Integrand size = 56, antiderivative size = 24 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=2+\frac {\left (1+\log \left (2+e^5\right )\right )^2}{\left (-1+\frac {e^3}{x^2}\right )^2} \]
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Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 6820, 270} \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {x^4 \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2} \]
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Rule 12
Rule 270
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \left (4 e^3 \left (1+\log \left (2+e^5\right )\right )^2\right ) \int \frac {1}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx \\ & = \left (4 e^3 \left (1+\log \left (2+e^5\right )\right )^2\right ) \int \frac {x^3}{\left (e^3-x^2\right )^3} \, dx \\ & = \frac {x^4 \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=-\frac {e^3 \left (e^3-2 x^2\right ) \left (1+\log \left (2+e^5\right )\right )^2}{\left (e^3-x^2\right )^2} \]
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Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96
method | result | size |
risch | \(\frac {\left (4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4\right ) {\mathrm e}^{3} \left (-\frac {{\mathrm e}^{3}}{4}+\frac {x^{2}}{2}\right )}{{\mathrm e}^{6}-2 x^{2} {\mathrm e}^{3}+x^{4}}\) | \(47\) |
parallelrisch | \(\frac {4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4}{\frac {4 \,{\mathrm e}^{6}}{x^{4}}-8 \,{\mathrm e}^{3-\ln \left (x^{2}\right )}+4}\) | \(47\) |
norman | \(\frac {-{\mathrm e}^{6} \left (\ln \left ({\mathrm e}^{5}+2\right )^{2}+2 \ln \left ({\mathrm e}^{5}+2\right )+1\right ) x +\left (2 \,{\mathrm e}^{3} \ln \left ({\mathrm e}^{5}+2\right )^{2}+4 \,{\mathrm e}^{3} \ln \left ({\mathrm e}^{5}+2\right )+2 \,{\mathrm e}^{3}\right ) x^{3}}{x \left (-x^{2}+{\mathrm e}^{3}\right )^{2}}\) | \(68\) |
default | \(\frac {\left (4 \ln \left ({\mathrm e}^{5}+2\right )^{2}+8 \ln \left ({\mathrm e}^{5}+2\right )+4\right ) {\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \left (x \right )} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (-3 \,{\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \left (x \right )} \textit {\_Z}^{2}+\textit {\_Z}^{3}+3 \,{\mathrm e}^{6-2 \ln \left (x^{2}\right )+4 \ln \left (x \right )} \textit {\_Z} -{\mathrm e}^{9-3 \ln \left (x^{2}\right )+6 \ln \left (x \right )}\right )}{\sum }\frac {\textit {\_R} \ln \left (x^{2}-\textit {\_R} \right )}{-\textit {\_R}^{2}+2 \,{\mathrm e}^{3-\ln \left (x^{2}\right )+2 \ln \left (x \right )} \textit {\_R} -{\mathrm e}^{6-2 \ln \left (x^{2}\right )+4 \ln \left (x \right )}}\right )}{6}\) | \(140\) |
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Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.79 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {2 \, x^{2} e^{3} + {\left (2 \, x^{2} e^{3} - e^{6}\right )} \log \left (e^{5} + 2\right )^{2} + 2 \, {\left (2 \, x^{2} e^{3} - e^{6}\right )} \log \left (e^{5} + 2\right ) - e^{6}}{x^{4} - 2 \, x^{2} e^{3} + e^{6}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (20) = 40\).
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.42 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {\left (- 2 x^{2} + e^{3}\right ) \left (- 4 e^{3} \log {\left (2 + e^{5} \right )}^{2} - 8 e^{3} \log {\left (2 + e^{5} \right )} - 4 e^{3}\right )}{4 x^{4} - 8 x^{2} e^{3} + 4 e^{6}} \]
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Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {{\left (2 \, x^{2} - e^{3}\right )} {\left (\log \left (e^{5} + 2\right )^{2} + 2 \, \log \left (e^{5} + 2\right ) + 1\right )} e^{3}}{x^{4} - 2 \, x^{2} e^{3} + e^{6}} \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {{\left (2 \, x^{2} - e^{3}\right )} {\left (\log \left (e^{5} + 2\right )^{2} + 2 \, \log \left (e^{5} + 2\right ) + 1\right )} e^{3}}{{\left (x^{2} - e^{3}\right )}^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^3 \left (4+8 \log \left (2+e^5\right )+4 \log ^2\left (2+e^5\right )\right )}{\left (\frac {e^9}{x^5}-\frac {3 e^6}{x^3}+\frac {3 e^3}{x}-x\right ) x^2} \, dx=\frac {x^4\,\left (2\,\ln \left ({\mathrm {e}}^5+2\right )+{\ln \left ({\mathrm {e}}^5+2\right )}^2+1\right )}{x^4-2\,{\mathrm {e}}^3\,x^2+{\mathrm {e}}^6} \]
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