Integrand size = 64, antiderivative size = 32 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=2 x^2-x^2 (-3-\log (2))-\log (x)-\log \left (\log \left ((3-x)^2\right )\right ) \]
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Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {1607, 6820, 2437, 2339, 29} \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=\frac {1}{2} x^2 (10+\log (4))-\log (x)-\log \left (\log \left ((x-3)^2\right )\right ) \]
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Rule 29
Rule 1607
Rule 2339
Rule 2437
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{(-3+x) x \log \left (9-6 x+x^2\right )} \, dx \\ & = \int \left (-\frac {1}{x}+x (10+\log (4))-\frac {2}{(-3+x) \log \left ((-3+x)^2\right )}\right ) \, dx \\ & = \frac {1}{2} x^2 (10+\log (4))-\log (x)-2 \int \frac {1}{(-3+x) \log \left ((-3+x)^2\right )} \, dx \\ & = \frac {1}{2} x^2 (10+\log (4))-\log (x)-2 \text {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-3+x\right ) \\ & = \frac {1}{2} x^2 (10+\log (4))-\log (x)-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-3+x)^2\right )\right ) \\ & = \frac {1}{2} x^2 (10+\log (4))-\log (x)-\log \left (\log \left ((-3+x)^2\right )\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=5 x^2+\frac {1}{2} x^2 \log (4)-\log (x)-\log \left (\log \left ((-3+x)^2\right )\right ) \]
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Time = 0.48 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81
method | result | size |
norman | \(\left (\ln \left (2\right )+5\right ) x^{2}-\ln \left (x \right )-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\) | \(26\) |
default | \(5 x^{2}-\ln \left (x \right )+x^{2} \ln \left (2\right )-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\) | \(29\) |
risch | \(5 x^{2}-\ln \left (x \right )+x^{2} \ln \left (2\right )-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\) | \(29\) |
parts | \(5 x^{2}-\ln \left (x \right )+x^{2} \ln \left (2\right )-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\) | \(29\) |
parallelrisch | \(x^{2} \ln \left (2\right )-45+5 x^{2}-9 \ln \left (2\right )-\ln \left (x \right )-\ln \left (\ln \left (x^{2}-6 x +9\right )\right )\) | \(34\) |
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=x^{2} \log \left (2\right ) + 5 \, x^{2} - \log \left (x\right ) - \log \left (\log \left (x^{2} - 6 \, x + 9\right )\right ) \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=x^{2} \left (\log {\left (2 \right )} + 5\right ) - \log {\left (x \right )} - \log {\left (\log {\left (x^{2} - 6 x + 9 \right )} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (27) = 54\).
Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.22 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=5 \, x^{2} + {\left (x^{2} + 6 \, x + 18 \, \log \left (x - 3\right )\right )} \log \left (2\right ) - 6 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} \log \left (2\right ) - \frac {1}{2} \, \log \left (x^{2} - 6 \, x + 9\right ) \log \left (\log \left (x - 3\right )\right ) + \log \left (x - 3\right ) \log \left (\log \left (x - 3\right )\right ) - \log \left (x\right ) - \log \left (\log \left (x - 3\right )\right ) \]
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Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=x^{2} {\left (\log \left (2\right ) + 5\right )} - \log \left (x\right ) - \log \left (\log \left (x^{2} - 6 \, x + 9\right )\right ) \]
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Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {-2 x+\left (3-x-30 x^2+10 x^3+\left (-6 x^2+2 x^3\right ) \log (2)\right ) \log \left (9-6 x+x^2\right )}{\left (-3 x+x^2\right ) \log \left (9-6 x+x^2\right )} \, dx=\frac {\left (\ln \left (2\right )+5\right )\,x^4+\left (-\ln \left (64\right )-30\right )\,x^3+\left (\ln \left (512\right )+45\right )\,x^2}{x^2-6\,x+9}-\ln \left (\ln \left (x^2-6\,x+9\right )\right )-\ln \left (x\right ) \]
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