Integrand size = 24, antiderivative size = 166 \[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {\left (2-3 x^2\right )^{3/4}}{8 x}+\frac {\sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{8\ 2^{3/4}}+\frac {\sqrt {3} \text {arctanh}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{8\ 2^{3/4}}-\frac {\sqrt {3} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4\ 2^{3/4}} \]
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Time = 0.04 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {451, 331, 234, 406} \[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {\sqrt {3} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4\ 2^{3/4}}+\frac {\sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{8\ 2^{3/4}}+\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{8\ 2^{3/4}}-\frac {\left (2-3 x^2\right )^{3/4}}{8 x} \]
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Rule 234
Rule 331
Rule 406
Rule 451
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4 x^2 \sqrt [4]{2-3 x^2}}-\frac {3}{4 \sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {1}{x^2 \sqrt [4]{2-3 x^2}} \, dx-\frac {3}{4} \int \frac {1}{\sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )} \, dx \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{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 )}{8\ 2^{3/4}}+\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 )}{8\ 2^{3/4}}-\frac {3}{16} \int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx \\ & = -\frac {\left (2-3 x^2\right )^{3/4}}{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 )}{8\ 2^{3/4}}+\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 )}{8\ 2^{3/4}}-\frac {\sqrt {3} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{4\ 2^{3/4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 11.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {\left (2-3 x^2\right )^{3/4}}{8 x}+\frac {3 x^3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{64 \sqrt [4]{2}} \]
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\[\int \frac {1}{x^{2} \left (-3 x^{2}+2\right )^{\frac {1}{4}} \left (-3 x^{2}+4\right )}d x\]
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\[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=- \int \frac {1}{3 x^{4} \sqrt [4]{2 - 3 x^{2}} - 4 x^{2} \sqrt [4]{2 - 3 x^{2}}}\, dx \]
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\[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {1}{x^2 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\int \frac {1}{x^2\,{\left (2-3\,x^2\right )}^{1/4}\,\left (3\,x^2-4\right )} \,d x \]
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