Integrand size = 51, antiderivative size = 17 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \]
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Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.137, Rules used = {6820, 12, 6874, 78, 2581, 30, 45} \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \]
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Rule 12
Rule 30
Rule 45
Rule 78
Rule 2581
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {3 x^2 \left (2 x+\log (2)+3 (x+\log (2)) \log \left (\frac {1}{4} x (x+\log (2))\right )\right )}{10 (x+\log (2))} \, dx \\ & = \frac {3}{10} \int \frac {x^2 \left (2 x+\log (2)+3 (x+\log (2)) \log \left (\frac {1}{4} x (x+\log (2))\right )\right )}{x+\log (2)} \, dx \\ & = \frac {3}{10} \int \left (\frac {x^2 (2 x+\log (2))}{x+\log (2)}+3 x^2 \log \left (\frac {1}{4} x (x+\log (2))\right )\right ) \, dx \\ & = \frac {3}{10} \int \frac {x^2 (2 x+\log (2))}{x+\log (2)} \, dx+\frac {9}{10} \int x^2 \log \left (\frac {1}{4} x (x+\log (2))\right ) \, dx \\ & = \frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right )-\frac {3 \int x^2 \, dx}{10}-\frac {3}{10} \int \frac {x^3}{x+\log (2)} \, dx+\frac {3}{10} \int \left (2 x^2-x \log (2)+\log ^2(2)-\frac {\log ^3(2)}{x+\log (2)}\right ) \, dx \\ & = \frac {x^3}{10}-\frac {3}{20} x^2 \log (2)+\frac {3}{10} x \log ^2(2)-\frac {3}{10} \log ^3(2) \log (x+\log (2))+\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right )-\frac {3}{10} \int \left (x^2-x \log (2)+\log ^2(2)-\frac {\log ^3(2)}{x+\log (2)}\right ) \, dx \\ & = \frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=\frac {3}{10} x^3 \log \left (\frac {1}{4} x (x+\log (2))\right ) \]
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Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \(\frac {3 x^{3} \ln \left (\frac {x \left (\ln \left (2\right )+x \right )}{4}\right )}{10}\) | \(14\) |
norman | \(\frac {3 x^{3} \ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x^{2}}{4}\right )}{10}\) | \(18\) |
risch | \(\frac {3 x^{3} \ln \left (\frac {x \ln \left (2\right )}{4}+\frac {x^{2}}{4}\right )}{10}\) | \(18\) |
default | \(\frac {3 x^{3} \ln \left (x \ln \left (2\right )+x^{2}\right )}{10}-\frac {3 x^{3} \ln \left (2\right )}{5}\) | \(23\) |
parts | \(\frac {3 x^{3} \ln \left (x \ln \left (2\right )+x^{2}\right )}{10}-\frac {3 x^{3} \ln \left (2\right )}{5}\) | \(23\) |
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=\frac {3}{10} \, x^{3} \log \left (\frac {1}{4} \, x^{2} + \frac {1}{4} \, x \log \left (2\right )\right ) \]
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Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=\frac {3 x^{3} \log {\left (\frac {x^{2}}{4} + \frac {x \log {\left (2 \right )}}{4} \right )}}{10} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (13) = 26\).
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 5.18 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=-\frac {1}{5} \, x^{3} {\left (3 \, \log \left (2\right ) + 1\right )} - \frac {3}{5} \, \log \left (2\right )^{3} \log \left (x + \log \left (2\right )\right ) + \frac {3}{10} \, x^{3} \log \left (x\right ) + \frac {1}{5} \, x^{3} - \frac {3}{20} \, x^{2} \log \left (2\right ) + \frac {3}{10} \, x \log \left (2\right )^{2} + \frac {3}{20} \, {\left (2 \, \log \left (2\right )^{2} \log \left (x + \log \left (2\right )\right ) + x^{2} - 2 \, x \log \left (2\right )\right )} \log \left (2\right ) + \frac {3}{10} \, {\left (x^{3} + \log \left (2\right )^{3}\right )} \log \left (x + \log \left (2\right )\right ) \]
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=-\frac {3}{5} \, x^{3} \log \left (2\right ) + \frac {3}{10} \, x^{3} \log \left (x^{2} + x \log \left (2\right )\right ) \]
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Time = 11.91 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {6 x^3+3 x^2 \log (2)+\left (9 x^3+9 x^2 \log (2)\right ) \log \left (\frac {1}{4} \left (x^2+x \log (2)\right )\right )}{10 x+10 \log (2)} \, dx=-\frac {3\,x^3\,\left (\ln \left (4\right )-\ln \left (x^2+\ln \left (2\right )\,x\right )\right )}{10} \]
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