\(\int \frac {1}{\sqrt {-3+2 x^4}} \, dx\) [60]

Optimal result

Integrand size = 11, antiderivative size = 112 \[ \int \frac {1}{\sqrt {-3+2 x^4}} \, dx=\frac {\sqrt {-3+\sqrt {6} x^2} \sqrt {\frac {3+\sqrt {6} x^2}{3-\sqrt {6} x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt {-3+\sqrt {6} x^2}}\right ),\frac {1}{2}\right )}{6^{3/4} \sqrt {\frac {1}{3-\sqrt {6} x^2}} \sqrt {-3+2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 112, 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} \[ \int \frac {1}{\sqrt {-3+2 x^4}} \, dx=\frac {\sqrt {\sqrt {6} x^2-3} \sqrt {\frac {\sqrt {6} x^2+3}{3-\sqrt {6} x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt {\sqrt {6} x^2-3}}\right ),\frac {1}{2}\right )}{6^{3/4} \sqrt {\frac {1}{3-\sqrt {6} x^2}} \sqrt {2 x^4-3}} \]

[In]

Int[1/Sqrt[-3 + 2*x^4],x]

[Out]

(Sqrt[-3 + Sqrt[6]*x^2]*Sqrt[(3 + Sqrt[6]*x^2)/(3 - Sqrt[6]*x^2)]*EllipticF[ArcSin[(2^(3/4)*3^(1/4)*x)/Sqrt[-3
+ Sqrt[6]*x^2]], 1/2])/(6^(3/4)*Sqrt[(3 - Sqrt[6]*x^2)^(-1)]*Sqrt[-3 + 2*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 {-3+\sqrt {6} x^2} \sqrt {\frac {3+\sqrt {6} x^2}{3-\sqrt {6} x^2}} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{3} x}{\sqrt {-3+\sqrt {6} x^2}}\right )|\frac {1}{2}\right )}{6^{3/4} \sqrt {\frac {1}{3-\sqrt {6} x^2}} \sqrt {-3+2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\sqrt {-3+2 x^4}} \, dx=\frac {\sqrt {3-2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt [4]{\frac {2}{3}} x\right ),-1\right )}{\sqrt [4]{6} \sqrt {-3+2 x^4}} \]

[In]

Integrate[1/Sqrt[-3 + 2*x^4],x]

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.66 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.36

method	result	size
meijerg	\(\frac {\sqrt {3}\, \sqrt {-\operatorname {signum}\left (-1+\frac {2 x^{4}}{3}\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\frac {2 x^{4}}{3}\right )}{3 \sqrt {\operatorname {signum}\left (-1+\frac {2 x^{4}}{3}\right )}}\)	\(40\)
default	\(\frac {\sqrt {9+3 x^{2} \sqrt {6}}\, \sqrt {9-3 x^{2} \sqrt {6}}\, F\left (\frac {\sqrt {-3 \sqrt {6}}\, x}{3}, i\right )}{3 \sqrt {-3 \sqrt {6}}\, \sqrt {2 x^{4}-3}}\)	\(56\)
elliptic	\(\frac {\sqrt {9+3 x^{2} \sqrt {6}}\, \sqrt {9-3 x^{2} \sqrt {6}}\, F\left (\frac {\sqrt {-3 \sqrt {6}}\, x}{3}, i\right )}{3 \sqrt {-3 \sqrt {6}}\, \sqrt {2 x^{4}-3}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.31 \[ \int \frac {1}{\sqrt {-3+2 x^4}} \, dx=-\frac {1}{6} \, \sqrt {2} \sqrt {-3} \sqrt {\sqrt {3} \sqrt {2}} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} \sqrt {\sqrt {3} \sqrt {2}} x\right )\,|\,-1) \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.30 \[ \int \frac {1}{\sqrt {-3+2 x^4}} \, dx=- \frac {\sqrt {3} i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {2 x^{4}}{3}} \right )}}{12 \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

-sqrt(3)*I*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), 2*x**4/3)/(12*gamma(5/4))

Maxima [F]

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

[In]

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

[Out]

integrate(1/sqrt(2*x^4 - 3), x)

Giac [F]

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

[In]

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

[Out]

integrate(1/sqrt(2*x^4 - 3), x)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\sqrt {-3+2 x^4}} \, dx=\frac {x\,\sqrt {9-6\,x^4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{2};\ \frac {5}{4};\ \frac {2\,x^4}{3}\right )}{3\,\sqrt {2\,x^4-3}} \]

[In]

int(1/(2*x^4 - 3)^(1/2),x)

[Out]

(x*(9 - 6*x^4)^(1/2)*hypergeom([1/4, 1/2], 5/4, (2*x^4)/3))/(3*(2*x^4 - 3)^(1/2))