Integrand size = 47, antiderivative size = 24 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=6+x^2+x \left (-x (2+x+\log (1-x))+\log ^2(x)\right ) \]
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Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {6820, 786, 2442, 45, 2332, 2333} \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=-x^3-x^2-x^2 \log (1-x)+x \log ^2(x) \]
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Rule 45
Rule 786
Rule 2332
Rule 2333
Rule 2442
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {x \left (2-3 x^2\right )}{-1+x}-2 x \log (1-x)+2 \log (x)+\log ^2(x)\right ) \, dx \\ & = -(2 \int x \log (1-x) \, dx)+2 \int \log (x) \, dx+\int \frac {x \left (2-3 x^2\right )}{-1+x} \, dx+\int \log ^2(x) \, dx \\ & = -2 x-x^2 \log (1-x)+2 x \log (x)+x \log ^2(x)-2 \int \log (x) \, dx-\int \frac {x^2}{1-x} \, dx+\int \left (-1+\frac {1}{1-x}-3 x-3 x^2\right ) \, dx \\ & = -x-\frac {3 x^2}{2}-x^3-\log (1-x)-x^2 \log (1-x)+x \log ^2(x)-\int \left (-1+\frac {1}{1-x}-x\right ) \, dx \\ & = -x^2-x^3-x^2 \log (1-x)+x \log ^2(x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=\frac {7}{2}-x^2-x^3-\left (-1+x^2\right ) \log (1-x)-\log (-1+x)+x \log ^2(x) \]
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Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21
method | result | size |
risch | \(-x^{3}-x^{2} \ln \left (1-x \right )+x \ln \left (x \right )^{2}-x^{2}\) | \(29\) |
parallelrisch | \(-x^{3}-x^{2} \ln \left (1-x \right )+x \ln \left (x \right )^{2}-x^{2}-\ln \left (-1+x \right )+\ln \left (1-x \right )\) | \(41\) |
default | \(x \ln \left (x \right )^{2}+2 \left (1-x \right ) \ln \left (1-x \right )-\frac {3}{2}-\left (1-x \right )^{2} \ln \left (1-x \right )-x^{2}-x^{3}-\ln \left (-1+x \right )\) | \(53\) |
parts | \(x \ln \left (x \right )^{2}+2 \left (1-x \right ) \ln \left (1-x \right )-\frac {3}{2}-\left (1-x \right )^{2} \ln \left (1-x \right )-x^{2}-x^{3}-\ln \left (-1+x \right )\) | \(53\) |
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Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=-x^{3} + x \log \left (x\right )^{2} - x^{2} \log \left (-x + 1\right ) - x^{2} \]
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Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=- x^{3} - x^{2} + x \log {\left (x \right )}^{2} + \left (\frac {1}{3} - x^{2}\right ) \log {\left (1 - x \right )} - \frac {\log {\left (x - 1 \right )}}{3} \]
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Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=-x^{3} + x \log \left (x\right )^{2} - x^{2} - {\left (x^{2} + 2 \, x + 2 \, \log \left (x - 1\right )\right )} \log \left (-x + 1\right ) + 2 \, {\left (x + \log \left (x - 1\right )\right )} \log \left (-x + 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=-x^{3} + x \log \left (x\right )^{2} - x^{2} \log \left (-x + 1\right ) - x^{2} \]
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Time = 9.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {2 x-3 x^3+\left (2 x-2 x^2\right ) \log (1-x)+(-2+2 x) \log (x)+(-1+x) \log ^2(x)}{-1+x} \, dx=x\,{\ln \left (x\right )}^2-x^2\,\left (\ln \left (1-x\right )+1\right )-x^3 \]
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