Integrand size = 89, antiderivative size = 35 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=e^{\frac {-x+x^2+(2+x)^2}{4 e^2+x}}+\frac {3 (5-x)}{x} \]
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\[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{x^2 \left (16 e^4+8 e^2 x+x^2\right )} \, dx \\ & = \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{x^2 \left (4 e^2+x\right )^2} \, dx \\ & = \int \left (-\frac {15}{x^2}+\frac {2 e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-2 \left (1-3 e^2\right )+8 e^2 x+x^2\right )}{\left (4 e^2+x\right )^2}\right ) \, dx \\ & = \frac {15}{x}+2 \int \frac {e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-2 \left (1-3 e^2\right )+8 e^2 x+x^2\right )}{\left (4 e^2+x\right )^2} \, dx \\ & = \frac {15}{x}+2 \int \left (e^{\frac {4+3 x+2 x^2}{4 e^2+x}}-\frac {2 e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (1-3 e^2+8 e^4\right )}{\left (4 e^2+x\right )^2}\right ) \, dx \\ & = \frac {15}{x}+2 \int e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \, dx-\left (4 \left (1-3 e^2+8 e^4\right )\right ) \int \frac {e^{\frac {4+3 x+2 x^2}{4 e^2+x}}}{\left (4 e^2+x\right )^2} \, dx \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=e^{\frac {4+3 x+2 x^2}{4 e^2+x}}+\frac {15}{x} \]
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Time = 1.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77
method | result | size |
risch | \(\frac {15}{x}+{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}} x +15}{x}\) | \(31\) |
parts | \(\frac {15}{x}+\frac {{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}} x +4 \,{\mathrm e}^{2} {\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}}{4 \,{\mathrm e}^{2}+x}\) | \(71\) |
norman | \(\frac {{\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}} x^{2}+15 x +60 \,{\mathrm e}^{2}+4 x \,{\mathrm e}^{2} {\mathrm e}^{\frac {2 x^{2}+3 x +4}{4 \,{\mathrm e}^{2}+x}}}{x \left (4 \,{\mathrm e}^{2}+x \right )}\) | \(80\) |
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\frac {x e^{\left (\frac {2 \, x^{2} + 3 \, x + 4}{x + 4 \, e^{2}}\right )} + 15}{x} \]
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Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.57 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=e^{\frac {2 x^{2} + 3 x + 4}{x + 4 e^{2}}} + \frac {15}{x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (31) = 62\).
Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 3.77 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=-\frac {15}{2} \, {\left (e^{\left (-6\right )} \log \left (x + 4 \, e^{2}\right ) - e^{\left (-6\right )} \log \left (x\right ) - \frac {4 \, {\left (x + 2 \, e^{2}\right )}}{x^{2} e^{4} + 4 \, x e^{6}}\right )} e^{4} + \frac {15}{2} \, {\left (e^{\left (-4\right )} \log \left (x + 4 \, e^{2}\right ) - e^{\left (-4\right )} \log \left (x\right ) - \frac {4}{x e^{2} + 4 \, e^{4}}\right )} e^{2} + \frac {15}{x + 4 \, e^{2}} + e^{\left (2 \, x + \frac {32 \, e^{4}}{x + 4 \, e^{2}} - \frac {12 \, e^{2}}{x + 4 \, e^{2}} + \frac {4}{x + 4 \, e^{2}} - 8 \, e^{2} + 3\right )} \]
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Time = 0.40 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\frac {{\left (x e^{\left ({\left (2 \, e^{2} + 1\right )} e^{\left (-2\right )} + \frac {2 \, x^{2} e^{2} + 3 \, x e^{2} - x}{x e^{2} + 4 \, e^{4}}\right )} + 15 \, e^{2}\right )} e^{\left (-2\right )}}{x} \]
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Time = 0.62 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {-240 e^4-120 e^2 x-15 x^2+e^{\frac {4+3 x+2 x^2}{4 e^2+x}} \left (-4 x^2+2 x^4+e^2 \left (12 x^2+16 x^3\right )\right )}{16 e^4 x^2+8 e^2 x^3+x^4} \, dx=\frac {15}{x}+{\mathrm {e}}^{\frac {3\,x}{x+4\,{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {2\,x^2}{x+4\,{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {4}{x+4\,{\mathrm {e}}^2}} \]
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