\(\int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx\) [895]

Optimal result

Integrand size = 15, antiderivative size = 221 \[ \int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx=\frac {\left (-2+3 x^2\right )^{3/4}}{2 x}-\frac {3 x \sqrt [4]{-2+3 x^2}}{2 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}+\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{2^{3/4} x}-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2\ 2^{3/4} x} \]

[Out]

1/2*(3*x^2-2)^(3/4)/x-3/2*x*(3*x^2-2)^(1/4)/(2^(1/2)+(3*x^2-2)^(1/2))+1/2*2^(1/4)*(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)))*EllipticE(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)-1/4*2^(1/4
)*(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)))*EllipticF(si
n(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 = 0.08 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {331, 236, 311, 226, 1210} \[ \int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx=-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2\ 2^{3/4} x}+\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{2^{3/4} x}-\frac {3 \sqrt [4]{3 x^2-2} x}{2 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}+\frac {\left (3 x^2-2\right )^{3/4}}{2 x} \]

[In]

Int[1/(x^2*(-2 + 3*x^2)^(1/4)),x]

[Out]

(-2 + 3*x^2)^(3/4)/(2*x) - (3*x*(-2 + 3*x^2)^(1/4))/(2*(Sqrt[2] + Sqrt[-2 + 3*x^2])) + (Sqrt[3]*Sqrt[x^2/(Sqrt
[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*EllipticE[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/
(2^(3/4)*x) - (Sqrt[3]*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*EllipticF[2*ArcTa
n[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(2*2^(3/4)*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 236

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

Rule 311

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

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (-2+3 x^2\right )^{3/4}}{2 x}-\frac {3}{4} \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{2 x}-\frac {\left (\sqrt {\frac {3}{2}} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{2 x} \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{2 x}-\frac {\left (\sqrt {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 )}{2 x}+\frac {\left (\sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{\sqrt {2}}}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{2 x} \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{2 x}-\frac {3 x \sqrt [4]{-2+3 x^2}}{2 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}+\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 ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{2^{3/4} 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 )}{2\ 2^{3/4} x} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 5.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.21 \[ \int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx=-\frac {\sqrt [4]{1-\frac {3 x^2}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},\frac {3 x^2}{2}\right )}{x \sqrt [4]{-2+3 x^2}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 2.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.19

method	result	size
meijerg	\(-\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\frac {3 x^{2}}{2}\right )}{2 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}} x}\)	\(42\)
risch	\(\frac {\left (3 x^{2}-2\right )^{\frac {3}{4}}}{2 x}-\frac {3 \,2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{8 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\)	\(55\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((3*x^2 - 2)^(3/4)/(3*x^4 - 2*x^2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.14 \[ \int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{2 x} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.55 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx=-\frac {2\,3^{3/4}\,{\left (3-\frac {2}{x^2}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ \frac {2}{3\,x^2}\right )}{9\,x\,{\left (3\,x^2-2\right )}^{1/4}} \]

[In]

int(1/(x^2*(3*x^2 - 2)^(1/4)),x)

[Out]

-(2*3^(3/4)*(3 - 2/x^2)^(1/4)*hypergeom([1/4, 3/4], 7/4, 2/(3*x^2)))/(9*x*(3*x^2 - 2)^(1/4))