Integrand size = 10, antiderivative size = 54 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2+x}{6 \left (1+x+x^2\right )^2}-\frac {1+2 x}{6 \left (1+x+x^2\right )}-\frac {2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \]
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Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {652, 628, 632, 210} \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {x+2}{6 \left (x^2+x+1\right )^2}-\frac {2 x+1}{6 \left (x^2+x+1\right )} \]
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Rule 210
Rule 628
Rule 632
Rule 652
Rubi steps \begin{align*} \text {integral}& = -\frac {2+x}{6 \left (1+x+x^2\right )^2}-\frac {1}{2} \int \frac {1}{\left (1+x+x^2\right )^2} \, dx \\ & = -\frac {2+x}{6 \left (1+x+x^2\right )^2}-\frac {1+2 x}{6 \left (1+x+x^2\right )}-\frac {1}{3} \int \frac {1}{1+x+x^2} \, dx \\ & = -\frac {2+x}{6 \left (1+x+x^2\right )^2}-\frac {1+2 x}{6 \left (1+x+x^2\right )}+\frac {2}{3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = -\frac {2+x}{6 \left (1+x+x^2\right )^2}-\frac {1+2 x}{6 \left (1+x+x^2\right )}-\frac {2 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=\frac {1}{18} \left (-\frac {3 \left (3+4 x+3 x^2+2 x^3\right )}{\left (1+x+x^2\right )^2}-4 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )\right ) \]
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Time = 0.38 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {-\frac {1}{3} x^{3}-\frac {1}{2} x^{2}-\frac {2}{3} x -\frac {1}{2}}{\left (x^{2}+x +1\right )^{2}}-\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\) | \(42\) |
default | \(\frac {-2-x}{6 \left (x^{2}+x +1\right )^{2}}-\frac {1+2 x}{6 \left (x^{2}+x +1\right )}-\frac {2 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{9}\) | \(48\) |
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Time = 0.37 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.31 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {6 \, x^{3} + 4 \, \sqrt {3} {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + 9 \, x^{2} + 12 \, x + 9}{18 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \]
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Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=\frac {- 2 x^{3} - 3 x^{2} - 4 x - 3}{6 x^{4} + 12 x^{3} + 18 x^{2} + 12 x + 6} - \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]
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Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {2 \, x^{3} + 3 \, x^{2} + 4 \, x + 3}{6 \, {\left (x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1\right )}} \]
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Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {2 \, x^{3} + 3 \, x^{2} + 4 \, x + 3}{6 \, {\left (x^{2} + x + 1\right )}^{2}} \]
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Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04 \[ \int \frac {x}{\left (1+x+x^2\right )^3} \, dx=-\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )}{9}-\frac {\frac {x^3}{3}+\frac {x^2}{2}+\frac {2\,x}{3}+\frac {1}{2}}{x^4+2\,x^3+3\,x^2+2\,x+1} \]
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