Integrand size = 21, antiderivative size = 56 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {5+3 x}{12 \left (2+3 x+x^2\right )}-\frac {1}{36} \log (1-x)+\frac {1}{144} \log (2-x)-\frac {7}{36} \log (1+x)+\frac {31}{144} \log (2+x) \]
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Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1600, 988, 1086, 646, 31} \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {3 x+5}{12 \left (x^2+3 x+2\right )}-\frac {1}{36} \log (1-x)+\frac {1}{144} \log (2-x)-\frac {7}{36} \log (x+1)+\frac {31}{144} \log (x+2) \]
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Rule 31
Rule 646
Rule 988
Rule 1086
Rule 1600
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (2-3 x+x^2\right ) \left (2+3 x+x^2\right )^2} \, dx \\ & = -\frac {5+3 x}{12 \left (2+3 x+x^2\right )}+\frac {1}{72} \int \frac {-18+48 x-18 x^2}{\left (2-3 x+x^2\right ) \left (2+3 x+x^2\right )} \, dx \\ & = -\frac {5+3 x}{12 \left (2+3 x+x^2\right )}+\frac {\int \frac {252-108 x}{2-3 x+x^2} \, dx}{5184}+\frac {\int \frac {-900+108 x}{2+3 x+x^2} \, dx}{5184} \\ & = -\frac {5+3 x}{12 \left (2+3 x+x^2\right )}+\frac {1}{144} \int \frac {1}{-2+x} \, dx-\frac {1}{36} \int \frac {1}{-1+x} \, dx-\frac {7}{36} \int \frac {1}{1+x} \, dx+\frac {31}{144} \int \frac {1}{2+x} \, dx \\ & = -\frac {5+3 x}{12 \left (2+3 x+x^2\right )}-\frac {1}{36} \log (1-x)+\frac {1}{144} \log (2-x)-\frac {7}{36} \log (1+x)+\frac {31}{144} \log (2+x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {1}{144} \left (-\frac {12 (5+3 x)}{2+3 x+x^2}-4 \log (1-x)+\log (2-x)-28 \log (1+x)+31 \log (2+x)\right ) \]
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Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.71
method | result | size |
default | \(-\frac {1}{12 \left (x +2\right )}+\frac {31 \ln \left (x +2\right )}{144}-\frac {1}{6 \left (x +1\right )}-\frac {7 \ln \left (x +1\right )}{36}-\frac {\ln \left (x -1\right )}{36}+\frac {\ln \left (x -2\right )}{144}\) | \(40\) |
risch | \(\frac {-\frac {x}{4}-\frac {5}{12}}{x^{2}+3 x +2}+\frac {\ln \left (x -2\right )}{144}-\frac {\ln \left (x -1\right )}{36}-\frac {7 \ln \left (x +1\right )}{36}+\frac {31 \ln \left (x +2\right )}{144}\) | \(42\) |
norman | \(\frac {\frac {1}{3} x^{2}+\frac {3}{4} x -\frac {1}{4} x^{3}-\frac {5}{6}}{x^{4}-5 x^{2}+4}+\frac {\ln \left (x -2\right )}{144}-\frac {\ln \left (x -1\right )}{36}-\frac {7 \ln \left (x +1\right )}{36}+\frac {31 \ln \left (x +2\right )}{144}\) | \(54\) |
parallelrisch | \(\frac {\ln \left (x -2\right ) x^{2}-4 \ln \left (x -1\right ) x^{2}-28 \ln \left (x +1\right ) x^{2}+31 \ln \left (x +2\right ) x^{2}-60+3 \ln \left (x -2\right ) x -12 \ln \left (x -1\right ) x -84 \ln \left (x +1\right ) x +93 \ln \left (x +2\right ) x +2 \ln \left (x -2\right )-8 \ln \left (x -1\right )-56 \ln \left (x +1\right )+62 \ln \left (x +2\right )-36 x}{144 x^{2}+432 x +288}\) | \(105\) |
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Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.29 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {31 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x + 2\right ) - 28 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x + 1\right ) - 4 \, {\left (x^{2} + 3 \, x + 2\right )} \log \left (x - 1\right ) + {\left (x^{2} + 3 \, x + 2\right )} \log \left (x - 2\right ) - 36 \, x - 60}{144 \, {\left (x^{2} + 3 \, x + 2\right )}} \]
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Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {- 3 x - 5}{12 x^{2} + 36 x + 24} + \frac {\log {\left (x - 2 \right )}}{144} - \frac {\log {\left (x - 1 \right )}}{36} - \frac {7 \log {\left (x + 1 \right )}}{36} + \frac {31 \log {\left (x + 2 \right )}}{144} \]
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Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {3 \, x + 5}{12 \, {\left (x^{2} + 3 \, x + 2\right )}} + \frac {31}{144} \, \log \left (x + 2\right ) - \frac {7}{36} \, \log \left (x + 1\right ) - \frac {1}{36} \, \log \left (x - 1\right ) + \frac {1}{144} \, \log \left (x - 2\right ) \]
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Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=-\frac {3 \, x + 5}{12 \, {\left (x + 2\right )} {\left (x + 1\right )}} + \frac {31}{144} \, \log \left ({\left | x + 2 \right |}\right ) - \frac {7}{36} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{36} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{144} \, \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.75 \[ \int \frac {2-3 x+x^2}{\left (4-5 x^2+x^4\right )^2} \, dx=\frac {\ln \left (x-2\right )}{144}-\frac {7\,\ln \left (x+1\right )}{36}-\frac {\ln \left (x-1\right )}{36}+\frac {31\,\ln \left (x+2\right )}{144}-\frac {\frac {x}{4}+\frac {5}{12}}{x^2+3\,x+2} \]
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