Integrand size = 14, antiderivative size = 31 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=x+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (1+x+x^2\right ) \]
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Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1671, 648, 632, 210, 642} \[ \int \frac {1+x^2}{1+x+x^2} \, dx=\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (x^2+x+1\right )+x \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1671
Rubi steps \begin{align*} \text {integral}& = \int \left (1-\frac {x}{1+x+x^2}\right ) \, dx \\ & = x-\int \frac {x}{1+x+x^2} \, dx \\ & = x+\frac {1}{2} \int \frac {1}{1+x+x^2} \, dx-\frac {1}{2} \int \frac {1+2 x}{1+x+x^2} \, dx \\ & = x-\frac {1}{2} \log \left (1+x+x^2\right )-\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = x+\frac {\tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (1+x+x^2\right ) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=x+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (1+x+x^2\right ) \]
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Time = 0.76 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90
method | result | size |
default | \(x -\frac {\ln \left (x^{2}+x +1\right )}{2}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\) | \(28\) |
risch | \(x -\frac {\ln \left (4 x^{2}+4 x +4\right )}{2}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\) | \(32\) |
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Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]
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Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=x - \frac {\log {\left (x^{2} + x + 1 \right )}}{2} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]
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Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) \]
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Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {1+x^2}{1+x+x^2} \, dx=x-\frac {\ln \left (x^2+x+1\right )}{2}+\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3}+\frac {\sqrt {3}}{3}\right )}{3} \]
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