Integrand size = 53, antiderivative size = 13 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log \left (\frac {x}{\log (-2+(-632+x) x)}\right ) \]
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Time = 0.17 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {1608, 6820, 6816} \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log (x)-\log \left (\log \left (x^2-632 x-2\right )\right ) \]
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Rule 1608
Rule 6816
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{x \left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )} \, dx \\ & = \int \left (\frac {1}{x}-\frac {2 (-316+x)}{\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}\right ) \, dx \\ & = \log (x)-2 \int \frac {-316+x}{\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )} \, dx \\ & = \log (x)-\log \left (\log \left (-2-632 x+x^2\right )\right ) \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log (x)-\log \left (\log \left (-2-632 x+x^2\right )\right ) \]
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Time = 0.38 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23
method | result | size |
default | \(\ln \left (x \right )-\ln \left (\ln \left (x^{2}-632 x -2\right )\right )\) | \(16\) |
norman | \(\ln \left (x \right )-\ln \left (\ln \left (x^{2}-632 x -2\right )\right )\) | \(16\) |
risch | \(\ln \left (x \right )-\ln \left (\ln \left (x^{2}-632 x -2\right )\right )\) | \(16\) |
parallelrisch | \(\ln \left (x \right )-\ln \left (\ln \left (x^{2}-632 x -2\right )\right )\) | \(16\) |
parts | \(\ln \left (x \right )-\ln \left (\ln \left (x^{2}-632 x -2\right )\right )\) | \(16\) |
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Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log \left (x\right ) - \log \left (\log \left (x^{2} - 632 \, x - 2\right )\right ) \]
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Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log {\left (x \right )} - \log {\left (\log {\left (x^{2} - 632 x - 2 \right )} \right )} \]
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Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log \left (x\right ) - \log \left (\log \left (x^{2} - 632 \, x - 2\right )\right ) \]
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Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\log \left (x\right ) - \log \left (\log \left (x^{2} - 632 \, x - 2\right )\right ) \]
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Time = 0.37 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {632 x-2 x^2+\left (-2-632 x+x^2\right ) \log \left (-2-632 x+x^2\right )}{\left (-2 x-632 x^2+x^3\right ) \log \left (-2-632 x+x^2\right )} \, dx=\ln \left (x\right )-\ln \left (\ln \left (x^2-632\,x-2\right )\right ) \]
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