Integrand size = 15, antiderivative size = 104 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{-2+3 x^2}}{2 x}+\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{2} x} \]
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Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {331, 240, 226} \[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{2} x}+\frac {\sqrt [4]{3 x^2-2}}{2 x} \]
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Rule 226
Rule 240
Rule 331
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-2+3 x^2}}{2 x}+\frac {3}{4} \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx \\ & = \frac {\sqrt [4]{-2+3 x^2}}{2 x}+\frac {\left (\sqrt {\frac {3}{2}} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{2 x} \\ & = \frac {\sqrt [4]{-2+3 x^2}}{2 x}+\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{2} x} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=-\frac {\left (1-\frac {3 x^2}{2}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {1}{2},\frac {3 x^2}{2}\right )}{x \left (-2+3 x^2\right )^{3/4}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.40
method | result | size |
meijerg | \(-\frac {2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{2},\frac {3}{4};\frac {1}{2};\frac {3 x^{2}}{2}\right )}{2 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}} x}\) | \(42\) |
risch | \(\frac {\left (3 x^{2}-2\right )^{\frac {1}{4}}}{2 x}+\frac {3 \,2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{8 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}}}\) | \(55\) |
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
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Result contains complex when optimal does not.
Time = 0.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.28 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{2} e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{2 x} \]
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
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\[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
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Time = 4.69 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx=-\frac {2\,3^{1/4}\,{\left (\frac {1}{x^2}\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{4},\frac {5}{4};\ \frac {9}{4};\ \frac {2}{3\,x^2}\right )}{15\,x} \]
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