Integrand size = 56, antiderivative size = 33 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=\frac {x-x^2+2 \left (-5+\frac {3}{x}-x^2-\log (2)+\log (1+x)\right )}{x} \]
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Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1607, 6874, 1634, 2442, 36, 29, 31} \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=\frac {6}{x^2}-3 x+\frac {2 \log (x+1)}{x}-\frac {10+\log (4)}{x} \]
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Rule 29
Rule 31
Rule 36
Rule 1607
Rule 1634
Rule 2442
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3 (1+x)} \, dx \\ & = \int \left (\frac {-12-3 x^3-3 x^4-x (2-\log (4))+x^2 (12+\log (4))}{x^3 (1+x)}-\frac {2 \log (1+x)}{x^2}\right ) \, dx \\ & = -\left (2 \int \frac {\log (1+x)}{x^2} \, dx\right )+\int \frac {-12-3 x^3-3 x^4-x (2-\log (4))+x^2 (12+\log (4))}{x^3 (1+x)} \, dx \\ & = \frac {2 \log (1+x)}{x}-2 \int \frac {1}{x (1+x)} \, dx+\int \left (-3-\frac {12}{x^3}+\frac {2}{x}-\frac {2}{1+x}+\frac {10+\log (4)}{x^2}\right ) \, dx \\ & = \frac {6}{x^2}-3 x-\frac {10+\log (4)}{x}+2 \log (x)-2 \log (1+x)+\frac {2 \log (1+x)}{x}-2 \int \frac {1}{x} \, dx+2 \int \frac {1}{1+x} \, dx \\ & = \frac {6}{x^2}-3 x-\frac {10+\log (4)}{x}+\frac {2 \log (1+x)}{x} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=\frac {6}{x^2}-\frac {10}{x}-3 x-\frac {2 \log (2)}{x}+\frac {2 \log (1+x)}{x} \]
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Time = 0.63 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82
method | result | size |
norman | \(\frac {6+\left (-10-2 \ln \left (2\right )\right ) x -3 x^{3}+2 \ln \left (1+x \right ) x}{x^{2}}\) | \(27\) |
risch | \(\frac {2 \ln \left (1+x \right )}{x}-\frac {3 x^{3}+2 x \ln \left (2\right )+10 x -6}{x^{2}}\) | \(31\) |
parallelrisch | \(\frac {-3 x^{3}+6-2 x \ln \left (2\right )+6 x^{2}+2 \ln \left (1+x \right ) x -10 x}{x^{2}}\) | \(32\) |
parts | \(-3 x -2 \ln \left (1+x \right )-\frac {10+2 \ln \left (2\right )}{x}+\frac {6}{x^{2}}+\frac {2 \ln \left (1+x \right ) \left (1+x \right )}{x}\) | \(39\) |
derivativedivides | \(-\frac {2 \ln \left (2\right )}{x}+\frac {2 \ln \left (1+x \right ) \left (1+x \right )}{x}-3-3 x +\frac {6}{x^{2}}-\frac {10}{x}-2 \ln \left (1+x \right )\) | \(41\) |
default | \(-\frac {2 \ln \left (2\right )}{x}+\frac {2 \ln \left (1+x \right ) \left (1+x \right )}{x}-3-3 x +\frac {6}{x^{2}}-\frac {10}{x}-2 \ln \left (1+x \right )\) | \(41\) |
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Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=-\frac {3 \, x^{3} + 2 \, x \log \left (2\right ) - 2 \, x \log \left (x + 1\right ) + 10 \, x - 6}{x^{2}} \]
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Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=- 3 x + \frac {2 \log {\left (x + 1 \right )}}{x} - \frac {x \left (2 \log {\left (2 \right )} + 10\right ) - 6}{x^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (30) = 60\).
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=-2 \, {\left (\frac {1}{x} - \log \left (x + 1\right ) + \log \left (x\right )\right )} \log \left (2\right ) - 2 \, {\left (\log \left (x + 1\right ) - \log \left (x\right )\right )} \log \left (2\right ) - 3 \, x + \frac {2 \, {\left (x + 1\right )} \log \left (x + 1\right )}{x} - \frac {6 \, {\left (2 \, x - 1\right )}}{x^{2}} + \frac {2}{x} - 2 \, \log \left (x + 1\right ) \]
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Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=-3 \, x + \frac {2 \, \log \left (x + 1\right )}{x} - \frac {2 \, {\left (x \log \left (2\right ) + 5 \, x - 3\right )}}{x^{2}} \]
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Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {-12-2 x+12 x^2-3 x^3-3 x^4+\left (2 x+2 x^2\right ) \log (2)+\left (-2 x-2 x^2\right ) \log (1+x)}{x^3+x^4} \, dx=-3\,x-\frac {x\,\left (2\,\ln \left (2\right )-2\,\ln \left (x+1\right )+10\right )-6}{x^2} \]
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