Integrand size = 122, antiderivative size = 31 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=\log (5) \left (x-x^2-\frac {1}{\log \left (\left (e^2-\frac {2-x}{x}\right )^2\right )}\right ) \]
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Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {6, 1607, 6820, 12, 2572, 2562, 2339, 30} \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=x^2 (-\log (5))-\frac {\log (5)}{\log \left (\frac {\left (2-\left (1+e^2\right ) x\right )^2}{x^2}\right )}+x \log (5) \]
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Rule 6
Rule 12
Rule 30
Rule 1607
Rule 2339
Rule 2562
Rule 2572
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+\left (1+e^2\right ) x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx \\ & = \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{x \left (-2+\left (1+e^2\right ) x\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx \\ & = \int \log (5) \left (1-2 x+\frac {4}{x \left (-2+x+e^2 x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \, dx \\ & = \log (5) \int \left (1-2 x+\frac {4}{x \left (-2+x+e^2 x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \, dx \\ & = x \log (5)-x^2 \log (5)+(4 \log (5)) \int \frac {1}{x \left (-2+x+e^2 x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )} \, dx \\ & = x \log (5)-x^2 \log (5)+(4 \log (5)) \int \frac {1}{x \left (-2+\left (1+e^2\right ) x\right ) \log ^2\left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )} \, dx \\ & = x \log (5)-x^2 \log (5)+(4 \log (5)) \int \frac {1}{x \left (-2+\left (1+e^2\right ) x\right ) \log ^2\left (\frac {\left (-2+\left (1+e^2\right ) x\right )^2}{x^2}\right )} \, dx \\ & = x \log (5)-x^2 \log (5)+(2 \log (5)) \text {Subst}\left (\int \frac {1}{x \log ^2\left (x^2\right )} \, dx,x,\frac {-2+\left (1+e^2\right ) x}{x}\right ) \\ & = x \log (5)-x^2 \log (5)+\log (5) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {\left (-2+\left (1+e^2\right ) x\right )^2}{x^2}\right )\right ) \\ & = x \log (5)-x^2 \log (5)-\frac {\log (5)}{\log \left (\frac {\left (2-\left (1+e^2\right ) x\right )^2}{x^2}\right )} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=\log (5) \left (x-x^2-\frac {1}{\log \left (\frac {\left (-2+x+e^2 x\right )^2}{x^2}\right )}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
method | result | size |
derivativedivides | \(-\ln \left (5\right ) \left (x^{2}-x +\frac {1}{\ln \left ({\mathrm e}^{4}-\frac {4 \,{\mathrm e}^{2}}{x}+\frac {4}{x^{2}}+2 \,{\mathrm e}^{2}-\frac {4}{x}+1\right )}\right )\) | \(42\) |
default | \(-\ln \left (5\right ) \left (x^{2}-x +\frac {1}{\ln \left ({\mathrm e}^{4}-\frac {4 \,{\mathrm e}^{2}}{x}+\frac {4}{x^{2}}+2 \,{\mathrm e}^{2}-\frac {4}{x}+1\right )}\right )\) | \(42\) |
risch | \(-\ln \left (5\right ) x \left (-1+x \right )-\frac {\ln \left (5\right )}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(47\) |
parts | \(-\ln \left (5\right ) \left (x^{2}-x \right )-\frac {\ln \left (5\right )}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(52\) |
norman | \(\frac {x \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-x^{2} \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-\ln \left (5\right )}{\ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(119\) |
parallelrisch | \(-\frac {4 x^{2} \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )-4 x \ln \left (5\right ) \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )+4 \ln \left (5\right )}{4 \ln \left (\frac {x^{2} {\mathrm e}^{4}+\left (2 x^{2}-4 x \right ) {\mathrm e}^{2}+x^{2}-4 x +4}{x^{2}}\right )}\) | \(121\) |
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Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.48 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=-\frac {{\left (x^{2} - x\right )} \log \left (5\right ) \log \left (\frac {x^{2} e^{4} + x^{2} + 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 4 \, x + 4}{x^{2}}\right ) + \log \left (5\right )}{\log \left (\frac {x^{2} e^{4} + x^{2} + 2 \, {\left (x^{2} - 2 \, x\right )} e^{2} - 4 \, x + 4}{x^{2}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (20) = 40\).
Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=- x^{2} \log {\left (5 \right )} + x \log {\left (5 \right )} - \frac {\log {\left (5 \right )}}{\log {\left (\frac {x^{2} + x^{2} e^{4} - 4 x + \left (2 x^{2} - 4 x\right ) e^{2} + 4}{x^{2}} \right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (26) = 52\).
Time = 0.31 (sec) , antiderivative size = 217, normalized size of antiderivative = 7.00 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=-{\left (\frac {x^{2} {\left (e^{2} + 1\right )} + 4 \, x}{e^{4} + 2 \, e^{2} + 1} + \frac {8 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{6} + 3 \, e^{4} + 3 \, e^{2} + 1}\right )} e^{2} \log \left (5\right ) + {\left (\frac {x}{e^{2} + 1} + \frac {2 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{4} + 2 \, e^{2} + 1}\right )} e^{2} \log \left (5\right ) - {\left (\frac {x^{2} {\left (e^{2} + 1\right )} + 4 \, x}{e^{4} + 2 \, e^{2} + 1} + \frac {8 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{6} + 3 \, e^{4} + 3 \, e^{2} + 1}\right )} \log \left (5\right ) + 5 \, {\left (\frac {x}{e^{2} + 1} + \frac {2 \, \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{4} + 2 \, e^{2} + 1}\right )} \log \left (5\right ) - \frac {2 \, \log \left (5\right ) \log \left (x {\left (e^{2} + 1\right )} - 2\right )}{e^{2} + 1} - \frac {\log \left (5\right )}{2 \, {\left (\log \left (x {\left (e^{2} + 1\right )} - 2\right ) - \log \left (x\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (26) = 52\).
Time = 0.78 (sec) , antiderivative size = 125, normalized size of antiderivative = 4.03 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=-\frac {x^{2} \log \left (5\right ) \log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) - x^{2} \log \left (5\right ) \log \left (x^{2}\right ) - x \log \left (5\right ) \log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) + x \log \left (5\right ) \log \left (x^{2}\right ) + \log \left (5\right )}{\log \left (x^{2} e^{4} + 2 \, x^{2} e^{2} + x^{2} - 4 \, x e^{2} - 4 \, x + 4\right ) - \log \left (x^{2}\right )} \]
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Time = 12.91 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {4 \log (5)+\left (-2 x+5 x^2-2 x^3+e^2 \left (x^2-2 x^3\right )\right ) \log (5) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )}{\left (-2 x+x^2+e^2 x^2\right ) \log ^2\left (\frac {4-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )}{x^2}\right )} \, dx=x\,\ln \left (5\right )-x^2\,\ln \left (5\right )-\frac {\ln \left (5\right )}{\ln \left (\frac {x^2\,{\mathrm {e}}^4-{\mathrm {e}}^2\,\left (4\,x-2\,x^2\right )-4\,x+x^2+4}{x^2}\right )} \]
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