Integrand size = 48, antiderivative size = 25 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=\frac {2 x \left (x+\frac {x^3}{8}+\frac {x}{\log (7)}\right )}{1-x} \]
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Leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(25)=50\).
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 27, 1864} \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=-\frac {x^3}{4}-\frac {x^2}{4}-\frac {x (8+9 \log (7))}{4 \log (7)}+\frac {8+9 \log (7)}{4 (1-x) \log (7)} \]
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Rule 12
Rule 27
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{4-8 x+4 x^2} \, dx}{\log (7)} \\ & = \frac {\int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{4 (-1+x)^2} \, dx}{\log (7)} \\ & = \frac {\int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{(-1+x)^2} \, dx}{4 \log (7)} \\ & = \frac {\int \left (-2 x \log (7)-3 x^2 \log (7)-8 \left (1+\frac {9 \log (7)}{8}\right )+\frac {8+9 \log (7)}{(-1+x)^2}\right ) \, dx}{4 \log (7)} \\ & = -\frac {x^2}{4}-\frac {x^3}{4}+\frac {8+9 \log (7)}{4 (1-x) \log (7)}-\frac {x (8+9 \log (7))}{4 \log (7)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=\frac {\frac {-8-9 \log (7)}{-1+x}-4 (-1+x)^2 \log (7)-(-1+x)^3 \log (7)-2 (-1+x) (4+7 \log (7))}{4 \log (7)} \]
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {-\frac {x^{4}}{4}-\frac {2 \left (\ln \left (7\right )+1\right ) x^{2}}{\ln \left (7\right )}}{-1+x}\) | \(26\) |
gosper | \(-\frac {x^{2} \left (x^{2} \ln \left (7\right )+8 \ln \left (7\right )+8\right )}{4 \ln \left (7\right ) \left (-1+x \right )}\) | \(27\) |
parallelrisch | \(-\frac {\ln \left (7\right ) x^{4}+8 x^{2} \ln \left (7\right )+8 x^{2}}{4 \ln \left (7\right ) \left (-1+x \right )}\) | \(31\) |
risch | \(-\frac {x^{3}}{4}-\frac {x^{2}}{4}-\frac {9 x}{4}-\frac {2 x}{\ln \left (7\right )}-\frac {9}{4 \left (-1+x \right )}-\frac {2}{\ln \left (7\right ) \left (-1+x \right )}\) | \(40\) |
default | \(\frac {-\ln \left (7\right ) x^{3}-x^{2} \ln \left (7\right )-9 x \ln \left (7\right )-8 x -\frac {9 \ln \left (7\right )+8}{-1+x}}{4 \ln \left (7\right )}\) | \(43\) |
meijerg | \(-\frac {x \left (-5 x^{3}-10 x^{2}-30 x +60\right )}{20 \left (1-x \right )}+\frac {x \left (-2 x^{2}-6 x +12\right )}{4-4 x}-\frac {\left (-2 \ln \left (7\right )-2\right ) \left (-\frac {x \left (-3 x +6\right )}{3 \left (1-x \right )}-2 \ln \left (1-x \right )\right )}{\ln \left (7\right )}+\frac {\left (4 \ln \left (7\right )+4\right ) \left (\frac {x}{1-x}+\ln \left (1-x \right )\right )}{\ln \left (7\right )}\) | \(110\) |
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=-\frac {8 \, x^{2} + {\left (x^{4} + 8 \, x^{2} - 9 \, x + 9\right )} \log \left (7\right ) - 8 \, x + 8}{4 \, {\left (x - 1\right )} \log \left (7\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=- \frac {x^{3}}{4} - \frac {x^{2}}{4} - x \left (\frac {2}{\log {\left (7 \right )}} + \frac {9}{4}\right ) - \frac {8 + 9 \log {\left (7 \right )}}{4 x \log {\left (7 \right )} - 4 \log {\left (7 \right )}} \]
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Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=-\frac {x^{3} \log \left (7\right ) + x^{2} \log \left (7\right ) + x {\left (9 \, \log \left (7\right ) + 8\right )} + \frac {9 \, \log \left (7\right ) + 8}{x - 1}}{4 \, \log \left (7\right )} \]
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=-\frac {x^{3} \log \left (7\right ) + x^{2} \log \left (7\right ) + 9 \, x \log \left (7\right ) + 8 \, x + \frac {9 \, \log \left (7\right ) + 8}{x - 1}}{4 \, \log \left (7\right )} \]
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Time = 8.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {16 x-8 x^2+\left (16 x-8 x^2+4 x^3-3 x^4\right ) \log (7)}{\left (4-8 x+4 x^2\right ) \log (7)} \, dx=\frac {9\,\ln \left (7\right )+8}{4\,\ln \left (7\right )-4\,x\,\ln \left (7\right )}-x\,\left (\frac {8\,\ln \left (7\right )+8}{4\,\ln \left (7\right )}+\frac {1}{4}\right )-\frac {x^2}{4}-\frac {x^3}{4} \]
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