Integrand size = 28, antiderivative size = 20 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=7-\frac {4}{15} x \left (-\frac {1}{x}+x\right ) \log \left (x+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {12, 14, 2581, 30, 45} \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=-\frac {4}{15} x^2 \log (x (x+1))+\frac {4 \log (x)}{15}+\frac {4}{15} \log (x+1) \]
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Rule 12
Rule 14
Rule 30
Rule 45
Rule 2581
Rubi steps \begin{align*} \text {integral}& = \frac {1}{15} \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{x} \, dx \\ & = \frac {1}{15} \int \left (-\frac {4 \left (-1-x+2 x^2\right )}{x}-8 x \log (x (1+x))\right ) \, dx \\ & = -\left (\frac {4}{15} \int \frac {-1-x+2 x^2}{x} \, dx\right )-\frac {8}{15} \int x \log (x (1+x)) \, dx \\ & = -\frac {4}{15} x^2 \log (x (1+x))+\frac {4 \int x \, dx}{15}+\frac {4}{15} \int \frac {x^2}{1+x} \, dx-\frac {4}{15} \int \left (-1-\frac {1}{x}+2 x\right ) \, dx \\ & = \frac {4 x}{15}-\frac {2 x^2}{15}+\frac {4 \log (x)}{15}-\frac {4}{15} x^2 \log (x (1+x))+\frac {4}{15} \int \left (-1+x+\frac {1}{1+x}\right ) \, dx \\ & = \frac {4 \log (x)}{15}+\frac {4}{15} \log (1+x)-\frac {4}{15} x^2 \log (x (1+x)) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=-\frac {4}{15} \left (-\log (x)-\log (1+x)+x^2 \log (x (1+x))\right ) \]
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Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
norman | \(-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (x^{2}+x \right )}{15}\) | \(21\) |
risch | \(-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (x^{2}+x \right )}{15}\) | \(21\) |
parallelrisch | \(-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (x^{2}+x \right )}{15}\) | \(21\) |
default | \(\frac {4 \ln \left (x \right )}{15}-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (1+x \right )}{15}\) | \(23\) |
parts | \(\frac {4 \ln \left (x \right )}{15}-\frac {4 x^{2} \ln \left (x^{2}+x \right )}{15}+\frac {4 \ln \left (1+x \right )}{15}\) | \(23\) |
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Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=-\frac {4}{15} \, {\left (x^{2} - 1\right )} \log \left (x^{2} + x\right ) \]
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Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=- \frac {4 x^{2} \log {\left (x^{2} + x \right )}}{15} + \frac {4 \log {\left (x^{2} + x \right )}}{15} \]
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Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=-\frac {4}{15} \, x^{2} \log \left (x^{2} + x\right ) + \frac {4}{15} \, \log \left (x + 1\right ) + \frac {4}{15} \, \log \left (x\right ) \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=-\frac {4}{15} \, x^{2} \log \left (x^{2} + x\right ) + \frac {4}{15} \, \log \left (x + 1\right ) + \frac {4}{15} \, \log \left (x\right ) \]
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Time = 9.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {4+4 x-8 x^2-8 x^2 \log \left (x+x^2\right )}{15 x} \, dx=-\frac {4\,\ln \left (x^2+x\right )\,\left (x^2-1\right )}{15} \]
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