Integrand size = 16, antiderivative size = 90 \[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-3 x^2+2 x^4}} \]
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Time = 0.01 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1117} \[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\frac {\left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2 x^4-3 x^2+3}} \]
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Rule 1117
Rubi steps \begin{align*} \text {integral}& = \frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-3 x^2+2 x^4}} \\ \end{align*}
Result contains complex when optimal does not.
Time = 10.07 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=-\frac {i \sqrt {1-\frac {4 x^2}{3-i \sqrt {15}}} \sqrt {1-\frac {4 x^2}{3+i \sqrt {15}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {-\frac {1}{3-i \sqrt {15}}} x\right ),\frac {3-i \sqrt {15}}{3+i \sqrt {15}}\right )}{2 \sqrt {-\frac {1}{3-i \sqrt {15}}} \sqrt {3-3 x^2+2 x^4}} \]
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Result contains complex when optimal does not.
Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97
method | result | size |
default | \(\frac {6 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )}{\sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}}\) | \(87\) |
elliptic | \(\frac {6 \sqrt {1-\left (\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1-i \sqrt {15}}}{2}\right )}{\sqrt {18+6 i \sqrt {15}}\, \sqrt {2 x^{4}-3 x^{2}+3}}\) | \(87\) |
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none
Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\frac {1}{24} \, \sqrt {6} \sqrt {\sqrt {3} \sqrt {-5} + 3} {\left (\sqrt {3} - \sqrt {-5}\right )} F(\arcsin \left (\frac {1}{6} \, \sqrt {6} \sqrt {\sqrt {3} \sqrt {-5} + 3} x\right )\,|\,-\frac {1}{4} \, \sqrt {3} \sqrt {-5} - \frac {1}{4}) \]
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\[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2 x^{4} - 3 x^{2} + 3}}\, dx \]
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\[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} - 3 \, x^{2} + 3}} \,d x } \]
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\[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} - 3 \, x^{2} + 3}} \,d x } \]
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Timed out. \[ \int \frac {1}{\sqrt {3-3 x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2\,x^4-3\,x^2+3}} \,d x \]
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