Integrand size = 56, antiderivative size = 26 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=4 (-1+x) x \left (5+\frac {5}{x}-\frac {7-x}{\log (3)}\right ) \log (x) \]
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Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {12, 14, 779, 2403, 2332, 2341} \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {4 x^3 \log (x)}{\log (3)}-\frac {4 x^2 (8-\log (243)) \log (x)}{\log (3)}+\frac {28 x \log (x)}{\log (3)}-\frac {4 \log (243) \log (x)}{\log (3)} \]
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Rule 12
Rule 14
Rule 779
Rule 2332
Rule 2341
Rule 2403
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x} \, dx}{\log (3)} \\ & = \frac {\int \left (\frac {4 (1-x) \left (-x^2+x (7-\log (243))-\log (243)\right )}{x}+4 \left (7+3 x^2-2 x (8-\log (243))\right ) \log (x)\right ) \, dx}{\log (3)} \\ & = \frac {4 \int \frac {(1-x) \left (-x^2+x (7-\log (243))-\log (243)\right )}{x} \, dx}{\log (3)}+\frac {4 \int \left (7+3 x^2-2 x (8-\log (243))\right ) \log (x) \, dx}{\log (3)} \\ & = \frac {4 \int \left (7+x^2-x (8-\log (243))-\frac {\log (243)}{x}\right ) \, dx}{\log (3)}+\frac {4 \int \left (7 \log (x)+3 x^2 \log (x)+2 x (-8+\log (243)) \log (x)\right ) \, dx}{\log (3)} \\ & = \frac {28 x}{\log (3)}+\frac {4 x^3}{3 \log (3)}-\frac {2 x^2 (8-\log (243))}{\log (3)}-\frac {4 \log (243) \log (x)}{\log (3)}+\frac {12 \int x^2 \log (x) \, dx}{\log (3)}+\frac {28 \int \log (x) \, dx}{\log (3)}-\frac {(8 (8-\log (243))) \int x \log (x) \, dx}{\log (3)} \\ & = \frac {28 x \log (x)}{\log (3)}+\frac {4 x^3 \log (x)}{\log (3)}-\frac {4 x^2 (8-\log (243)) \log (x)}{\log (3)}-\frac {4 \log (243) \log (x)}{\log (3)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {4 (-1+x) \left (x^2+x (-7+\log (243))+\log (243)\right ) \log (x)}{\log (3)} \]
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Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\frac {\left (20 x^{2} \ln \left (3\right )+4 x^{3}-32 x^{2}+28 x \right ) \ln \left (x \right )}{\ln \left (3\right )}-20 \ln \left (x \right )\) | \(34\) |
parallelrisch | \(\frac {20 x^{2} \ln \left (3\right ) \ln \left (x \right )+4 x^{3} \ln \left (x \right )-32 x^{2} \ln \left (x \right )-20 \ln \left (3\right ) \ln \left (x \right )+28 x \ln \left (x \right )}{\ln \left (3\right )}\) | \(41\) |
norman | \(-20 \ln \left (x \right )+\frac {28 x \ln \left (x \right )}{\ln \left (3\right )}+\frac {4 x^{3} \ln \left (x \right )}{\ln \left (3\right )}+\frac {4 \left (5 \ln \left (3\right )-8\right ) x^{2} \ln \left (x \right )}{\ln \left (3\right )}\) | \(43\) |
default | \(\frac {40 \ln \left (3\right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+4 x^{3} \ln \left (x \right )+10 x^{2} \ln \left (3\right )-32 x^{2} \ln \left (x \right )+28 x \ln \left (x \right )-20 \ln \left (3\right ) \ln \left (x \right )}{\ln \left (3\right )}\) | \(56\) |
parts | \(\frac {40 \ln \left (3\right ) \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+4 x^{3} \ln \left (x \right )-\frac {4 x^{3}}{3}-32 x^{2} \ln \left (x \right )+16 x^{2}+28 x \ln \left (x \right )-28 x}{\ln \left (3\right )}+10 x^{2}+\frac {4 x^{3}}{3 \ln \left (3\right )}-\frac {16 x^{2}}{\ln \left (3\right )}+\frac {28 x}{\ln \left (3\right )}-20 \ln \left (x \right )\) | \(91\) |
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Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {4 \, {\left (x^{3} - 8 \, x^{2} + 5 \, {\left (x^{2} - 1\right )} \log \left (3\right ) + 7 \, x\right )} \log \left (x\right )}{\log \left (3\right )} \]
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Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {\left (4 x^{3} - 32 x^{2} + 20 x^{2} \log {\left (3 \right )} + 28 x\right ) \log {\left (x \right )}}{\log {\left (3 \right )}} - 20 \log {\left (x \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (23) = 46\).
Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {2 \, {\left (2 \, x^{3} \log \left (x\right ) + 5 \, x^{2} \log \left (3\right ) - 16 \, x^{2} \log \left (x\right ) + 5 \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (3\right ) + 14 \, x \log \left (x\right ) - 10 \, \log \left (3\right ) \log \left (x\right )\right )}}{\log \left (3\right )} \]
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Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {4 \, {\left ({\left (x^{3} + x^{2} {\left (5 \, \log \left (3\right ) - 8\right )} + 7 \, x\right )} \log \left (x\right ) - 5 \, \log \left (3\right ) \log \left (x\right )\right )}}{\log \left (3\right )} \]
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Time = 9.79 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {28 x-32 x^2+4 x^3+\left (-20+20 x^2\right ) \log (3)+\left (28 x-64 x^2+12 x^3+40 x^2 \log (3)\right ) \log (x)}{x \log (3)} \, dx=\frac {28\,x\,\ln \left (x\right )}{\ln \left (3\right )}-20\,\ln \left (x\right )+\frac {4\,x^3\,\ln \left (x\right )}{\ln \left (3\right )}+\frac {x^2\,\ln \left (x\right )\,\left (20\,\ln \left (3\right )-32\right )}{\ln \left (3\right )} \]
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