Integrand size = 24, antiderivative size = 149 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{-1+3 x^2}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{2 x} \]
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Time = 0.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {454, 331, 240, 226, 409, 453} \[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {1}{4} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{2 x}-\frac {1}{4} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {\sqrt [4]{3 x^2-1}}{2 x} \]
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Rule 226
Rule 240
Rule 331
Rule 409
Rule 453
Rule 454
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{2 x^2 \left (-1+3 x^2\right )^{3/4}}+\frac {3}{2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{x^2 \left (-1+3 x^2\right )^{3/4}} \, dx\right )+\frac {3}{2} \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{2 x}-2 \left (\frac {3}{4} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx\right )+\frac {9}{4} \int \frac {x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-2 \frac {\left (\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 )}{2 x} \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {\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 )}{2 x} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\left (1-3 x^2\right )^{3/4} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {3}{4},1,\frac {1}{2},3 x^2,\frac {3 x^2}{2}\right )}{2 x \left (-1+3 x^2\right )^{3/4}} \]
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\[\int \frac {1}{x^{2} \left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}}d x\]
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{x^{2} \cdot \left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \]
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{x^2\,{\left (3\,x^2-1\right )}^{3/4}\,\left (3\,x^2-2\right )} \,d x \]
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