Integrand size = 22, antiderivative size = 45 \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {10}}+\frac {\arctan \left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {10}} \]
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Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1177, 209} \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {10}}+\frac {\arctan \left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {10}} \]
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Rule 209
Rule 1177
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \left (5-\sqrt {5}\right ) \int \frac {1}{3-\sqrt {5}+4 x^2} \, dx+\frac {1}{5} \left (5+\sqrt {5}\right ) \int \frac {1}{3+\sqrt {5}+4 x^2} \, dx \\ & = \frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{\sqrt {10}}+\frac {\tan ^{-1}\left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{\sqrt {10}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.84 \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\left (-1+\sqrt {5}\right ) \arctan \left (\frac {2 x}{\sqrt {3-\sqrt {5}}}\right )}{2 \sqrt {5 \left (3-\sqrt {5}\right )}}+\frac {\left (1+\sqrt {5}\right ) \arctan \left (\frac {2 x}{\sqrt {3+\sqrt {5}}}\right )}{2 \sqrt {5 \left (3+\sqrt {5}\right )}} \]
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Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\sqrt {10}\, \arctan \left (\frac {\sqrt {10}\, x}{5}\right )}{10}+\frac {\sqrt {10}\, \arctan \left (\frac {2 \sqrt {10}\, x^{3}}{5}+\frac {4 \sqrt {10}\, x}{5}\right )}{10}\) | \(35\) |
default | \(\frac {2 \left (\sqrt {5}+1\right ) \sqrt {5}\, \arctan \left (\frac {8 x}{2 \sqrt {10}+2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}+2 \sqrt {2}\right )}+\frac {2 \left (\sqrt {5}-1\right ) \sqrt {5}\, \arctan \left (\frac {8 x}{2 \sqrt {10}-2 \sqrt {2}}\right )}{5 \left (2 \sqrt {10}-2 \sqrt {2}\right )}\) | \(82\) |
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Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.69 \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {1}{10} \, \sqrt {10} \arctan \left (\frac {2}{5} \, \sqrt {10} {\left (x^{3} + 2 \, x\right )}\right ) + \frac {1}{10} \, \sqrt {10} \arctan \left (\frac {1}{5} \, \sqrt {10} x\right ) \]
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Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93 \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\sqrt {10} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {10} x}{5} \right )} + 2 \operatorname {atan}{\left (\frac {2 \sqrt {10} x^{3}}{5} + \frac {4 \sqrt {10} x}{5} \right )}\right )}{20} \]
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\[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\int { \frac {2 \, x^{2} + 1}{4 \, x^{4} + 6 \, x^{2} + 1} \,d x } \]
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Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.87 \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {1}{10} \, \sqrt {10} \arctan \left (\frac {4 \, x}{\sqrt {10} + \sqrt {2}}\right ) + \frac {1}{10} \, \sqrt {10} \arctan \left (\frac {4 \, x}{\sqrt {10} - \sqrt {2}}\right ) \]
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Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.64 \[ \int \frac {1+2 x^2}{1+6 x^2+4 x^4} \, dx=\frac {\sqrt {10}\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {10}\,x^3}{5}+\frac {4\,\sqrt {10}\,x}{5}\right )+\mathrm {atan}\left (\frac {\sqrt {10}\,x}{5}\right )\right )}{10} \]
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