Integrand size = 52, antiderivative size = 28 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=\frac {5 x^2+\log (x)}{\frac {2}{5 \left (2-\frac {15}{x}\right )}-x} \]
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Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.173, Rules used = {1608, 27, 6874, 46, 45, 2404, 2341, 2351, 31} \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-5 x+\frac {77}{77-10 x}+\frac {200 x \log (x)}{5929 (77-10 x)}+\frac {20 \log (x)}{5929}-\frac {75 \log (x)}{77 x} \]
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Rule 27
Rule 31
Rule 45
Rule 46
Rule 1608
Rule 2341
Rule 2351
Rule 2404
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{x^2 \left (5929-1540 x+100 x^2\right )} \, dx \\ & = \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{x^2 (-77+10 x)^2} \, dx \\ & = \int \left (-\frac {28975}{(-77+10 x)^2}-\frac {5775}{x^2 (-77+10 x)^2}+\frac {1520}{x (-77+10 x)^2}+\frac {7700 x}{(-77+10 x)^2}-\frac {500 x^2}{(-77+10 x)^2}+\frac {25 \left (231-60 x+4 x^2\right ) \log (x)}{x^2 (-77+10 x)^2}\right ) \, dx \\ & = -\frac {5795}{2 (77-10 x)}+25 \int \frac {\left (231-60 x+4 x^2\right ) \log (x)}{x^2 (-77+10 x)^2} \, dx-500 \int \frac {x^2}{(-77+10 x)^2} \, dx+1520 \int \frac {1}{x (-77+10 x)^2} \, dx-5775 \int \frac {1}{x^2 (-77+10 x)^2} \, dx+7700 \int \frac {x}{(-77+10 x)^2} \, dx \\ & = -\frac {5795}{2 (77-10 x)}+25 \int \left (\frac {3 \log (x)}{77 x^2}+\frac {8 \log (x)}{77 (-77+10 x)^2}\right ) \, dx-500 \int \left (\frac {1}{100}+\frac {5929}{100 (-77+10 x)^2}+\frac {77}{50 (-77+10 x)}\right ) \, dx+1520 \int \left (\frac {1}{5929 x}+\frac {10}{77 (-77+10 x)^2}-\frac {10}{5929 (-77+10 x)}\right ) \, dx-5775 \int \left (\frac {1}{5929 x^2}+\frac {20}{456533 x}+\frac {100}{5929 (-77+10 x)^2}-\frac {200}{456533 (-77+10 x)}\right ) \, dx+7700 \int \left (\frac {77}{10 (-77+10 x)^2}+\frac {1}{10 (-77+10 x)}\right ) \, dx \\ & = \frac {77}{77-10 x}+\frac {75}{77 x}-5 x-\frac {20 \log (77-10 x)}{5929}+\frac {20 \log (x)}{5929}+\frac {75}{77} \int \frac {\log (x)}{x^2} \, dx+\frac {200}{77} \int \frac {\log (x)}{(-77+10 x)^2} \, dx \\ & = \frac {77}{77-10 x}-5 x-\frac {20 \log (77-10 x)}{5929}+\frac {20 \log (x)}{5929}-\frac {75 \log (x)}{77 x}+\frac {200 x \log (x)}{5929 (77-10 x)}+\frac {200 \int \frac {1}{-77+10 x} \, dx}{5929} \\ & = \frac {77}{77-10 x}-5 x+\frac {20 \log (x)}{5929}-\frac {75 \log (x)}{77 x}+\frac {200 x \log (x)}{5929 (77-10 x)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-5 \left (-\frac {77}{5 (77-10 x)}+x-\frac {4 \log (x)}{77 (77-10 x)}+\frac {15 \log (x)}{77 x}\right ) \]
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Time = 0.59 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
norman | \(\frac {\frac {5775 x}{2}-10 x \ln \left (x \right )-50 x^{3}+75 \ln \left (x \right )}{x \left (10 x -77\right )}\) | \(30\) |
parallelrisch | \(\frac {-500 x^{3}-100 x \ln \left (x \right )+28875 x +750 \ln \left (x \right )}{10 x \left (10 x -77\right )}\) | \(31\) |
default | \(-5 x -\frac {77}{10 x -77}+\frac {20 \ln \left (x \right )}{5929}-\frac {200 \ln \left (x \right ) x}{5929 \left (10 x -77\right )}-\frac {75 \ln \left (x \right )}{77 x}\) | \(37\) |
parts | \(-5 x -\frac {77}{10 x -77}+\frac {20 \ln \left (x \right )}{5929}-\frac {200 \ln \left (x \right ) x}{5929 \left (10 x -77\right )}-\frac {75 \ln \left (x \right )}{77 x}\) | \(37\) |
risch | \(-\frac {5 \left (2 x -15\right ) \ln \left (x \right )}{x \left (10 x -77\right )}-\frac {50 x^{2}-385 x +77}{10 x -77}\) | \(40\) |
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Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-\frac {50 \, x^{3} - 385 \, x^{2} + 5 \, {\left (2 \, x - 15\right )} \log \left (x\right ) + 77 \, x}{10 \, x^{2} - 77 \, x} \]
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Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=- 5 x + \frac {\left (75 - 10 x\right ) \log {\left (x \right )}}{10 x^{2} - 77 x} - \frac {77}{10 x - 77} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).
Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-5 \, x - \frac {25 \, {\left ({\left (8 \, x^{2} + 2310 \, x - 17787\right )} \log \left (x\right ) + 2310 \, x - 17787\right )}}{5929 \, {\left (10 \, x^{2} - 77 \, x\right )}} + \frac {75 \, {\left (20 \, x - 77\right )}}{77 \, {\left (10 \, x^{2} - 77 \, x\right )}} - \frac {6679}{77 \, {\left (10 \, x - 77\right )}} + \frac {20}{5929} \, \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-\frac {5}{77} \, {\left (\frac {4}{10 \, x - 77} + \frac {15}{x}\right )} \log \left (x\right ) - 5 \, x - \frac {77}{10 \, x - 77} \]
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Time = 14.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-5775+1520 x-28975 x^2+7700 x^3-500 x^4+\left (5775-1500 x+100 x^2\right ) \log (x)}{5929 x^2-1540 x^3+100 x^4} \, dx=-\frac {5\,\left (2\,x-15\right )\,\left (\ln \left (x\right )+5\,x^2\right )}{x\,\left (10\,x-77\right )} \]
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