Integrand size = 63, antiderivative size = 34 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=e^4-\frac {4+x}{5 x}+\frac {\log \left (\left (-\frac {3}{x^2}+\frac {20+x}{x}\right )^2\right )}{x} \]
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Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1608, 6820, 12, 14, 1642, 646, 31, 2605, 814} \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log \left (\frac {\left (-x^2-20 x+3\right )^2}{x^4}\right )}{x}-\frac {4}{5 x} \]
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Rule 12
Rule 14
Rule 31
Rule 646
Rule 814
Rule 1608
Rule 1642
Rule 2605
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{x^2 \left (-15+100 x+5 x^2\right )} \, dx \\ & = \int \frac {\frac {4 \left (12-30 x+x^2\right )}{-3+20 x+x^2}-5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{5 x^2} \, dx \\ & = \frac {1}{5} \int \frac {\frac {4 \left (12-30 x+x^2\right )}{-3+20 x+x^2}-5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {4 \left (12-30 x+x^2\right )}{x^2 \left (-3+20 x+x^2\right )}-\frac {5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x^2}\right ) \, dx \\ & = \frac {4}{5} \int \frac {12-30 x+x^2}{x^2 \left (-3+20 x+x^2\right )} \, dx-\int \frac {\log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x^2} \, dx \\ & = \frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}+\frac {4}{5} \int \left (-\frac {4}{x^2}-\frac {50}{3 x}+\frac {5 (203+10 x)}{3 \left (-3+20 x+x^2\right )}\right ) \, dx-\int \frac {-12+40 x}{x^2 \left (3-20 x-x^2\right )} \, dx \\ & = \frac {16}{5 x}-\frac {40 \log (x)}{3}+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}+\frac {4}{3} \int \frac {203+10 x}{-3+20 x+x^2} \, dx-\int \left (-\frac {4}{x^2}-\frac {40}{3 x}+\frac {4 (203+10 x)}{3 \left (-3+20 x+x^2\right )}\right ) \, dx \\ & = -\frac {4}{5 x}+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}-\frac {4}{3} \int \frac {203+10 x}{-3+20 x+x^2} \, dx+\frac {1}{3} \left (2 \left (10-\sqrt {103}\right )\right ) \int \frac {1}{10+\sqrt {103}+x} \, dx+\frac {1}{3} \left (2 \left (10+\sqrt {103}\right )\right ) \int \frac {1}{10-\sqrt {103}+x} \, dx \\ & = -\frac {4}{5 x}+\frac {2}{3} \left (10+\sqrt {103}\right ) \log \left (10-\sqrt {103}+x\right )+\frac {2}{3} \left (10-\sqrt {103}\right ) \log \left (10+\sqrt {103}+x\right )+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x}-\frac {1}{3} \left (2 \left (10-\sqrt {103}\right )\right ) \int \frac {1}{10+\sqrt {103}+x} \, dx-\frac {1}{3} \left (2 \left (10+\sqrt {103}\right )\right ) \int \frac {1}{10-\sqrt {103}+x} \, dx \\ & = -\frac {4}{5 x}+\frac {\log \left (\frac {\left (3-20 x-x^2\right )^2}{x^4}\right )}{x} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {1}{5} \left (-\frac {4}{x}+\frac {5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x}\right ) \]
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Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {-\frac {4}{5}+\ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{x}\) | \(30\) |
parallelrisch | \(-\frac {8-10 \ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{10 x}\) | \(33\) |
derivativedivides | \(-\frac {4}{5 x}+\frac {\ln \left (1+\frac {394}{x^{2}}-\frac {120}{x^{3}}+\frac {40}{x}+\frac {9}{x^{4}}\right )}{x}\) | \(34\) |
default | \(-\frac {4}{5 x}+\frac {\ln \left (1+\frac {394}{x^{2}}-\frac {120}{x^{3}}+\frac {40}{x}+\frac {9}{x^{4}}\right )}{x}\) | \(34\) |
risch | \(\frac {\ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{x}-\frac {4}{5 x}\) | \(34\) |
parts | \(\frac {\ln \left (1+\frac {394}{x^{2}}-\frac {120}{x^{3}}+\frac {40}{x}+\frac {9}{x^{4}}\right )}{x}-\frac {4}{5 x}-\frac {20 \ln \left (\frac {3}{x^{2}}-\frac {20}{x}-1\right )}{3}+\frac {4 \sqrt {103}\, \operatorname {arctanh}\left (\frac {\left (\frac {6}{x}-20\right ) \sqrt {103}}{206}\right )}{3}-\frac {40 \ln \left (x \right )}{3}+\frac {20 \ln \left (x^{2}+20 x -3\right )}{3}-\frac {4 \sqrt {103}\, \operatorname {arctanh}\left (\frac {\left (2 x +20\right ) \sqrt {103}}{206}\right )}{3}\) | \(98\) |
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Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {5 \, \log \left (\frac {x^{4} + 40 \, x^{3} + 394 \, x^{2} - 120 \, x + 9}{x^{4}}\right ) - 4}{5 \, x} \]
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Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log {\left (\frac {x^{4} + 40 x^{3} + 394 x^{2} - 120 x + 9}{x^{4}} \right )}}{x} - \frac {4}{5 x} \]
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Time = 0.33 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=-\frac {2 \, {\left ({\left (16 \, x - 3\right )} \log \left (x^{2} + 20 \, x - 3\right ) - 2 \, {\left (16 \, x - 3\right )} \log \left (x\right ) + 6\right )}}{3 \, x} + \frac {16}{5 \, x} + \frac {32}{3} \, \log \left (x^{2} + 20 \, x - 3\right ) - \frac {64}{3} \, \log \left (x\right ) \]
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Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log \left (\frac {x^{4} + 40 \, x^{3} + 394 \, x^{2} - 120 \, x + 9}{x^{4}}\right )}{x} - \frac {4}{5 \, x} \]
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Time = 16.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\ln \left (\frac {x^4+40\,x^3+394\,x^2-120\,x+9}{x^4}\right )-\frac {4}{5}}{x} \]
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