Integrand size = 86, antiderivative size = 28 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 x \log \left (\frac {4 \left (-3+\frac {x^2}{4}\right )}{e^2 \left (2 x+\log \left (\frac {5}{2}\right )\right )}\right ) \]
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Time = 0.43 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6873, 6820, 12, 6857, 1643, 213, 2603} \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )-4 x \]
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Rule 12
Rule 213
Rule 1643
Rule 2603
Rule 6820
Rule 6857
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-48 x-4 x^3-4 x^2 \log \left (\frac {5}{2}\right )-\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{24 x-2 x^3+12 \log \left (\frac {5}{2}\right )-x^2 \log \left (\frac {5}{2}\right )} \, dx \\ & = \int \frac {2 \left (-72 x+2 x^3-24 \log \left (\frac {5}{2}\right )-\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx \\ & = 2 \int \frac {-72 x+2 x^3-24 \log \left (\frac {5}{2}\right )-\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx \\ & = 2 \int \left (-\frac {2 \left (-36 x+x^3-12 \log \left (\frac {5}{2}\right )\right )}{\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )}+\log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )\right ) \, dx \\ & = 2 \int \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right ) \, dx-4 \int \frac {-36 x+x^3-12 \log \left (\frac {5}{2}\right )}{\left (-12+x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx \\ & = 2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )-2 \int \frac {2 x \left (-12-x^2-x \log \left (\frac {5}{2}\right )\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx-4 \int \left (\frac {1}{2}-\frac {12}{-12+x^2}-\frac {\log \left (\frac {5}{2}\right )}{4 x+\log \left (\frac {25}{4}\right )}\right ) \, dx \\ & = -2 x+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )-4 \int \frac {x \left (-12-x^2-x \log \left (\frac {5}{2}\right )\right )}{\left (12-x^2\right ) \left (2 x+\log \left (\frac {5}{2}\right )\right )} \, dx+48 \int \frac {1}{-12+x^2} \, dx \\ & = -2 x-8 \sqrt {3} \text {arctanh}\left (\frac {x}{2 \sqrt {3}}\right )+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )-4 \int \left (\frac {1}{2}+\frac {12}{-12+x^2}+\frac {\log \left (\frac {5}{2}\right )}{4 x+\log \left (\frac {25}{4}\right )}\right ) \, dx \\ & = -4 x-8 \sqrt {3} \text {arctanh}\left (\frac {x}{2 \sqrt {3}}\right )+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )-48 \int \frac {1}{-12+x^2} \, dx \\ & = -4 x+2 x \log \left (-\frac {12-x^2}{2 x+\log \left (\frac {5}{2}\right )}\right ) \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(28)=56\).
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 \left (-2 x+x \log \left (\frac {-12+x^2}{2 x+\log \left (\frac {5}{2}\right )}\right )+\frac {1}{2} \log \left (\frac {5}{2}\right ) \log \left (2 x+\log \left (\frac {5}{2}\right )\right )-\frac {1}{2} \log \left (\frac {5}{2}\right ) \log \left (4 x+\log \left (\frac {25}{4}\right )\right )\right ) \]
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Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \(2 \ln \left (\frac {\left (x^{2}-12\right ) {\mathrm e}^{-2}}{\ln \left (\frac {5}{2}\right )+2 x}\right ) x\) | \(23\) |
norman | \(2 x \ln \left (\frac {x^{2}-12}{{\mathrm e}^{2} \ln \left (\frac {5}{2}\right )+2 \,{\mathrm e}^{2} x}\right )\) | \(24\) |
default | \(-4 x +2 x \ln \left (\frac {x^{2}-12}{\ln \left (5\right )-\ln \left (2\right )+2 x}\right )\) | \(27\) |
risch | \(2 x \ln \left (\frac {x^{2}-12}{{\mathrm e}^{2} \left (\ln \left (5\right )-\ln \left (2\right )\right )+2 \,{\mathrm e}^{2} x}\right )\) | \(29\) |
parts | \(-4 x +\ln \left (\frac {5}{2}\right ) \ln \left (\ln \left (\frac {5}{2}\right )+2 x \right )+2 x \ln \left (\frac {x^{2}-12}{\ln \left (5\right )-\ln \left (2\right )+2 x}\right )-4 \left (\frac {\ln \left (5\right )}{4}-\frac {\ln \left (2\right )}{4}\right ) \ln \left (\ln \left (5\right )-\ln \left (2\right )+2 x \right )\) | \(59\) |
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Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 \, x \log \left (\frac {x^{2} - 12}{2 \, x e^{2} + e^{2} \log \left (\frac {5}{2}\right )}\right ) \]
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Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 x \log {\left (\frac {x^{2} - 12}{2 x e^{2} + e^{2} \log {\left (\frac {5}{2} \right )}} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 233, normalized size of antiderivative = 8.32 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=-\frac {\log \left (\frac {5}{2}\right )^{3} \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} + 2 \, {\left (\frac {\log \left (\frac {5}{2}\right )^{2} \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} + \frac {2 \, \sqrt {3} \log \left (\frac {5}{2}\right ) \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {24 \, \log \left (x^{2} - 12\right )}{\log \left (\frac {5}{2}\right )^{2} - 48}\right )} \log \left (\frac {5}{2}\right ) + 2 \, x \log \left (x^{2} - 12\right ) - 2 \, x \log \left (2 \, x + \log \left (5\right ) - \log \left (2\right )\right ) - {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (2 \, x + \log \left (5\right ) - \log \left (2\right )\right ) - 4 \, \sqrt {3} \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right ) - 4 \, x + \frac {48 \, \log \left (\frac {5}{2}\right ) \log \left (x^{2} - 12\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {48 \, \log \left (\frac {5}{2}\right ) \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} - \frac {192 \, \sqrt {3} \log \left (\frac {x - 2 \, \sqrt {3}}{x + 2 \, \sqrt {3}}\right )}{\log \left (\frac {5}{2}\right )^{2} - 48} \]
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Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=2 \, x \log \left (x^{2} - 12\right ) + {\left (\log \left (5\right ) - \log \left (2\right )\right )} \log \left (2 \, x + \log \left (5\right ) - \log \left (2\right )\right ) - 2 \, x \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right ) - \log \left (\frac {5}{2}\right ) \log \left (2 \, x + \log \left (\frac {5}{2}\right )\right ) - 4 \, x \]
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Timed out. \[ \int \frac {48 x+4 x^3+4 x^2 \log \left (\frac {5}{2}\right )+\left (-48 x+4 x^3+\left (-24+2 x^2\right ) \log \left (\frac {5}{2}\right )\right ) \log \left (\frac {-12+x^2}{2 e^2 x+e^2 \log \left (\frac {5}{2}\right )}\right )}{-24 x+2 x^3+\left (-12+x^2\right ) \log \left (\frac {5}{2}\right )} \, dx=\text {Hanged} \]
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