Integrand size = 29, antiderivative size = 16 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=\frac {1}{9} (-1+4 x (-1+2 x)) \log (x) \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2350} \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=\frac {8}{9} x^2 \log (x)-\frac {4}{9} x \log (x)-\frac {\log (x)}{9} \]
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Rule 12
Rule 14
Rule 2350
Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{x} \, dx \\ & = \frac {1}{9} \int \left (\frac {-1-4 x+8 x^2}{x}+4 (-1+4 x) \log (x)\right ) \, dx \\ & = \frac {1}{9} \int \frac {-1-4 x+8 x^2}{x} \, dx+\frac {4}{9} \int (-1+4 x) \log (x) \, dx \\ & = -\frac {4}{9} x \log (x)+\frac {8}{9} x^2 \log (x)+\frac {1}{9} \int \left (-4-\frac {1}{x}+8 x\right ) \, dx-\frac {4}{9} \int (-1+2 x) \, dx \\ & = -\frac {\log (x)}{9}-\frac {4}{9} x \log (x)+\frac {8}{9} x^2 \log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=-\frac {\log (x)}{9}-\frac {4}{9} x \log (x)+\frac {8}{9} x^2 \log (x) \]
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Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {8 x^{2} \ln \left (x \right )}{9}-\frac {4 x \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\) | \(18\) |
norman | \(\frac {8 x^{2} \ln \left (x \right )}{9}-\frac {4 x \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\) | \(18\) |
parallelrisch | \(\frac {8 x^{2} \ln \left (x \right )}{9}-\frac {4 x \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\) | \(18\) |
parts | \(\frac {8 x^{2} \ln \left (x \right )}{9}-\frac {4 x \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\) | \(18\) |
risch | \(\frac {\left (8 x^{2}-4 x \right ) \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\) | \(19\) |
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Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=\frac {1}{9} \, {\left (8 \, x^{2} - 4 \, x - 1\right )} \log \left (x\right ) \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=\left (\frac {8 x^{2}}{9} - \frac {4 x}{9}\right ) \log {\left (x \right )} - \frac {\log {\left (x \right )}}{9} \]
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Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=\frac {8}{9} \, x^{2} \log \left (x\right ) - \frac {4}{9} \, x \log \left (x\right ) - \frac {1}{9} \, \log \left (x\right ) \]
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Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=\frac {4}{9} \, {\left (2 \, x^{2} - x\right )} \log \left (x\right ) - \frac {1}{9} \, \log \left (x\right ) \]
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Time = 14.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{9 x} \, dx=-\frac {\ln \left (x\right )\,\left (-8\,x^2+4\,x+1\right )}{9} \]
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