Integrand size = 21, antiderivative size = 127 \[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{2 \sqrt {3} x} \]
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Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {409, 240, 226, 453} \[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt {6}}-\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{2 \sqrt {3} x}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{2 \sqrt {6}} \]
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Rule 226
Rule 240
Rule 409
Rule 453
Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx\right )+\frac {3}{2} \int \frac {x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\sqrt {x^2} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{\sqrt {3} x} \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\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 \sqrt {3} x} \\ \end{align*}
Result contains complex when optimal does not.
Time = 5.49 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{-1} \sqrt {x^2} \left (\operatorname {EllipticPi}\left (-i,\arcsin \left ((-1)^{3/4} \sqrt [4]{-1+3 x^2}\right ),-1\right )+\operatorname {EllipticPi}\left (i,\arcsin \left ((-1)^{3/4} \sqrt [4]{-1+3 x^2}\right ),-1\right )\right )}{\sqrt {3} x} \]
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\[\int \frac {1}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}}d x\]
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\[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )}} \,d x } \]
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\[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{\left (3 x^{2} - 2\right ) \left (3 x^{2} - 1\right )^{\frac {3}{4}}}\, dx \]
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\[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )}} \,d x } \]
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\[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} {\left (3 \, x^{2} - 2\right )}} \,d x } \]
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Timed out. \[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{{\left (3\,x^2-1\right )}^{3/4}\,\left (3\,x^2-2\right )} \,d x \]
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