Integrand size = 22, antiderivative size = 24 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=\frac {2 \left (3+\frac {5}{x}\right ) (-5+x) (x-x \log (x))}{5 x} \]
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Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 14, 45, 2372} \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=\frac {6 x}{5}-\frac {10}{x}-\frac {6}{5} x \log (x)+4 \log (x)+\frac {10 \log (x)}{x} \]
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Rule 12
Rule 14
Rule 45
Rule 2372
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{x^2} \, dx \\ & = \frac {1}{5} \int \left (\frac {20 (5+x)}{x^2}-\frac {2 \left (25+3 x^2\right ) \log (x)}{x^2}\right ) \, dx \\ & = -\left (\frac {2}{5} \int \frac {\left (25+3 x^2\right ) \log (x)}{x^2} \, dx\right )+4 \int \frac {5+x}{x^2} \, dx \\ & = \frac {10 \log (x)}{x}-\frac {6}{5} x \log (x)+\frac {2}{5} \int \left (3-\frac {25}{x^2}\right ) \, dx+4 \int \left (\frac {5}{x^2}+\frac {1}{x}\right ) \, dx \\ & = -\frac {10}{x}+\frac {6 x}{5}+4 \log (x)+\frac {10 \log (x)}{x}-\frac {6}{5} x \log (x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=-\frac {10}{x}+\frac {6 x}{5}+4 \log (x)+\frac {10 \log (x)}{x}-\frac {6}{5} x \log (x) \]
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Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
default | \(-\frac {6 x \ln \left (x \right )}{5}+\frac {6 x}{5}+\frac {10 \ln \left (x \right )}{x}-\frac {10}{x}+4 \ln \left (x \right )\) | \(26\) |
parts | \(-\frac {6 x \ln \left (x \right )}{5}+\frac {6 x}{5}+\frac {10 \ln \left (x \right )}{x}-\frac {10}{x}+4 \ln \left (x \right )\) | \(26\) |
norman | \(\frac {-10+4 x \ln \left (x \right )+\frac {6 x^{2}}{5}-\frac {6 x^{2} \ln \left (x \right )}{5}+10 \ln \left (x \right )}{x}\) | \(28\) |
parallelrisch | \(\frac {-6 x^{2} \ln \left (x \right )+6 x^{2}+20 x \ln \left (x \right )-50+50 \ln \left (x \right )}{5 x}\) | \(29\) |
risch | \(-\frac {2 \left (3 x^{2}-25\right ) \ln \left (x \right )}{5 x}+\frac {4 x \ln \left (x \right )+\frac {6 x^{2}}{5}-10}{x}\) | \(33\) |
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Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=\frac {2 \, {\left (3 \, x^{2} - {\left (3 \, x^{2} - 10 \, x - 25\right )} \log \left (x\right ) - 25\right )}}{5 \, x} \]
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Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=\frac {6 x}{5} + 4 \log {\left (x \right )} + \frac {\left (50 - 6 x^{2}\right ) \log {\left (x \right )}}{5 x} - \frac {10}{x} \]
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Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=-\frac {6}{5} \, x \log \left (x\right ) + \frac {6}{5} \, x + \frac {10 \, \log \left (x\right )}{x} - \frac {10}{x} + 4 \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=-\frac {2}{5} \, {\left (3 \, x - \frac {25}{x}\right )} \log \left (x\right ) + \frac {6}{5} \, x - \frac {10}{x} + 4 \, \log \left (x\right ) \]
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Time = 8.66 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {100+20 x+\left (-50-6 x^2\right ) \log (x)}{5 x^2} \, dx=4\,\ln \left (x\right )-x\,\left (\frac {6\,\ln \left (x\right )}{5}-\frac {6}{5}\right )+\frac {10\,\ln \left (x\right )-10}{x} \]
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