Integrand size = 24, antiderivative size = 166 \[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\sqrt [4]{2-3 x^2}}{8 x}+\frac {\sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \text {arctanh}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}+\frac {\sqrt {3} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{4 \sqrt [4]{2}} \]
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Time = 0.06 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {454, 331, 238, 409, 452} \[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {\sqrt {3} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{4 \sqrt [4]{2}}+\frac {\sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt [4]{2-3 x^2}}{8 x} \]
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Rule 238
Rule 331
Rule 409
Rule 452
Rule 454
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4 x^2 \left (2-3 x^2\right )^{3/4}}-\frac {3}{4 \left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4}} \, dx-\frac {3}{4} \int \frac {1}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx \\ & = -\frac {\sqrt [4]{2-3 x^2}}{8 x}+2 \left (\frac {3}{16} \int \frac {1}{\left (2-3 x^2\right )^{3/4}} \, dx\right )-\frac {9}{16} \int \frac {x^2}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx \\ & = -\frac {\sqrt [4]{2-3 x^2}}{8 x}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}+\frac {\sqrt {3} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4 \sqrt [4]{2}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 11.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {1}{2},\frac {3}{4},1,\frac {1}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{4\ 2^{3/4} x} \]
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\[\int \frac {1}{x^{2} \left (-3 x^{2}+2\right )^{\frac {3}{4}} \left (-3 x^{2}+4\right )}d x\]
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\[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=- \int \frac {1}{3 x^{4} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 x^{2} \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \]
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\[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\int \frac {1}{x^2\,{\left (2-3\,x^2\right )}^{3/4}\,\left (3\,x^2-4\right )} \,d x \]
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