\(\int \frac {1}{\sqrt {-2+3 x^4}} \, dx\) [47]

Optimal result

Integrand size = 11, antiderivative size = 115 \[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=\frac {\sqrt {-2+\sqrt {6} x^2} \sqrt {\frac {2+\sqrt {6} x^2}{2-\sqrt {6} x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt {-2+\sqrt {6} x^2}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{6} \sqrt {\frac {1}{2-\sqrt {6} x^2}} \sqrt {-2+3 x^4}} \]

[Out]

1/12*EllipticF(2^(3/4)*3^(1/4)*x/(-2+x^2*6^(1/2))^(1/2),1/2*2^(1/2))*(-2+x^2*6^(1/2))^(1/2)*((2+x^2*6^(1/2))/(
2-x^2*6^(1/2)))^(1/2)*6^(3/4)/(3*x^4-2)^(1/2)/(1/(2-x^2*6^(1/2)))^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {229} \[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=\frac {\sqrt {\sqrt {6} x^2-2} \sqrt {\frac {\sqrt {6} x^2+2}{2-\sqrt {6} x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt {\sqrt {6} x^2-2}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{6} \sqrt {\frac {1}{2-\sqrt {6} x^2}} \sqrt {3 x^4-2}} \]

[In]

Int[1/Sqrt[-2 + 3*x^4],x]

[Out]

(Sqrt[-2 + Sqrt[6]*x^2]*Sqrt[(2 + Sqrt[6]*x^2)/(2 - Sqrt[6]*x^2)]*EllipticF[ArcSin[(2^(3/4)*3^(1/4)*x)/Sqrt[-2
+ Sqrt[6]*x^2]], 1/2])/(2*6^(1/4)*Sqrt[(2 - Sqrt[6]*x^2)^(-1)]*Sqrt[-2 + 3*x^4])

Rule 229

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*b, 2]}, Simp[Sqrt[(a - q*x^2)/(a + q*x^2)]*(Sq
rt[(a + q*x^2)/q]/(Sqrt[2]*Sqrt[a + b*x^4]*Sqrt[a/(a + q*x^2)]))*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]],
1/2], x]] /; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-2+\sqrt {6} x^2} \sqrt {\frac {2+\sqrt {6} x^2}{2-\sqrt {6} x^2}} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt {-2+\sqrt {6} x^2}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{6} \sqrt {\frac {1}{2-\sqrt {6} x^2}} \sqrt {-2+3 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=\frac {\sqrt {2-3 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt [4]{\frac {3}{2}} x\right ),-1\right )}{\sqrt [4]{6} \sqrt {-2+3 x^4}} \]

[In]

Integrate[1/Sqrt[-2 + 3*x^4],x]

[Out]

(Sqrt[2 - 3*x^4]*EllipticF[ArcSin[(3/2)^(1/4)*x], -1])/(6^(1/4)*Sqrt[-2 + 3*x^4])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.54 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.35

method	result	size
meijerg	\(\frac {\sqrt {2}\, \sqrt {-\operatorname {signum}\left (-1+\frac {3 x^{4}}{2}\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {3 x^{4}}{2}\right )}{2 \sqrt {\operatorname {signum}\left (-1+\frac {3 x^{4}}{2}\right )}}\)	\(40\)
default	\(\frac {\sqrt {4+2 x^{2} \sqrt {6}}\, \sqrt {4-2 x^{2} \sqrt {6}}\, F\left (\frac {\sqrt {-2 \sqrt {6}}\, x}{2}, i\right )}{2 \sqrt {-2 \sqrt {6}}\, \sqrt {3 x^{4}-2}}\)	\(56\)
elliptic	\(\frac {\sqrt {4+2 x^{2} \sqrt {6}}\, \sqrt {4-2 x^{2} \sqrt {6}}\, F\left (\frac {\sqrt {-2 \sqrt {6}}\, x}{2}, i\right )}{2 \sqrt {-2 \sqrt {6}}\, \sqrt {3 x^{4}-2}}\)	\(56\)

[In]

int(1/(3*x^4-2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(1/2)/signum(-1+3/2*x^4)^(1/2)*(-signum(-1+3/2*x^4))^(1/2)*x*hypergeom([1/4,1/2],[5/4],3/2*x^4)

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.20 \[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=-\frac {1}{12} \cdot 6^{\frac {3}{4}} \sqrt {2} \sqrt {-2} F(\arcsin \left (\frac {1}{2} \cdot 6^{\frac {1}{4}} \sqrt {2} x\right )\,|\,-1) \]

[In]

integrate(1/(3*x^4-2)^(1/2),x, algorithm="fricas")

[Out]

-1/12*6^(3/4)*sqrt(2)*sqrt(-2)*elliptic_f(arcsin(1/2*6^(1/4)*sqrt(2)*x), -1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=- \frac {\sqrt {2} i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {3 x^{4}}{2}} \right )}}{8 \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(1/(3*x**4-2)**(1/2),x)

[Out]

-sqrt(2)*I*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), 3*x**4/2)/(8*gamma(5/4))

Maxima [F]

\[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{4} - 2}} \,d x } \]

[In]

integrate(1/(3*x^4-2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(3*x^4 - 2), x)

Giac [F]

\[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{4} - 2}} \,d x } \]

[In]

integrate(1/(3*x^4-2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(3*x^4 - 2), x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.27 \[ \int \frac {1}{\sqrt {-2+3 x^4}} \, dx=\frac {x\,\sqrt {4-6\,x^4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ \frac {3\,x^4}{2}\right )}{2\,\sqrt {3\,x^4-2}} \]

[In]

int(1/(3*x^4 - 2)^(1/2),x)

[Out]

(x*(4 - 6*x^4)^(1/2)*hypergeom([1/4, 1/2], 5/4, (3*x^4)/2))/(2*(3*x^4 - 2)^(1/2))