right )}{x^3 \left (5+x+e^{8 x} x^2\right )^2}+\frac {50+5 x+x^3}{x^3}-\frac {10 \left (50+115 x+41 x^2+4 x^3\right )}{x^3 \left (5+x+e^{8 x} x^2\right )}\right ) \, dx \\ & = 5 \int \frac {(5+x)^2 \left (10+41 x+8 x^2\right )}{x^3 \left (5+x+e^{8 x} x^2\right )^2} \, dx-10 \int \frac {50+115 x+41 x^2+4 x^3}{x^3 \left (5+x+e^{8 x} x^2\right )} \, dx+\int \frac {50+5 x+x^3}{x^3} \, dx \\ & = 5 \int \left (\frac {121}{\left (5+x+e^{8 x} x^2\right )^2}+\frac {250}{x^3 \left (5+x+e^{8 x} x^2\right )^2}+\frac {1125}{x^2 \left (5+x+e^{8 x} x^2\right )^2}+\frac {620}{x \left (5+x+e^{8 x} x^2\right )^2}+\frac {8 x}{\left (5+x+e^{8 x} x^2\right )^2}\right ) \, dx-10 \int \left (\frac {4}{5+x+e^{8 x} x^2}+\frac {50}{x^3 \left (5+x+e^{8 x} x^2\right )}+\frac {115}{x^2 \left (5+x+e^{8 x} x^2\right )}+\frac {41}{x \left (5+x+e^{8 x} x^2\right )}\right ) \, dx+\int \left (1+\frac {50}{x^3}+\frac {5}{x^2}\right ) \, dx \\ & = -\frac {25}{x^2}-\frac {5}{x}+x+40 \int \frac {x}{\left (5+x+e^{8 x} x^2\right )^2} \, dx-40 \int \frac {1}{5+x+e^{8 x} x^2} \, dx-410 \int \frac {1}{x \left (5+x+e^{8 x} x^2\right )} \, dx-500 \int \frac {1}{x^3 \left (5+x+e^{8 x} x^2\right )} \, dx+605 \int \frac {1}{\left (5+x+e^{8 x} x^2\right )^2} \, dx-1150 \int \frac {1}{x^2 \left (5+x+e^{8 x} x^2\right )} \, dx+1250 \int \frac {1}{x^3 \left (5+x+e^{8 x} x^2\right )^2} \, dx+3100 \int \frac {1}{x \left (5+x+e^{8 x} x^2\right )^2} \, dx+5625 \int \frac {1}{x^2 \left (5+x+e^{8 x} x^2\right )^2} \, dx \\ \end{align*}
Time = 4.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x (5+x)+e^{8 x} \left (-25-5 x+x^3\right )}{5+x+e^{8 x} x^2} \]
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Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
risch | \(x -\frac {5 \left (5+x \right )}{{\mathrm e}^{-8 x} x +x^{2}+5 \,{\mathrm e}^{-8 x}}\) | \(26\) |
norman | \(\frac {-25+x^{3}+{\mathrm e}^{-8 x} x^{2}+5 \,{\mathrm e}^{-8 x} x -5 x}{{\mathrm e}^{-8 x} x +x^{2}+5 \,{\mathrm e}^{-8 x}}\) | \(43\) |
parallelrisch | \(\frac {x^{3}+{\mathrm e}^{-8 x} x^{2}-25-5 x^{2}-5 x -25 \,{\mathrm e}^{-8 x}}{{\mathrm e}^{-8 x} x +x^{2}+5 \,{\mathrm e}^{-8 x}}\) | \(47\) |
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x^{3} + {\left (x^{2} + 5 \, x\right )} e^{\left (-8 \, x\right )} - 5 \, x - 25}{x^{2} + {\left (x + 5\right )} e^{\left (-8 \, x\right )}} \]
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Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=x + \frac {- 5 x - 25}{x^{2} + \left (x + 5\right ) e^{- 8 x}} \]
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Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x^{2} + {\left (x^{3} - 5 \, x - 25\right )} e^{\left (8 \, x\right )} + 5 \, x}{x^{2} e^{\left (8 \, x\right )} + x + 5} \]
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Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=\frac {x^{3} + x^{2} e^{\left (-8 \, x\right )} + 5 \, x e^{\left (-8 \, x\right )} - 5 \, x - 25}{x^{2} + x e^{\left (-8 \, x\right )} + 5 \, e^{\left (-8 \, x\right )}} \]
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Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {50 x+5 x^2+x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (-1000-400 x-30 x^2+2 x^3\right )}{x^4+e^{-16 x} \left (25+10 x+x^2\right )+e^{-8 x} \left (10 x^2+2 x^3\right )} \, dx=x-\frac {25\,{\mathrm {e}}^{8\,x}+5\,x\,{\mathrm {e}}^{8\,x}}{x+x^2\,{\mathrm {e}}^{8\,x}+5} \]
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