Integrand size = 39, antiderivative size = 26 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=3-\log \left (\frac {5}{2}\right )-\log \left (\log \left (2 \left (2-\frac {5}{2} (-2+x) x^2\right )\right )\right ) \]
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Time = 0.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {1607, 6816} \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 x^3+10 x^2+4\right )\right ) \]
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Rule 1607
Rule 6816
Rubi steps \begin{align*} \text {integral}& = \int \frac {(20-15 x) x}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx \\ & = -\log \left (\log \left (4+10 x^2-5 x^3\right )\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (4+10 x^2-5 x^3\right )\right ) \]
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Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65
method | result | size |
default | \(-\ln \left (\ln \left (-5 x^{3}+10 x^{2}+4\right )\right )\) | \(17\) |
norman | \(-\ln \left (\ln \left (-5 x^{3}+10 x^{2}+4\right )\right )\) | \(17\) |
risch | \(-\ln \left (\ln \left (-5 x^{3}+10 x^{2}+4\right )\right )\) | \(17\) |
parallelrisch | \(-\ln \left (\ln \left (-5 x^{3}+10 x^{2}+4\right )\right )\) | \(17\) |
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none
Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \]
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Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=- \log {\left (\log {\left (- 5 x^{3} + 10 x^{2} + 4 \right )} \right )} \]
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none
Time = 0.32 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \]
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none
Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\log \left (\log \left (-5 \, x^{3} + 10 \, x^{2} + 4\right )\right ) \]
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Time = 14.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {20 x-15 x^2}{\left (-4-10 x^2+5 x^3\right ) \log \left (4+10 x^2-5 x^3\right )} \, dx=-\ln \left (\ln \left (-5\,x^3+10\,x^2+4\right )\right ) \]
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