Integrand size = 49, antiderivative size = 22 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {5 \left (\frac {4 x}{5}+\log (x)\right )}{4 x^2 \left (-5+x^2\right )^2} \]
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Leaf count is larger than twice the leaf count of optimal. \(116\) vs. \(2(22)=44\).
Time = 0.49 (sec) , antiderivative size = 116, normalized size of antiderivative = 5.27, number of steps used = 28, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6873, 12, 6874, 205, 213, 272, 46, 296, 331, 2404, 2341, 2376, 2373, 266} \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {3 x}{40 \left (5-x^2\right )}+\frac {x}{4 \left (5-x^2\right )^2}-\frac {1}{8 \left (5-x^2\right ) x}-\frac {1}{4 \left (5-x^2\right )^2 x}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {\log (x)}{20 x^2}+\frac {3}{40 x}+\frac {\log (x)}{100} \]
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Rule 12
Rule 46
Rule 205
Rule 213
Rule 266
Rule 272
Rule 296
Rule 331
Rule 2341
Rule 2373
Rule 2376
Rule 2404
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {25-20 x-5 x^2+20 x^3-\left (50-30 x^2\right ) \log (x)}{4 x^3 \left (5-x^2\right )^3} \, dx \\ & = \frac {1}{4} \int \frac {25-20 x-5 x^2+20 x^3-\left (50-30 x^2\right ) \log (x)}{x^3 \left (5-x^2\right )^3} \, dx \\ & = \frac {1}{4} \int \frac {5 \left (5-4 x-x^2+4 x^3-10 \log (x)+6 x^2 \log (x)\right )}{x^3 \left (5-x^2\right )^3} \, dx \\ & = \frac {5}{4} \int \frac {5-4 x-x^2+4 x^3-10 \log (x)+6 x^2 \log (x)}{x^3 \left (5-x^2\right )^3} \, dx \\ & = \frac {5}{4} \int \left (-\frac {4}{\left (-5+x^2\right )^3}-\frac {5}{x^3 \left (-5+x^2\right )^3}+\frac {4}{x^2 \left (-5+x^2\right )^3}+\frac {1}{x \left (-5+x^2\right )^3}-\frac {2 \left (-5+3 x^2\right ) \log (x)}{x^3 \left (-5+x^2\right )^3}\right ) \, dx \\ & = \frac {5}{4} \int \frac {1}{x \left (-5+x^2\right )^3} \, dx-\frac {5}{2} \int \frac {\left (-5+3 x^2\right ) \log (x)}{x^3 \left (-5+x^2\right )^3} \, dx-5 \int \frac {1}{\left (-5+x^2\right )^3} \, dx+5 \int \frac {1}{x^2 \left (-5+x^2\right )^3} \, dx-\frac {25}{4} \int \frac {1}{x^3 \left (-5+x^2\right )^3} \, dx \\ & = -\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {5}{8} \text {Subst}\left (\int \frac {1}{(-5+x)^3 x} \, dx,x,x^2\right )+\frac {3}{4} \int \frac {1}{\left (-5+x^2\right )^2} \, dx-\frac {5}{4} \int \frac {1}{x^2 \left (-5+x^2\right )^2} \, dx-\frac {5}{2} \int \left (\frac {\log (x)}{25 x^3}+\frac {2 x \log (x)}{5 \left (-5+x^2\right )^3}-\frac {x \log (x)}{25 \left (-5+x^2\right )^2}\right ) \, dx-\frac {25}{8} \text {Subst}\left (\int \frac {1}{(-5+x)^3 x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}-\frac {3}{40} \int \frac {1}{-5+x^2} \, dx-\frac {1}{10} \int \frac {\log (x)}{x^3} \, dx+\frac {1}{10} \int \frac {x \log (x)}{\left (-5+x^2\right )^2} \, dx+\frac {3}{8} \int \frac {1}{x^2 \left (-5+x^2\right )} \, dx+\frac {5}{8} \text {Subst}\left (\int \left (\frac {1}{5 (-5+x)^3}-\frac {1}{25 (-5+x)^2}+\frac {1}{125 (-5+x)}-\frac {1}{125 x}\right ) \, dx,x,x^2\right )-\frac {25}{8} \text {Subst}\left (\int \left (\frac {1}{25 (-5+x)^3}-\frac {2}{125 (-5+x)^2}+\frac {3}{625 (-5+x)}-\frac {1}{125 x^2}-\frac {3}{625 x}\right ) \, dx,x,x^2\right )-\int \frac {x \log (x)}{\left (-5+x^2\right )^3} \, dx \\ & = \frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {1}{40 \left (5-x^2\right )}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {3 \text {arctanh}\left (\frac {x}{\sqrt {5}}\right )}{40 \sqrt {5}}+\frac {\log (x)}{50}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}-\frac {1}{100} \log \left (5-x^2\right )+\frac {1}{100} \int \frac {x}{-5+x^2} \, dx+\frac {3}{40} \int \frac {1}{-5+x^2} \, dx-\frac {1}{4} \int \frac {1}{x \left (-5+x^2\right )^2} \, dx \\ & = \frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {1}{40 \left (5-x^2\right )}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {\log (x)}{50}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}-\frac {1}{200} \log \left (5-x^2\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1}{(-5+x)^2 x} \, dx,x,x^2\right ) \\ & = \frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}+\frac {1}{40 \left (5-x^2\right )}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {\log (x)}{50}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )}-\frac {1}{200} \log \left (5-x^2\right )-\frac {1}{8} \text {Subst}\left (\int \left (\frac {1}{5 (-5+x)^2}-\frac {1}{25 (-5+x)}+\frac {1}{25 x}\right ) \, dx,x,x^2\right ) \\ & = \frac {3}{40 x}-\frac {1}{4 x \left (5-x^2\right )^2}+\frac {x}{4 \left (5-x^2\right )^2}-\frac {1}{8 x \left (5-x^2\right )}+\frac {3 x}{40 \left (5-x^2\right )}+\frac {\log (x)}{100}+\frac {\log (x)}{20 x^2}+\frac {\log (x)}{4 \left (5-x^2\right )^2}+\frac {x^2 \log (x)}{100 \left (5-x^2\right )} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {4 x+5 \log (x)}{4 x^2 \left (-5+x^2\right )^2} \]
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Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {x +\frac {5 \ln \left (x \right )}{4}}{\left (x^{2}-5\right )^{2} x^{2}}\) | \(18\) |
parallelrisch | \(\frac {40 x +50 \ln \left (x \right )}{40 x^{2} \left (x^{4}-10 x^{2}+25\right )}\) | \(26\) |
risch | \(\frac {5 \ln \left (x \right )}{4 x^{2} \left (x^{4}-10 x^{2}+25\right )}+\frac {1}{x \left (x^{4}-10 x^{2}+25\right )}\) | \(37\) |
default | \(\frac {1}{25 x}+\frac {\ln \left (x \right )}{50}-\frac {4 x^{3}+\frac {5}{2} x^{2}-40 x -\frac {25}{2}}{100 \left (x^{2}-5\right )^{2}}+\frac {1}{40 x^{2}-200}-\frac {\ln \left (x \right ) x^{2} \left (x^{2}-10\right )}{100 \left (x^{2}-5\right )^{2}}-\frac {\ln \left (x \right ) x^{2}}{100 \left (x^{2}-5\right )}+\frac {\ln \left (x \right )}{20 x^{2}}\) | \(84\) |
parts | \(\frac {1}{25 x}+\frac {\ln \left (x \right )}{50}-\frac {4 x^{3}+\frac {5}{2} x^{2}-40 x -\frac {25}{2}}{100 \left (x^{2}-5\right )^{2}}+\frac {1}{40 x^{2}-200}-\frac {\ln \left (x \right ) x^{2} \left (x^{2}-10\right )}{100 \left (x^{2}-5\right )^{2}}-\frac {\ln \left (x \right ) x^{2}}{100 \left (x^{2}-5\right )}+\frac {\ln \left (x \right )}{20 x^{2}}\) | \(84\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {4 \, x + 5 \, \log \left (x\right )}{4 \, {\left (x^{6} - 10 \, x^{4} + 25 \, x^{2}\right )}} \]
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Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {5 \log {\left (x \right )}}{4 x^{6} - 40 x^{4} + 100 x^{2}} + \frac {1}{x^{5} - 10 x^{3} + 25 x} \]
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Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (20) = 40\).
Time = 0.33 (sec) , antiderivative size = 158, normalized size of antiderivative = 7.18 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {10 \, x^{4} - 75 \, x^{2} - 2 \, {\left (2 \, x^{6} - 20 \, x^{4} + 50 \, x^{2} - 125\right )} \log \left (x\right ) + 125}{200 \, {\left (x^{6} - 10 \, x^{4} + 25 \, x^{2}\right )}} - \frac {6 \, x^{4} - 45 \, x^{2} + 50}{80 \, {\left (x^{6} - 10 \, x^{4} + 25 \, x^{2}\right )}} + \frac {3 \, x^{4} - 25 \, x^{2} + 40}{40 \, {\left (x^{5} - 10 \, x^{3} + 25 \, x\right )}} - \frac {3 \, x^{3} - 25 \, x}{40 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {2 \, x^{2} - 15}{80 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {1}{50} \, \log \left (x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.50 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=-\frac {1}{20} \, {\left (\frac {x^{2} - 10}{x^{4} - 10 \, x^{2} + 25} - \frac {1}{x^{2}}\right )} \log \left (x\right ) - \frac {x^{3} - 10 \, x}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} + \frac {1}{25 \, x} \]
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Time = 12.45 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-25+20 x+5 x^2-20 x^3+\left (50-30 x^2\right ) \log (x)}{-500 x^3+300 x^5-60 x^7+4 x^9} \, dx=\frac {4\,x+5\,\ln \left (x\right )}{4\,x^2\,{\left (x^2-5\right )}^2} \]
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