Integrand size = 58, antiderivative size = 20 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \log (-1+\log (x (2+2 x (4+3 x)) \log (5))) \]
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Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6873, 6816} \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \log \left (1-\log \left (2 x \left (3 x^2+4 x+1\right ) \log (5)\right )\right ) \]
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Rule 6816
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {-10-80 x-90 x^2}{x \left (1+4 x+3 x^2\right ) \left (1-\log \left (2 x \left (1+4 x+3 x^2\right ) \log (5)\right )\right )} \, dx \\ & = 10 \log \left (1-\log \left (2 x \left (1+4 x+3 x^2\right ) \log (5)\right )\right ) \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \log \left (-1+\log \left (2 x \left (1+4 x+3 x^2\right ) \log (5)\right )\right ) \]
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Time = 1.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20
method | result | size |
default | \(10 \ln \left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (x \left (3 x^{2}+4 x +1\right )\right )-1\right )\) | \(24\) |
norman | \(10 \ln \left (\ln \left (\left (6 x^{3}+8 x^{2}+2 x \right ) \ln \left (5\right )\right )-1\right )\) | \(24\) |
risch | \(10 \ln \left (\ln \left (\left (6 x^{3}+8 x^{2}+2 x \right ) \ln \left (5\right )\right )-1\right )\) | \(24\) |
parallelrisch | \(10 \ln \left (\ln \left (\left (6 x^{3}+8 x^{2}+2 x \right ) \ln \left (5\right )\right )-1\right )\) | \(24\) |
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none
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \, \log \left (\log \left (2 \, {\left (3 \, x^{3} + 4 \, x^{2} + x\right )} \log \left (5\right )\right ) - 1\right ) \]
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Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \log {\left (\log {\left (\left (6 x^{3} + 8 x^{2} + 2 x\right ) \log {\left (5 \right )} \right )} - 1 \right )} \]
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none
Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \, \log \left (\log \left (2\right ) + \log \left (3 \, x + 1\right ) + \log \left (x + 1\right ) + \log \left (x\right ) + \log \left (\log \left (5\right )\right ) - 1\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10 \, \log \left (\log \left (6 \, x^{3} \log \left (5\right ) + 8 \, x^{2} \log \left (5\right ) + 2 \, x \log \left (5\right )\right ) - 1\right ) \]
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Time = 11.87 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {10+80 x+90 x^2}{-x-4 x^2-3 x^3+\left (x+4 x^2+3 x^3\right ) \log \left (\left (2 x+8 x^2+6 x^3\right ) \log (5)\right )} \, dx=10\,\ln \left (\ln \left (\ln \left (5\right )\,\left (6\,x^3+8\,x^2+2\,x\right )\right )-1\right ) \]
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