Integrand size = 25, antiderivative size = 29 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {x^2}{3}-2 x \left (i \pi -x+\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right ) \]
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Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {9} \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {1}{21} \left (7 x-3 \left (\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )+i \pi \right )\right )^2 \]
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Rule 9
Rubi steps \begin{align*} \text {integral}& = \frac {1}{21} \left (7 x-3 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right )^2 \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=-2 i \pi x+\frac {7 x^2}{3}-2 x \log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right ) \]
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Time = 0.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(-\frac {x \left (-7 x +6 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right )\right )}{3}\) | \(15\) |
default | \(-2 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) x +\frac {7 x^{2}}{3}\) | \(15\) |
parallelrisch | \(-2 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) x +\frac {7 x^{2}}{3}\) | \(15\) |
parts | \(-2 \ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) x +\frac {7 x^{2}}{3}\) | \(15\) |
risch | \(-2 \ln \left (-\frac {\ln \left (5\right )}{10}+\frac {\ln \left (2\right )}{5}\right ) x +\frac {7 x^{2}}{3}\) | \(20\) |
norman | \(\left (-2 i \pi +2 \ln \left (5\right )+2 \ln \left (10\right )-2 \ln \left (\ln \left (5\right )\right )-2 \ln \left (\ln \left (2\right )\right )\right ) x +\frac {7 x^{2}}{3}\) | \(32\) |
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Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7}{3} \, x^{2} - 2 \, x \log \left (\frac {1}{20} \, \log \left (\frac {16}{25}\right )\right ) \]
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Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7 x^{2}}{3} + x \left (- 2 \log {\left (- \frac {\log {\left (2 \right )}}{5} + \frac {\log {\left (5 \right )}}{10} \right )} - 2 i \pi \right ) \]
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Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7}{3} \, x^{2} - 2 \, x \log \left (\frac {1}{20} \, \log \left (\frac {16}{25}\right )\right ) \]
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Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7}{3} \, x^{2} - 2 \, x \log \left (\frac {1}{20} \, \log \left (\frac {16}{25}\right )\right ) \]
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Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.48 \[ \int \frac {1}{3} \left (14 x-6 \left (i \pi +\log \left (\frac {1}{20} \log \left (\frac {25}{16}\right )\right )\right )\right ) \, dx=\frac {7\,x^2}{3}-2\,x\,\ln \left (\frac {\ln \left (\frac {16}{25}\right )}{20}\right ) \]
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