Integrand size = 14, antiderivative size = 36 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {13+7 x}{9 \left (4+5 x+x^2\right )}+\frac {7}{27} \log (1+x)-\frac {7}{27} \log (4+x) \]
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Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {652, 630, 31} \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {7 x+13}{9 \left (x^2+5 x+4\right )}+\frac {7}{27} \log (x+1)-\frac {7}{27} \log (x+4) \]
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Rule 31
Rule 630
Rule 652
Rubi steps \begin{align*} \text {integral}& = \frac {13+7 x}{9 \left (4+5 x+x^2\right )}+\frac {7}{9} \int \frac {1}{4+5 x+x^2} \, dx \\ & = \frac {13+7 x}{9 \left (4+5 x+x^2\right )}+\frac {7}{27} \int \frac {1}{1+x} \, dx-\frac {7}{27} \int \frac {1}{4+x} \, dx \\ & = \frac {13+7 x}{9 \left (4+5 x+x^2\right )}+\frac {7}{27} \log (1+x)-\frac {7}{27} \log (4+x) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.92 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {1}{27} \left (\frac {39+21 x}{4+5 x+x^2}+7 \log (1+x)-7 \log (4+x)\right ) \]
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Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {5}{9 \left (4+x \right )}-\frac {7 \ln \left (4+x \right )}{27}+\frac {2}{9 \left (1+x \right )}+\frac {7 \ln \left (1+x \right )}{27}\) | \(28\) |
norman | \(\frac {\frac {7 x}{9}+\frac {13}{9}}{x^{2}+5 x +4}+\frac {7 \ln \left (1+x \right )}{27}-\frac {7 \ln \left (4+x \right )}{27}\) | \(30\) |
risch | \(\frac {\frac {7 x}{9}+\frac {13}{9}}{x^{2}+5 x +4}+\frac {7 \ln \left (1+x \right )}{27}-\frac {7 \ln \left (4+x \right )}{27}\) | \(30\) |
parallelrisch | \(\frac {7 \ln \left (1+x \right ) x^{2}-7 \ln \left (4+x \right ) x^{2}+39+35 \ln \left (1+x \right ) x -35 \ln \left (4+x \right ) x +28 \ln \left (1+x \right )-28 \ln \left (4+x \right )+21 x}{27 x^{2}+135 x +108}\) | \(62\) |
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Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=-\frac {7 \, {\left (x^{2} + 5 \, x + 4\right )} \log \left (x + 4\right ) - 7 \, {\left (x^{2} + 5 \, x + 4\right )} \log \left (x + 1\right ) - 21 \, x - 39}{27 \, {\left (x^{2} + 5 \, x + 4\right )}} \]
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Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.86 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {7 x + 13}{9 x^{2} + 45 x + 36} + \frac {7 \log {\left (x + 1 \right )}}{27} - \frac {7 \log {\left (x + 4 \right )}}{27} \]
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Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {7 \, x + 13}{9 \, {\left (x^{2} + 5 \, x + 4\right )}} - \frac {7}{27} \, \log \left (x + 4\right ) + \frac {7}{27} \, \log \left (x + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {7 \, x + 13}{9 \, {\left (x^{2} + 5 \, x + 4\right )}} - \frac {7}{27} \, \log \left ({\left | x + 4 \right |}\right ) + \frac {7}{27} \, \log \left ({\left | x + 1 \right |}\right ) \]
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Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69 \[ \int \frac {-1+x}{\left (4+5 x+x^2\right )^2} \, dx=\frac {\frac {7\,x}{9}+\frac {13}{9}}{x^2+5\,x+4}-\frac {14\,\mathrm {atanh}\left (\frac {2\,x}{3}+\frac {5}{3}\right )}{27} \]
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