Integrand size = 12, antiderivative size = 43 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3 \arctan \left (\frac {2+3 x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]
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Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {628, 632, 210} \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {3 \arctan \left (\frac {3 x+2}{\sqrt {2}}\right )}{4 \sqrt {2}}+\frac {3 x+2}{4 \left (3 x^2+4 x+2\right )} \]
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Rule 210
Rule 628
Rule 632
Rubi steps \begin{align*} \text {integral}& = \frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3}{4} \int \frac {1}{2+4 x+3 x^2} \, dx \\ & = \frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}-\frac {3}{2} \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,4+6 x\right ) \\ & = \frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3 \tan ^{-1}\left (\frac {2+3 x}{\sqrt {2}}\right )}{4 \sqrt {2}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {2+3 x}{4 \left (2+4 x+3 x^2\right )}+\frac {3 \arctan \left (\frac {2+3 x}{\sqrt {2}}\right )}{4 \sqrt {2}} \]
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Time = 2.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {\frac {x}{4}+\frac {1}{6}}{x^{2}+\frac {4}{3} x +\frac {2}{3}}+\frac {3 \arctan \left (\frac {\left (2+3 x \right ) \sqrt {2}}{2}\right ) \sqrt {2}}{8}\) | \(34\) |
default | \(\frac {4+6 x}{24 x^{2}+32 x +16}+\frac {3 \sqrt {2}\, \arctan \left (\frac {\left (4+6 x \right ) \sqrt {2}}{4}\right )}{8}\) | \(37\) |
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Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {3 \, \sqrt {2} {\left (3 \, x^{2} + 4 \, x + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) + 6 \, x + 4}{8 \, {\left (3 \, x^{2} + 4 \, x + 2\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {3 x + 2}{12 x^{2} + 16 x + 8} + \frac {3 \sqrt {2} \operatorname {atan}{\left (\frac {3 \sqrt {2} x}{2} + \sqrt {2} \right )}}{8} \]
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Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) + \frac {3 \, x + 2}{4 \, {\left (3 \, x^{2} + 4 \, x + 2\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {3}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (3 \, x + 2\right )}\right ) + \frac {3 \, x + 2}{4 \, {\left (3 \, x^{2} + 4 \, x + 2\right )}} \]
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Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (2+4 x+3 x^2\right )^2} \, dx=\frac {\frac {x}{4}+\frac {1}{6}}{x^2+\frac {4\,x}{3}+\frac {2}{3}}+\frac {3\,\sqrt {2}\,\mathrm {atan}\left (\frac {3\,\sqrt {2}\,x}{2}+\sqrt {2}\right )}{8} \]
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