Integrand size = 73, antiderivative size = 27 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=-5+\frac {e^{4+2 e^{\frac {1}{4-x}}+16 \log ^2(x)}}{x^2} \]
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Leaf count is larger than twice the leaf count of optimal. \(83\) vs. \(2(27)=54\).
Time = 0.67 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {1608, 27, 2326} \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{2 e^{\frac {1}{4-x}}+16 \log ^2(x)+4} \left (16 \left (x^2-8 x+16\right ) \log (x)+e^{\frac {1}{4-x}} x\right )}{(4-x)^2 x^3 \left (\frac {e^{\frac {1}{4-x}}}{(4-x)^2}+\frac {16 \log (x)}{x}\right )} \]
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Rule 27
Rule 1608
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{x^3 \left (16-8 x+x^2\right )} \, dx \\ & = \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{(-4+x)^2 x^3} \, dx \\ & = \frac {e^{4+2 e^{\frac {1}{4-x}}+16 \log ^2(x)} \left (e^{\frac {1}{4-x}} x+16 \left (16-8 x+x^2\right ) \log (x)\right )}{(4-x)^2 x^3 \left (\frac {e^{\frac {1}{4-x}}}{(4-x)^2}+\frac {16 \log (x)}{x}\right )} \\ \end{align*}
Time = 3.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{2 \left (2+e^{\frac {1}{4-x}}+8 \log ^2(x)\right )}}{x^2} \]
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Time = 190.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
risch | \(\frac {{\mathrm e}^{2 \,{\mathrm e}^{-\frac {1}{x -4}}+4+16 \ln \left (x \right )^{2}}}{x^{2}}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{16 \ln \left (x \right )^{2}} {\mathrm e}^{2 \,{\mathrm e}^{-\frac {1}{x -4}}+4}}{x^{2}}\) | \(27\) |
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Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{\left (16 \, \log \left (x\right )^{2} + 2 \, e^{\left (-\frac {1}{x - 4}\right )} + 4\right )}}{x^{2}} \]
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Time = 0.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{4 + 2 e^{- \frac {1}{x - 4}}} e^{16 \log {\left (x \right )}^{2}}}{x^{2}} \]
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Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {e^{\left (16 \, \log \left (x\right )^{2} + 2 \, e^{\left (-\frac {1}{x - 4}\right )} + 4\right )}}{x^{2}} \]
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\[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\int { -\frac {2 \, {\left (x^{2} - x e^{\left (-\frac {1}{x - 4}\right )} - 16 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (x\right ) - 8 \, x + 16\right )} e^{\left (16 \, \log \left (x\right )^{2} + 2 \, e^{\left (-\frac {1}{x - 4}\right )} + 4\right )}}{x^{5} - 8 \, x^{4} + 16 \, x^{3}} \,d x } \]
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Time = 14.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{4+2 e^{-\frac {1}{-4+x}}+16 \log ^2(x)} \left (-32+16 x+2 e^{-\frac {1}{-4+x}} x-2 x^2+\left (512-256 x+32 x^2\right ) \log (x)\right )}{16 x^3-8 x^4+x^5} \, dx=\frac {{\mathrm {e}}^{16\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^4\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-\frac {1}{x-4}}}}{x^2} \]
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