Integrand size = 12, antiderivative size = 42 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=-\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (1+3 x) \]
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Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {628, 630, 31} \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=-\frac {5-6 x}{49 \left (-3 x^2+5 x+2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (3 x+1) \]
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Rule 31
Rule 628
Rule 630
Rubi steps \begin{align*} \text {integral}& = -\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}+\frac {6}{49} \int \frac {1}{2+5 x-3 x^2} \, dx \\ & = -\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}-\frac {18}{343} \int \frac {1}{-1-3 x} \, dx+\frac {18}{343} \int \frac {1}{6-3 x} \, dx \\ & = -\frac {5-6 x}{49 \left (2+5 x-3 x^2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (1+3 x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=\frac {5-6 x}{49 \left (-2-5 x+3 x^2\right )}-\frac {6}{343} \log (2-x)+\frac {6}{343} \log (1+3 x) \]
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Time = 2.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76
method | result | size |
default | \(-\frac {1}{49 \left (-2+x \right )}-\frac {6 \ln \left (-2+x \right )}{343}-\frac {3}{49 \left (3 x +1\right )}+\frac {6 \ln \left (3 x +1\right )}{343}\) | \(32\) |
risch | \(\frac {-\frac {2 x}{49}+\frac {5}{147}}{x^{2}-\frac {5}{3} x -\frac {2}{3}}-\frac {6 \ln \left (-2+x \right )}{343}+\frac {6 \ln \left (3 x +1\right )}{343}\) | \(32\) |
norman | \(\frac {\frac {15}{98} x^{2}-\frac {37}{98} x}{3 x^{2}-5 x -2}-\frac {6 \ln \left (-2+x \right )}{343}+\frac {6 \ln \left (3 x +1\right )}{343}\) | \(38\) |
parallelrisch | \(-\frac {36 \ln \left (-2+x \right ) x^{2}-36 \ln \left (x +\frac {1}{3}\right ) x^{2}-60 \ln \left (-2+x \right ) x +60 \ln \left (x +\frac {1}{3}\right ) x -105 x^{2}-24 \ln \left (-2+x \right )+24 \ln \left (x +\frac {1}{3}\right )+259 x}{686 \left (3 x^{2}-5 x -2\right )}\) | \(68\) |
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Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=\frac {6 \, {\left (3 \, x^{2} - 5 \, x - 2\right )} \log \left (3 \, x + 1\right ) - 6 \, {\left (3 \, x^{2} - 5 \, x - 2\right )} \log \left (x - 2\right ) - 42 \, x + 35}{343 \, {\left (3 \, x^{2} - 5 \, x - 2\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=\frac {5 - 6 x}{147 x^{2} - 245 x - 98} - \frac {6 \log {\left (x - 2 \right )}}{343} + \frac {6 \log {\left (x + \frac {1}{3} \right )}}{343} \]
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Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=-\frac {6 \, x - 5}{49 \, {\left (3 \, x^{2} - 5 \, x - 2\right )}} + \frac {6}{343} \, \log \left (3 \, x + 1\right ) - \frac {6}{343} \, \log \left (x - 2\right ) \]
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=-\frac {6 \, x - 5}{49 \, {\left (3 \, x^{2} - 5 \, x - 2\right )}} + \frac {6}{343} \, \log \left ({\left | 3 \, x + 1 \right |}\right ) - \frac {6}{343} \, \log \left ({\left | x - 2 \right |}\right ) \]
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Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (2+5 x-3 x^2\right )^2} \, dx=\frac {6\,\ln \left (\frac {3\,x+1}{x-2}\right )}{343}+\frac {2\,\left (3\,x-\frac {5}{2}\right )}{49\,\left (-3\,x^2+5\,x+2\right )} \]
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