Integrand size = 58, antiderivative size = 19 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log (4+\log ((-3+2 x) (4+(-2-x) x))) \]
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Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6873, 6816} \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log \left (\log \left (-2 x^3-x^2+14 x-12\right )+4\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-14+2 x+6 x^2}{\left (12-14 x+x^2+2 x^3\right ) \left (4+\log \left (-12+14 x-x^2-2 x^3\right )\right )} \, dx \\ & = \log \left (4+\log \left (-12+14 x-x^2-2 x^3\right )\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log \left (4+\log \left (-12+14 x-x^2-2 x^3\right )\right ) \]
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Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
default | \(\ln \left (\ln \left (-2 x^{3}-x^{2}+14 x -12\right )+4\right )\) | \(20\) |
norman | \(\ln \left (\ln \left (-2 x^{3}-x^{2}+14 x -12\right )+4\right )\) | \(20\) |
risch | \(\ln \left (\ln \left (-2 x^{3}-x^{2}+14 x -12\right )+4\right )\) | \(20\) |
parallelrisch | \(\ln \left (\ln \left (-2 x^{3}-x^{2}+14 x -12\right )+4\right )\) | \(20\) |
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none
Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log \left (\log \left (-2 \, x^{3} - x^{2} + 14 \, x - 12\right ) + 4\right ) \]
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Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log {\left (\log {\left (- 2 x^{3} - x^{2} + 14 x - 12 \right )} + 4 \right )} \]
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none
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log \left (\log \left (-x^{2} - 2 \, x + 4\right ) + \log \left (2 \, x - 3\right ) + 4\right ) \]
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none
Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\log \left (\log \left (-2 \, x^{3} - x^{2} + 14 \, x - 12\right ) + 4\right ) \]
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Time = 0.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-14+2 x+6 x^2}{48-56 x+4 x^2+8 x^3+\left (12-14 x+x^2+2 x^3\right ) \log \left (-12+14 x-x^2-2 x^3\right )} \, dx=\ln \left (\ln \left (-\left (2\,x-3\right )\,\left (x^2+2\,x-4\right )\right )+4\right ) \]
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