Integrand size = 146, antiderivative size = 29 \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=\frac {1+x}{-x \left (4 e^x+x\right )+\frac {3}{x^2 (4+4 x)}} \]
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\[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=\int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {8 x (1+x) \left (3+6 x+8 e^x x^3+4 \left (1+4 e^x\right ) x^4+2 \left (3+8 e^x\right ) x^5+\left (2+8 e^x\right ) x^6\right )}{\left (3-4 x^4-4 x^5-16 e^x x^3 (1+x)\right )^2} \, dx \\ & = 8 \int \frac {x (1+x) \left (3+6 x+8 e^x x^3+4 \left (1+4 e^x\right ) x^4+2 \left (3+8 e^x\right ) x^5+\left (2+8 e^x\right ) x^6\right )}{\left (3-4 x^4-4 x^5-16 e^x x^3 (1+x)\right )^2} \, dx \\ & = 8 \int \left (\frac {x \left (1+2 x+2 x^2+x^3\right )}{2 \left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )}-\frac {x \left (-9-24 x-18 x^2-3 x^3-4 x^4-8 x^5+8 x^7+4 x^8\right )}{2 \left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}\right ) \, dx \\ & = 4 \int \frac {x \left (1+2 x+2 x^2+x^3\right )}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5} \, dx-4 \int \frac {x \left (-9-24 x-18 x^2-3 x^3-4 x^4-8 x^5+8 x^7+4 x^8\right )}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx \\ & = -\left (4 \int \left (-\frac {9 x}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}-\frac {24 x^2}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}-\frac {18 x^3}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}-\frac {3 x^4}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}-\frac {4 x^5}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}-\frac {8 x^6}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}+\frac {8 x^8}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}+\frac {4 x^9}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2}\right ) \, dx\right )+4 \int \left (\frac {x}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5}+\frac {2 x^2}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5}+\frac {2 x^3}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5}+\frac {x^4}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5}\right ) \, dx \\ & = 4 \int \frac {x}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5} \, dx+4 \int \frac {x^4}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5} \, dx+8 \int \frac {x^2}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5} \, dx+8 \int \frac {x^3}{-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5} \, dx+12 \int \frac {x^4}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx+16 \int \frac {x^5}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx-16 \int \frac {x^9}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx+32 \int \frac {x^6}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx-32 \int \frac {x^8}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx+36 \int \frac {x}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx+72 \int \frac {x^3}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx+96 \int \frac {x^2}{\left (-3+16 e^x x^3+4 x^4+16 e^x x^4+4 x^5\right )^2} \, dx \\ \end{align*}
Time = 5.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=-\frac {8 x^2 (1+x)^2}{-6+8 x^4+8 x^5+32 e^x x^3 (1+x)} \]
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Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34
method | result | size |
risch | \(-\frac {4 x^{2} \left (1+x \right )^{2}}{16 \,{\mathrm e}^{x} x^{4}+4 x^{5}+16 \,{\mathrm e}^{x} x^{3}+4 x^{4}-3}\) | \(39\) |
norman | \(\frac {-4 x^{4}-8 x^{3}-4 x^{2}}{16 \,{\mathrm e}^{x} x^{4}+4 x^{5}+16 \,{\mathrm e}^{x} x^{3}+4 x^{4}-3}\) | \(46\) |
parallelrisch | \(\frac {-64 x^{4}-128 x^{3}-64 x^{2}}{256 \,{\mathrm e}^{x} x^{4}+64 x^{5}+256 \,{\mathrm e}^{x} x^{3}+64 x^{4}-48}\) | \(47\) |
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=-\frac {4 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}{4 \, x^{5} + 4 \, x^{4} + 16 \, {\left (x^{4} + x^{3}\right )} e^{x} - 3} \]
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Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=\frac {- 4 x^{4} - 8 x^{3} - 4 x^{2}}{4 x^{5} + 4 x^{4} + \left (16 x^{4} + 16 x^{3}\right ) e^{x} - 3} \]
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Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=-\frac {4 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}{4 \, x^{5} + 4 \, x^{4} + 16 \, {\left (x^{4} + x^{3}\right )} e^{x} - 3} \]
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Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=-\frac {4 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}}{4 \, x^{5} + 16 \, x^{4} e^{x} + 4 \, x^{4} + 16 \, x^{3} e^{x} - 3} \]
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Timed out. \[ \int \frac {24 x+72 x^2+48 x^3+32 x^5+80 x^6+64 x^7+16 x^8+e^x \left (64 x^4+192 x^5+256 x^6+192 x^7+64 x^8\right )}{9-24 x^4-24 x^5+16 x^8+32 x^9+16 x^{10}+e^{2 x} \left (256 x^6+512 x^7+256 x^8\right )+e^x \left (-96 x^3-96 x^4+128 x^7+256 x^8+128 x^9\right )} \, dx=\int \frac {24\,x+{\mathrm {e}}^x\,\left (64\,x^8+192\,x^7+256\,x^6+192\,x^5+64\,x^4\right )+72\,x^2+48\,x^3+32\,x^5+80\,x^6+64\,x^7+16\,x^8}{{\mathrm {e}}^{2\,x}\,\left (256\,x^8+512\,x^7+256\,x^6\right )+{\mathrm {e}}^x\,\left (128\,x^9+256\,x^8+128\,x^7-96\,x^4-96\,x^3\right )-24\,x^4-24\,x^5+16\,x^8+32\,x^9+16\,x^{10}+9} \,d x \]
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