Integrand size = 37, antiderivative size = 18 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=12+\frac {5 x}{3}+\frac {\log (x)}{10 (2+x)} \]
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Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1608, 27, 12, 6874, 46, 45, 2351, 31} \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {5 x}{3}-\frac {x \log (x)}{20 (x+2)}+\frac {\log (x)}{20} \]
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Rule 12
Rule 27
Rule 31
Rule 45
Rule 46
Rule 1608
Rule 2351
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{x \left (120+120 x+30 x^2\right )} \, dx \\ & = \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{30 x (2+x)^2} \, dx \\ & = \frac {1}{30} \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{x (2+x)^2} \, dx \\ & = \frac {1}{30} \int \left (\frac {203}{(2+x)^2}+\frac {6}{x (2+x)^2}+\frac {200 x}{(2+x)^2}+\frac {50 x^2}{(2+x)^2}-\frac {3 \log (x)}{(2+x)^2}\right ) \, dx \\ & = -\frac {203}{30 (2+x)}-\frac {1}{10} \int \frac {\log (x)}{(2+x)^2} \, dx+\frac {1}{5} \int \frac {1}{x (2+x)^2} \, dx+\frac {5}{3} \int \frac {x^2}{(2+x)^2} \, dx+\frac {20}{3} \int \frac {x}{(2+x)^2} \, dx \\ & = -\frac {203}{30 (2+x)}-\frac {x \log (x)}{20 (2+x)}+\frac {1}{20} \int \frac {1}{2+x} \, dx+\frac {1}{5} \int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx+\frac {5}{3} \int \left (1+\frac {4}{(2+x)^2}-\frac {4}{2+x}\right ) \, dx+\frac {20}{3} \int \left (-\frac {2}{(2+x)^2}+\frac {1}{2+x}\right ) \, dx \\ & = \frac {5 x}{3}+\frac {\log (x)}{20}-\frac {x \log (x)}{20 (2+x)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {1}{30} \left (50 x+\frac {3 \log (x)}{2+x}\right ) \]
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Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\ln \left (x \right )}{10 x +20}+\frac {5 x}{3}\) | \(14\) |
norman | \(\frac {\frac {\ln \left (x \right )}{10}+\frac {5 x^{2}}{3}-\frac {20}{3}}{2+x}\) | \(18\) |
default | \(\frac {5 x}{3}+\frac {\ln \left (x \right )}{20}-\frac {\ln \left (x \right ) x}{20 \left (2+x \right )}\) | \(19\) |
parallelrisch | \(\frac {50 x^{2}-200+3 \ln \left (x \right )}{60+30 x}\) | \(19\) |
parts | \(\frac {5 x}{3}+\frac {\ln \left (x \right )}{20}-\frac {\ln \left (x \right ) x}{20 \left (2+x \right )}\) | \(19\) |
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Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {50 \, x^{2} + 100 \, x + 3 \, \log \left (x\right )}{30 \, {\left (x + 2\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {5 x}{3} + \frac {\log {\left (x \right )}}{10 x + 20} \]
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Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {5}{3} \, x + \frac {\log \left (x\right )}{10 \, {\left (x + 2\right )}} \]
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Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {5}{3} \, x + \frac {\log \left (x\right )}{10 \, {\left (x + 2\right )}} \]
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Time = 12.88 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx=\frac {5\,x}{3}+\frac {\ln \left (x\right )}{10\,\left (x+2\right )} \]
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