Integrand size = 26, antiderivative size = 20 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^3 (-4+i \pi +\log (3))}{180 \log ^2(5)} \]
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Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6, 12, 30} \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=-\frac {x^3 (4-i \pi -\log (3))}{180 \log ^2(5)} \]
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Rule 6
Rule 12
Rule 30
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (-4+i \pi +\log (3))}{60 \log ^2(5)} \, dx \\ & = -\frac {(4-i \pi -\log (3)) \int x^2 \, dx}{60 \log ^2(5)} \\ & = -\frac {x^3 (4-i \pi -\log (3))}{180 \log ^2(5)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^3 (-4+i \pi +\log (3))}{180 \log ^2(5)} \]
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Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
method | result | size |
gosper | \(\frac {x^{3} \left (\ln \left (3\right )+i \pi -4\right )}{180 \ln \left (5\right )^{2}}\) | \(18\) |
norman | \(\frac {x^{3} \left (\ln \left (3\right )+i \pi -4\right )}{180 \ln \left (5\right )^{2}}\) | \(18\) |
default | \(\frac {i \left (-i \ln \left (3\right )+\pi +4 i\right ) x^{3}}{180 \ln \left (5\right )^{2}}\) | \(20\) |
parallelrisch | \(\frac {\frac {x^{3} \ln \left (3\right )}{3}+\frac {i x^{3} \pi }{3}-\frac {4 x^{3}}{3}}{60 \ln \left (5\right )^{2}}\) | \(27\) |
risch | \(\frac {i x^{3} \pi }{180 \ln \left (5\right )^{2}}+\frac {x^{3} \ln \left (3\right )}{180 \ln \left (5\right )^{2}}-\frac {x^{3}}{45 \ln \left (5\right )^{2}}\) | \(33\) |
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {{\left (i \, \pi - 4\right )} x^{3} + x^{3} \log \left (3\right )}{180 \, \log \left (5\right )^{2}} \]
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Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^{3} \left (-4 + \log {\left (3 \right )} + i \pi \right )}{180 \log {\left (5 \right )}^{2}} \]
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Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=-\frac {{\left (-i \, \pi - \log \left (3\right )\right )} x^{3} + 4 \, x^{3}}{180 \, \log \left (5\right )^{2}} \]
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Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {{\left (i \, \pi + \log \left (3\right )\right )} x^{3} - 4 \, x^{3}}{180 \, \log \left (5\right )^{2}} \]
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Time = 9.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x^2+x^2 (i \pi +\log (3))}{60 \log ^2(5)} \, dx=\frac {x^3\,\left (\Pi -\ln \left (3\right )\,1{}\mathrm {i}+4{}\mathrm {i}\right )\,1{}\mathrm {i}}{180\,{\ln \left (5\right )}^2} \]
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