Integrand size = 27, antiderivative size = 16 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=6-x-\frac {x}{(1-4 \log (5))^2} \]
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Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {8} \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 x \left (1+8 \log ^2(5)-4 \log (5)\right )}{(1-\log (625))^2} \]
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Rule 8
Rubi steps \begin{align*} \text {integral}& = -\frac {2 x \left (1-4 \log (5)+8 \log ^2(5)\right )}{(1-\log (625))^2} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(16)=32\).
Time = 0.00 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.62 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 x}{1-8 \log (5)+16 \log ^2(5)}+\frac {8 x \log (5)}{1-8 \log (5)+16 \log ^2(5)}-\frac {16 x \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \]
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Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50
method | result | size |
norman | \(-\frac {2 \left (8 \ln \left (5\right )^{2}-4 \ln \left (5\right )+1\right ) x}{\left (4 \ln \left (5\right )-1\right )^{2}}\) | \(24\) |
parallelrisch | \(\frac {\left (-16 \ln \left (5\right )^{2}+8 \ln \left (5\right )-2\right ) x}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}\) | \(29\) |
default | \(\frac {2 \left (-8 \ln \left (5\right )^{2}+4 \ln \left (5\right )-1\right ) x}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}\) | \(30\) |
risch | \(-\frac {16 x \ln \left (5\right )^{2}}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}+\frac {8 x \ln \left (5\right )}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}-\frac {2 x}{16 \ln \left (5\right )^{2}-8 \ln \left (5\right )+1}\) | \(59\) |
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Time = 0.23 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.88 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 \, {\left (8 \, x \log \left (5\right )^{2} - 4 \, x \log \left (5\right ) + x\right )}}{16 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 1} \]
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Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=\frac {x \left (- 16 \log {\left (5 \right )}^{2} - 2 + 8 \log {\left (5 \right )}\right )}{- 8 \log {\left (5 \right )} + 1 + 16 \log {\left (5 \right )}^{2}} \]
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Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 \, {\left (8 \, \log \left (5\right )^{2} - 4 \, \log \left (5\right ) + 1\right )} x}{16 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 1} \]
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Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {2 \, {\left (8 \, \log \left (5\right )^{2} - 4 \, \log \left (5\right ) + 1\right )} x}{16 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 1} \]
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Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \frac {-2+8 \log (5)-16 \log ^2(5)}{1-8 \log (5)+16 \log ^2(5)} \, dx=-\frac {x\,\left (16\,{\ln \left (5\right )}^2-8\,\ln \left (5\right )+2\right )}{16\,{\ln \left (5\right )}^2-8\,\ln \left (5\right )+1} \]
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