Integrand size = 24, antiderivative size = 164 \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {2}{45} x \left (2-3 x^2\right )^{3/4}+\frac {4 \sqrt [4]{2} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \text {arctanh}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}} \]
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Time = 0.05 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {451, 234, 327, 406} \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}}+\frac {4 \sqrt [4]{2} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {2}{45} \left (2-3 x^2\right )^{3/4} x \]
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Rule 234
Rule 327
Rule 406
Rule 451
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {4}{9 \sqrt [4]{2-3 x^2}}-\frac {x^2}{3 \sqrt [4]{2-3 x^2}}+\frac {16}{9 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right ) \, dx \\ & = -\left (\frac {1}{3} \int \frac {x^2}{\sqrt [4]{2-3 x^2}} \, dx\right )-\frac {4}{9} \int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx+\frac {16}{9} \int \frac {1}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx \\ & = \frac {2}{45} x \left (2-3 x^2\right )^{3/4}+\frac {4 \sqrt [4]{2} \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 )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \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 )}{9 \sqrt {3}}-\frac {8 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{9 \sqrt {3}}-\frac {4}{45} \int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx \\ & = \frac {2}{45} x \left (2-3 x^2\right )^{3/4}+\frac {4 \sqrt [4]{2} \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 )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \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 )}{9 \sqrt {3}}-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 5.79 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {1}{45} x \left (3\ 2^{3/4} x^2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\frac {2 \left (2-3 x^2+\frac {32 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{\left (-4+3 x^2\right ) \left (4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )\right )}\right )}{\sqrt [4]{2-3 x^2}}\right ) \]
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\[\int \frac {x^{4}}{\left (-3 x^{2}+2\right )^{\frac {1}{4}} \left (-3 x^{2}+4\right )}d x\]
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\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=- \int \frac {x^{4}}{3 x^{2} \sqrt [4]{2 - 3 x^{2}} - 4 \sqrt [4]{2 - 3 x^{2}}}\, dx \]
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\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\int { -\frac {x^{4}}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\int \frac {x^4}{{\left (2-3\,x^2\right )}^{1/4}\,\left (3\,x^2-4\right )} \,d x \]
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