Integrand size = 61, antiderivative size = 25 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=\frac {5 x \left (x^2-2 x (5+x)\right )}{16 \left (10+4 e^x+x\right )} \]
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\[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=\int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {5 x \left (-(10+x)^2+2 e^x \left (-20+7 x+x^2\right )\right )}{8 \left (10+4 e^x+x\right )^2} \, dx \\ & = \frac {5}{8} \int \frac {x \left (-(10+x)^2+2 e^x \left (-20+7 x+x^2\right )\right )}{\left (10+4 e^x+x\right )^2} \, dx \\ & = \frac {5}{8} \int \left (\frac {x \left (-20+7 x+x^2\right )}{2 \left (10+4 e^x+x\right )}-\frac {x^2 \left (90+19 x+x^2\right )}{2 \left (10+4 e^x+x\right )^2}\right ) \, dx \\ & = \frac {5}{16} \int \frac {x \left (-20+7 x+x^2\right )}{10+4 e^x+x} \, dx-\frac {5}{16} \int \frac {x^2 \left (90+19 x+x^2\right )}{\left (10+4 e^x+x\right )^2} \, dx \\ & = -\left (\frac {5}{16} \int \left (\frac {90 x^2}{\left (10+4 e^x+x\right )^2}+\frac {19 x^3}{\left (10+4 e^x+x\right )^2}+\frac {x^4}{\left (10+4 e^x+x\right )^2}\right ) \, dx\right )+\frac {5}{16} \int \left (-\frac {20 x}{10+4 e^x+x}+\frac {7 x^2}{10+4 e^x+x}+\frac {x^3}{10+4 e^x+x}\right ) \, dx \\ & = -\left (\frac {5}{16} \int \frac {x^4}{\left (10+4 e^x+x\right )^2} \, dx\right )+\frac {5}{16} \int \frac {x^3}{10+4 e^x+x} \, dx+\frac {35}{16} \int \frac {x^2}{10+4 e^x+x} \, dx-\frac {95}{16} \int \frac {x^3}{\left (10+4 e^x+x\right )^2} \, dx-\frac {25}{4} \int \frac {x}{10+4 e^x+x} \, dx-\frac {225}{8} \int \frac {x^2}{\left (10+4 e^x+x\right )^2} \, dx \\ \end{align*}
Time = 0.66 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=-\frac {5 x^2 (10+x)}{16 \left (10+4 e^x+x\right )} \]
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Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72
method | result | size |
risch | \(-\frac {5 \left (x +10\right ) x^{2}}{16 \left (4 \,{\mathrm e}^{x}+10+x \right )}\) | \(18\) |
norman | \(\frac {-\frac {25}{8} x^{2}-\frac {5}{16} x^{3}}{4 \,{\mathrm e}^{x}+10+x}\) | \(22\) |
parallelrisch | \(-\frac {10 x^{3}+100 x^{2}}{32 \left (4 \,{\mathrm e}^{x}+10+x \right )}\) | \(23\) |
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Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=-\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=\frac {- 5 x^{3} - 50 x^{2}}{16 x + 64 e^{x} + 160} \]
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Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=-\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=-\frac {5 \, {\left (x^{3} + 10 \, x^{2}\right )}}{16 \, {\left (x + 4 \, e^{x} + 10\right )}} \]
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Time = 15.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-500 x-100 x^2-5 x^3+e^x \left (-200 x+70 x^2+10 x^3\right )}{800+128 e^{2 x}+160 x+8 x^2+e^x (640+64 x)} \, dx=-\frac {\frac {5\,x^3}{2}+25\,x^2}{8\,x+32\,{\mathrm {e}}^x+80} \]
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