Integrand size = 26, antiderivative size = 38 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=\frac {1-x}{5-4 x+x^2}-2 \arctan (2-x)+\frac {5}{2} \log \left (5-4 x+x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1674, 648, 632, 210, 642} \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=-2 \arctan (2-x)+\frac {1-x}{x^2-4 x+5}+\frac {5}{2} \log \left (x^2-4 x+5\right ) \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1674
Rubi steps \begin{align*} \text {integral}& = \frac {1-x}{5-4 x+x^2}+\frac {1}{4} \int \frac {-32+20 x}{5-4 x+x^2} \, dx \\ & = \frac {1-x}{5-4 x+x^2}+2 \int \frac {1}{5-4 x+x^2} \, dx+\frac {5}{2} \int \frac {-4+2 x}{5-4 x+x^2} \, dx \\ & = \frac {1-x}{5-4 x+x^2}+\frac {5}{2} \log \left (5-4 x+x^2\right )-4 \text {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,-4+2 x\right ) \\ & = \frac {1-x}{5-4 x+x^2}-2 \arctan (2-x)+\frac {5}{2} \log \left (5-4 x+x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=\frac {1-x}{5-4 x+x^2}-2 \arctan (2-x)+\frac {5}{2} \log \left (5-4 x+x^2\right ) \]
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Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {1-x}{x^{2}-4 x +5}+2 \arctan \left (-2+x \right )+\frac {5 \ln \left (x^{2}-4 x +5\right )}{2}\) | \(35\) |
risch | \(\frac {1-x}{x^{2}-4 x +5}+2 \arctan \left (-2+x \right )+\frac {5 \ln \left (x^{2}-4 x +5\right )}{2}\) | \(35\) |
parallelrisch | \(-\frac {-40 i \ln \left (x -2-i\right ) x -10 i \ln \left (x -2+i\right ) x^{2}-50 i \ln \left (x -2+i\right )-25 \ln \left (x -2-i\right ) x^{2}+10 i \ln \left (x -2-i\right ) x^{2}-25 \ln \left (x -2+i\right ) x^{2}+40 i \ln \left (x -2+i\right ) x +100 \ln \left (x -2-i\right ) x +50 i \ln \left (x -2-i\right )+100 \ln \left (x -2+i\right ) x +2 x^{2}-125 \ln \left (x -2-i\right )-125 \ln \left (x -2+i\right )+2 x}{10 \left (x^{2}-4 x +5\right )}\) | \(140\) |
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Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=\frac {4 \, {\left (x^{2} - 4 \, x + 5\right )} \arctan \left (x - 2\right ) + 5 \, {\left (x^{2} - 4 \, x + 5\right )} \log \left (x^{2} - 4 \, x + 5\right ) - 2 \, x + 2}{2 \, {\left (x^{2} - 4 \, x + 5\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=\frac {1 - x}{x^{2} - 4 x + 5} + \frac {5 \log {\left (x^{2} - 4 x + 5 \right )}}{2} + 2 \operatorname {atan}{\left (x - 2 \right )} \]
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Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=-\frac {x - 1}{x^{2} - 4 \, x + 5} + 2 \, \arctan \left (x - 2\right ) + \frac {5}{2} \, \log \left (x^{2} - 4 \, x + 5\right ) \]
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Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=-\frac {x - 1}{x^{2} - 4 \, x + 5} + 2 \, \arctan \left (x - 2\right ) + \frac {5}{2} \, \log \left (x^{2} - 4 \, x + 5\right ) \]
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Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {-41+55 x-27 x^2+5 x^3}{\left (5-4 x+x^2\right )^2} \, dx=2\,\mathrm {atan}\left (x-2\right )+\frac {5\,\ln \left (x^2-4\,x+5\right )}{2}-\frac {x}{x^2-4\,x+5}+\frac {1}{x^2-4\,x+5} \]
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