Integrand size = 26, antiderivative size = 35 \[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=-\frac {E\left (\arcsin (2 x)\left |\frac {3}{8}\right .\right )}{3 \sqrt {2}}+\frac {\operatorname {EllipticF}\left (\arcsin (2 x),\frac {3}{8}\right )}{3 \sqrt {2}} \]
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Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {507, 435, 430} \[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\frac {\operatorname {EllipticF}\left (\arcsin (2 x),\frac {3}{8}\right )}{3 \sqrt {2}}-\frac {E\left (\arcsin (2 x)\left |\frac {3}{8}\right .\right )}{3 \sqrt {2}} \]
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Rule 430
Rule 435
Rule 507
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{3} \int \frac {\sqrt {2-3 x^2}}{\sqrt {1-4 x^2}} \, dx\right )+\frac {2}{3} \int \frac {1}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx \\ & = -\frac {E\left (\sin ^{-1}(2 x)|\frac {3}{8}\right )}{3 \sqrt {2}}+\frac {F\left (\sin ^{-1}(2 x)|\frac {3}{8}\right )}{3 \sqrt {2}} \\ \end{align*}
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\frac {-E\left (\arcsin (2 x)\left |\frac {3}{8}\right .\right )+\operatorname {EllipticF}\left (\arcsin (2 x),\frac {3}{8}\right )}{3 \sqrt {2}} \]
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Time = 3.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77
method | result | size |
default | \(\frac {\sqrt {2}\, \left (F\left (2 x , \frac {\sqrt {6}}{4}\right )-E\left (2 x , \frac {\sqrt {6}}{4}\right )\right )}{6}\) | \(27\) |
elliptic | \(\frac {\sqrt {\left (3 x^{2}-2\right ) \left (4 x^{2}-1\right )}\, \sqrt {-6 x^{2}+4}\, \left (F\left (2 x , \frac {\sqrt {6}}{4}\right )-E\left (2 x , \frac {\sqrt {6}}{4}\right )\right )}{6 \sqrt {-3 x^{2}+2}\, \sqrt {12 x^{4}-11 x^{2}+2}}\) | \(73\) |
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.89 \[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\frac {4 \, \sqrt {2} x E(\arcsin \left (\frac {\sqrt {3} \sqrt {2}}{3 \, x}\right )\,|\,\frac {3}{8}) - 4 \, \sqrt {2} x F(\arcsin \left (\frac {\sqrt {3} \sqrt {2}}{3 \, x}\right )\,|\,\frac {3}{8}) + 3 \, \sqrt {-3 \, x^{2} + 2} \sqrt {-4 \, x^{2} + 1}}{36 \, x} \]
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\[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\int \frac {x^{2}}{\sqrt {- \left (2 x - 1\right ) \left (2 x + 1\right )} \sqrt {2 - 3 x^{2}}}\, dx \]
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\[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\int { \frac {x^{2}}{\sqrt {-3 \, x^{2} + 2} \sqrt {-4 \, x^{2} + 1}} \,d x } \]
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\[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\int { \frac {x^{2}}{\sqrt {-3 \, x^{2} + 2} \sqrt {-4 \, x^{2} + 1}} \,d x } \]
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Timed out. \[ \int \frac {x^2}{\sqrt {1-4 x^2} \sqrt {2-3 x^2}} \, dx=\int \frac {x^2}{\sqrt {2-3\,x^2}\,\sqrt {1-4\,x^2}} \,d x \]
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