Integrand size = 12, antiderivative size = 23 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {2}{3} (4+2 (-2+x)) \left (4-\frac {3}{x}\right ) x^2 \log (3) \]
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Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12} \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {16}{3} x^3 \log (3)-4 x^2 \log (3) \]
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Rule 12
Rubi steps \begin{align*} \text {integral}& = \log (3) \int \left (-8 x+16 x^2\right ) \, dx \\ & = -4 x^2 \log (3)+\frac {16}{3} x^3 \log (3) \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=8 \left (-\frac {x^2}{2}+\frac {2 x^3}{3}\right ) \log (3) \]
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Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(\frac {4 \ln \left (3\right ) \left (-3+4 x \right ) x^{2}}{3}\) | \(13\) |
parallelrisch | \(\ln \left (3\right ) \left (\frac {16}{3} x^{3}-4 x^{2}\right )\) | \(15\) |
default | \(8 \ln \left (3\right ) \left (\frac {2}{3} x^{3}-\frac {1}{2} x^{2}\right )\) | \(16\) |
norman | \(-4 x^{2} \ln \left (3\right )+\frac {16 x^{3} \ln \left (3\right )}{3}\) | \(16\) |
risch | \(-4 x^{2} \ln \left (3\right )+\frac {16 x^{3} \ln \left (3\right )}{3}\) | \(16\) |
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Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {4}{3} \, {\left (4 \, x^{3} - 3 \, x^{2}\right )} \log \left (3\right ) \]
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Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {16 x^{3} \log {\left (3 \right )}}{3} - 4 x^{2} \log {\left (3 \right )} \]
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Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {4}{3} \, {\left (4 \, x^{3} - 3 \, x^{2}\right )} \log \left (3\right ) \]
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Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {4}{3} \, {\left (4 \, x^{3} - 3 \, x^{2}\right )} \log \left (3\right ) \]
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Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \left (-8 x+16 x^2\right ) \log (3) \, dx=\frac {4\,x^2\,\ln \left (3\right )\,\left (4\,x-3\right )}{3} \]
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