Integrand size = 35, antiderivative size = 20 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {1}{2} x \left (3 e (-2+x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 2369, 2350} \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {3}{2} e x^2 \log \left (x^3\right )-\frac {1}{2} x \left (6 e-\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \]
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Rule 12
Rule 2350
Rule 2369
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx \\ & = \frac {9}{4} e (2-x)^2+\frac {3}{2} x \log \left (\frac {5}{4}\right )+\frac {1}{2} \int \left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \, dx \\ & = \frac {9}{4} e (2-x)^2+\frac {3}{2} x \log \left (\frac {5}{4}\right )+\frac {1}{2} \int \left (-6 e+6 e x+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \, dx \\ & = \frac {9}{4} e (2-x)^2+\frac {3}{2} x \log \left (\frac {5}{4}\right )+\frac {3}{2} e x^2 \log \left (x^3\right )-\frac {1}{2} x \left (6 e-\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )-\frac {3}{2} \int \left (3 e (-2+x)+\log \left (\frac {5}{4}\right )\right ) \, dx \\ & = \frac {3}{2} e x^2 \log \left (x^3\right )-\frac {1}{2} x \left (6 e-\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=-3 e x \log \left (x^3\right )+\frac {3}{2} e x^2 \log \left (x^3\right )+\frac {1}{2} x \log \left (\frac {5}{4}\right ) \log \left (x^3\right ) \]
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Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55
method | result | size |
parallelrisch | \(\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-3 \,{\mathrm e} \ln \left (x^{3}\right ) x +\frac {\ln \left (x^{3}\right ) x \ln \left (\frac {5}{4}\right )}{2}\) | \(31\) |
norman | \(\left (-3 \,{\mathrm e}+\frac {\ln \left (5\right )}{2}-\ln \left (2\right )\right ) x \ln \left (x^{3}\right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}\) | \(32\) |
risch | \(\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}+\frac {\ln \left (5\right ) \ln \left (x^{3}\right ) x}{2}-3 \,{\mathrm e} \ln \left (x^{3}\right ) x -\ln \left (2\right ) \ln \left (x^{3}\right ) x\) | \(40\) |
parts | \(\frac {\ln \left (5\right ) \left (x \ln \left (x^{3}\right )-3 x \right )}{2}-3 \,{\mathrm e} \left (x \ln \left (x^{3}\right )-3 x \right )-\ln \left (2\right ) \left (x \ln \left (x^{3}\right )-3 x \right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-9 x \,{\mathrm e}+\frac {3 x \ln \left (\frac {5}{4}\right )}{2}\) | \(65\) |
default | \(\frac {9 \,{\mathrm e} \left (\frac {1}{2} x^{2}-2 x \right )}{2}+\frac {\ln \left (5\right ) \left (x \ln \left (x^{3}\right )-3 x \right )}{2}-3 \,{\mathrm e} \left (x \ln \left (x^{3}\right )-3 x \right )-\ln \left (2\right ) \left (x \ln \left (x^{3}\right )-3 x \right )+\frac {3 \,{\mathrm e} \ln \left (x^{3}\right ) x^{2}}{2}-\frac {9 x^{2} {\mathrm e}}{4}+\frac {3 x \ln \left (\frac {5}{4}\right )}{2}\) | \(80\) |
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Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\left (\frac {3 e x^{2}}{2} - 3 e x - x \log {\left (2 \right )} + \frac {x \log {\left (5 \right )}}{2}\right ) \log {\left (x^{3} \right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (17) = 34\).
Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.90 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=-\frac {9}{4} \, x^{2} e + \frac {3}{2} \, x {\left (6 \, e - \log \left (\frac {5}{4}\right )\right )} + \frac {9}{4} \, {\left (x^{2} - 4 \, x\right )} e + \frac {3}{2} \, x \log \left (\frac {5}{4}\right ) + \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (17) = 34\).
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.30 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=-\frac {9}{4} \, x^{2} e + \frac {9}{4} \, {\left (x^{2} - 4 \, x\right )} e + 9 \, x e + \frac {1}{2} \, {\left (3 \, {\left (x^{2} - 2 \, x\right )} e + x \log \left (\frac {5}{4}\right )\right )} \log \left (x^{3}\right ) \]
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Time = 13.76 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {1}{2} \left (e (-18+9 x)+3 \log \left (\frac {5}{4}\right )+\left (e (-6+6 x)+\log \left (\frac {5}{4}\right )\right ) \log \left (x^3\right )\right ) \, dx=\frac {x\,\ln \left (x^3\right )\,\left (\ln \left (\frac {5}{4}\right )-6\,\mathrm {e}+3\,x\,\mathrm {e}\right )}{2} \]
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