Integrand size = 103, antiderivative size = 32 \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=\frac {4}{e^x-\frac {x^2}{2}+x^2 \left (-x+\frac {x}{5-x}\right )} \]
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\[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=\int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-16 e^x (-5+x)^2+16 x \left (25+50 x-27 x^2+3 x^3\right )}{\left (2 e^x (-5+x)+x^2 \left (5+7 x-2 x^2\right )\right )^2} \, dx \\ & = \int \left (-\frac {8 x \left (-50-75 x+84 x^2-23 x^3+2 x^4\right )}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2}+\frac {8 (-5+x)}{10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4}\right ) \, dx \\ & = -\left (8 \int \frac {x \left (-50-75 x+84 x^2-23 x^3+2 x^4\right )}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2} \, dx\right )+8 \int \frac {-5+x}{10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4} \, dx \\ & = -\left (8 \int \left (-\frac {50 x}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2}-\frac {75 x^2}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2}+\frac {84 x^3}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2}-\frac {23 x^4}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2}+\frac {2 x^5}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2}\right ) \, dx\right )+8 \int \left (\frac {5}{-10 e^x+2 e^x x+5 x^2+7 x^3-2 x^4}+\frac {x}{10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4}\right ) \, dx \\ & = 8 \int \frac {x}{10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4} \, dx-16 \int \frac {x^5}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2} \, dx+40 \int \frac {1}{-10 e^x+2 e^x x+5 x^2+7 x^3-2 x^4} \, dx+184 \int \frac {x^4}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2} \, dx+400 \int \frac {x}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2} \, dx+600 \int \frac {x^2}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2} \, dx-672 \int \frac {x^3}{\left (10 e^x-2 e^x x-5 x^2-7 x^3+2 x^4\right )^2} \, dx \\ \end{align*}
Time = 3.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=-\frac {8 (-5+x)}{-2 e^x (-5+x)+x^2 \left (-5-7 x+2 x^2\right )} \]
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Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {8 \left (-5+x \right )}{2 x^{4}-7 x^{3}-5 x^{2}-2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}\) | \(33\) |
norman | \(\frac {40-8 x}{2 x^{4}-7 x^{3}-5 x^{2}-2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}}\) | \(34\) |
parallelrisch | \(-\frac {16 x -80}{2 \left (2 x^{4}-7 x^{3}-5 x^{2}-2 \,{\mathrm e}^{x} x +10 \,{\mathrm e}^{x}\right )}\) | \(35\) |
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Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=-\frac {8 \, {\left (x - 5\right )}}{2 \, x^{4} - 7 \, x^{3} - 5 \, x^{2} - 2 \, {\left (x - 5\right )} e^{x}} \]
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Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=\frac {8 x - 40}{- 2 x^{4} + 7 x^{3} + 5 x^{2} + \left (2 x - 10\right ) e^{x}} \]
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Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=-\frac {8 \, {\left (x - 5\right )}}{2 \, x^{4} - 7 \, x^{3} - 5 \, x^{2} - 2 \, {\left (x - 5\right )} e^{x}} \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=-\frac {8 \, {\left (x - 5\right )}}{2 \, x^{4} - 7 \, x^{3} - 5 \, x^{2} - 2 \, x e^{x} + 10 \, e^{x}} \]
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Timed out. \[ \int \frac {400 x+800 x^2-432 x^3+48 x^4+e^x \left (-400+160 x-16 x^2\right )}{25 x^4+70 x^5+29 x^6-28 x^7+4 x^8+e^{2 x} \left (100-40 x+4 x^2\right )+e^x \left (-100 x^2-120 x^3+68 x^4-8 x^5\right )} \, dx=\int \frac {400\,x-{\mathrm {e}}^x\,\left (16\,x^2-160\,x+400\right )+800\,x^2-432\,x^3+48\,x^4}{{\mathrm {e}}^{2\,x}\,\left (4\,x^2-40\,x+100\right )-{\mathrm {e}}^x\,\left (8\,x^5-68\,x^4+120\,x^3+100\,x^2\right )+25\,x^4+70\,x^5+29\,x^6-28\,x^7+4\,x^8} \,d x \]
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