Integrand size = 24, antiderivative size = 264 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=-\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{2 x}+\frac {9 x \sqrt [4]{-1+3 x^2}}{2 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {3}{8} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{2 x}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{4 x} \]
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Time = 0.15 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {451, 331, 236, 311, 226, 1210, 407} \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=-\frac {3}{8} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{4 x}-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) E\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{2 x}-\frac {3}{8} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {9 \sqrt [4]{3 x^2-1} x}{2 \left (\sqrt {3 x^2-1}+1\right )}-\frac {3 \left (3 x^2-1\right )^{3/4}}{2 x}-\frac {\left (3 x^2-1\right )^{3/4}}{6 x^3} \]
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Rule 226
Rule 236
Rule 311
Rule 331
Rule 407
Rule 451
Rule 1210
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2 x^4 \sqrt [4]{-1+3 x^2}}-\frac {3}{4 x^2 \sqrt [4]{-1+3 x^2}}+\frac {9}{4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{x^4 \sqrt [4]{-1+3 x^2}} \, dx\right )-\frac {3}{4} \int \frac {1}{x^2 \sqrt [4]{-1+3 x^2}} \, dx+\frac {9}{4} \int \frac {1}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx \\ & = -\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{4 x}-\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{4} \int \frac {1}{x^2 \sqrt [4]{-1+3 x^2}} \, dx+\frac {9}{8} \int \frac {1}{\sqrt [4]{-1+3 x^2}} \, dx \\ & = -\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{2 x}-\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {9}{8} \int \frac {1}{\sqrt [4]{-1+3 x^2}} \, dx+\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x} \\ & = -\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{2 x}-\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x}+\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x}-\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x} \\ & = -\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{2 x}+\frac {9 x \sqrt [4]{-1+3 x^2}}{4 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{4 x}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{8 x}+\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x}-\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x} \\ & = -\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{2 x}+\frac {9 x \sqrt [4]{-1+3 x^2}}{2 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{2 x}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{4 x} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {1}{2} \left (-1+3 x^2\right )^{3/4} \left (-\frac {1+9 x^2}{3 x^3}+\frac {9 x \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )}{\left (-2+3 x^2\right ) \left (2 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{4},2,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )-3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )\right )\right )}\right ) \]
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\[\int \frac {1}{x^{4} \left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}}d x\]
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\[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} {\left (3 \, x^{2} - 2\right )} x^{4}} \,d x } \]
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\[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\int \frac {1}{x^{4} \cdot \left (3 x^{2} - 2\right ) \sqrt [4]{3 x^{2} - 1}}\, dx \]
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\[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} {\left (3 \, x^{2} - 2\right )} x^{4}} \,d x } \]
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\[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} {\left (3 \, x^{2} - 2\right )} x^{4}} \,d x } \]
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Timed out. \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\int \frac {1}{x^4\,{\left (3\,x^2-1\right )}^{1/4}\,\left (3\,x^2-2\right )} \,d x \]
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