Integrand size = 23, antiderivative size = 23 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \arctan (x)+2 \log (1-x)+\frac {1}{2} \log \left (1+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1643, 649, 209, 266} \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \arctan (x)+\frac {1}{2} \log \left (x^2+1\right )+2 \log (1-x) \]
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Rule 209
Rule 266
Rule 649
Rule 1643
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{-1+x}+\frac {-3+x}{1+x^2}\right ) \, dx \\ & = 2 \log (1-x)+\int \frac {-3+x}{1+x^2} \, dx \\ & = 2 \log (1-x)-3 \int \frac {1}{1+x^2} \, dx+\int \frac {x}{1+x^2} \, dx \\ & = -3 \tan ^{-1}(x)+2 \log (1-x)+\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \arctan (x)+\frac {1}{2} \log \left (2+2 (-1+x)+(-1+x)^2\right )+2 \log (-1+x) \]
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Time = 0.76 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\ln \left (x^{2}+1\right )}{2}-3 \arctan \left (x \right )+2 \ln \left (x -1\right )\) | \(20\) |
risch | \(2 \ln \left (x -1\right )+\frac {\ln \left (9 x^{2}+9\right )}{2}-3 \arctan \left (x \right )\) | \(22\) |
parallelrisch | \(2 \ln \left (x -1\right )+\frac {\ln \left (x -i\right )}{2}+\frac {3 i \ln \left (x -i\right )}{2}+\frac {\ln \left (x +i\right )}{2}-\frac {3 i \ln \left (x +i\right )}{2}\) | \(38\) |
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none
Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) + 2 \, \log \left (x - 1\right ) \]
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Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=2 \log {\left (x - 1 \right )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} - 3 \operatorname {atan}{\left (x \right )} \]
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none
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) + 2 \, \log \left (x - 1\right ) \]
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none
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=-3 \, \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {5-4 x+3 x^2}{(-1+x) \left (1+x^2\right )} \, dx=2\,\ln \left (x-1\right )+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {3}{2}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{2}-\frac {3}{2}{}\mathrm {i}\right ) \]
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