\(\int \frac {1}{(-2+3 x^2)^{3/4}} \, dx\) [908]

Optimal result

Integrand size = 11, antiderivative size = 82 \[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{\sqrt [4]{2} \sqrt {3} x} \]

[Out]

1/6*2^(3/4)*(cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))*El
lipticF(sin(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*(x^2/(2^(1/2)+(3*x^2
-2)^(1/2))^2)^(1/2)/x*3^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {240, 226} \[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{\sqrt [4]{2} \sqrt {3} x} \]

[In]

Int[(-2 + 3*x^2)^(-3/4),x]

[Out]

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

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 240

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[2*(Sqrt[(-b)*(x^2/a)]/(b*x)), Subst[Int[1/Sqrt[1 - x^4/a],
x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {\frac {2}{3}} \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 )}{x} \\ & = \frac {\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 )}{\sqrt [4]{2} \sqrt {3} x} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.83 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {x \left (1-\frac {3 x^2}{2}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {3 x^2}{2}\right )}{\left (-2+3 x^2\right )^{3/4}} \]

[In]

Integrate[(-2 + 3*x^2)^(-3/4),x]

[Out]

(x*(1 - (3*x^2)/2)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, (3*x^2)/2])/(-2 + 3*x^2)^(3/4)

Maple [C] (warning: unable to verify)

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

Time = 2.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.49

method	result	size
meijerg	\(\frac {2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{2 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}}}\)	\(40\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}}} \,d x } \]

[In]

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

[Out]

integral((3*x^2 - 2)^(-3/4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{2} x e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{2} \]

[In]

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

[Out]

2**(1/4)*x*exp(-3*I*pi/4)*hyper((1/2, 3/4), (3/2,), 3*x**2/2)/2

Maxima [F]

\[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}}} \,d x } \]

[In]

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

[Out]

integrate((3*x^2 - 2)^(-3/4), x)

Giac [F]

\[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}}} \,d x } \]

[In]

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

[Out]

integrate((3*x^2 - 2)^(-3/4), x)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {2^{1/4}\,x\,{\left (2-3\,x^2\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {3}{4};\ \frac {3}{2};\ \frac {3\,x^2}{2}\right )}{2\,{\left (3\,x^2-2\right )}^{3/4}} \]

[In]

int(1/(3*x^2 - 2)^(3/4),x)

[Out]

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