\(\int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx\) [966]

Optimal result

Integrand size = 15, antiderivative size = 223 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\frac {\left (-2-3 x^2\right )^{3/4}}{2 x}+\frac {3 x \sqrt [4]{-2-3 x^2}}{2 \left (\sqrt {2}+\sqrt {-2-3 x^2}\right )}+\frac {\sqrt {3} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{2^{3/4} x}-\frac {\sqrt {3} \sqrt {-\frac {x^2}{\left (\sqrt {2}+\sqrt {-2-3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2-3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2-3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2\ 2^{3/4} x} \] Output:

1/2*(-3*x^2-2)^(3/4)/x+3*x*(-3*x^2-2)^(1/4)/(2*2^(1/2)+2*(-3*x^2-2)^(1/2)) 
+1/2*2^(1/4)*(-x^2/(2^(1/2)+(-3*x^2-2)^(1/2))^2)^(1/2)*(2^(1/2)+(-3*x^2-2) 
^(1/2))*EllipticE(sin(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2)) 
*3^(1/2)/x-1/4*2^(1/4)*(-x^2/(2^(1/2)+(-3*x^2-2)^(1/2))^2)^(1/2)*(2^(1/2)+ 
(-3*x^2-2)^(1/2))*InverseJacobiAM(2*arctan(1/2*(-3*x^2-2)^(1/4)*2^(3/4)),1 
/2*2^(1/2))*3^(1/2)/x
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 4.77 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.21 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=-\frac {\sqrt [4]{1+\frac {3 x^2}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {1}{2},-\frac {3 x^2}{2}\right )}{x \sqrt [4]{-2-3 x^2}} \] Input:

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

Output:

-(((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)))
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {264, 228, 27, 834, 27, 761, 1510}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 228

\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {\frac {3}{2}} \sqrt {-x^2} \int \frac {\sqrt {\frac {2}{3}} \sqrt {-3 x^2-2}}{\sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}}{2 x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \int \frac {\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}}{2 x}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}-\sqrt {2} \int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {6} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{2 x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}-\int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{2 x}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\frac {\sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {-x^2}}-\int \frac {\sqrt {2}-\sqrt {-3 x^2-2}}{\sqrt {3} \sqrt {-x^2}}d\sqrt [4]{-3 x^2-2}\right )}{2 x}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {\left (-3 x^2-2\right )^{3/4}}{2 x}-\frac {\sqrt {3} \sqrt {-x^2} \left (\frac {\sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {-x^2}}-\frac {\sqrt [4]{2} \sqrt {-\frac {x^2}{\left (\sqrt {-3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {-3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {-x^2}}+\frac {\sqrt {3} \sqrt {-x^2} \sqrt [4]{-3 x^2-2}}{\sqrt {-3 x^2-2}+\sqrt {2}}\right )}{2 x}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 228

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

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 834

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

rule 1510

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]))*E 
llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e 
}, x] && PosQ[c/a]
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.10

method	result	size
meijerg	\(\frac {\left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {1}{2}\right ], -\frac {3 x^{2}}{2}\right )}{2 x}\)	\(23\)
risch	\(-\frac {3 x^{2}+2}{2 x \left (-3 x^{2}-2\right )^{\frac {1}{4}}}-\frac {3 \left (-1\right )^{\frac {3}{4}} 2^{\frac {3}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -\frac {3 x^{2}}{2}\right )}{8}\)	\(43\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=\frac {2^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2} e^{i \pi }}{2}} \right )}}{2 x} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.16 \[ \int \frac {1}{x^2 \sqrt [4]{-2-3 x^2}} \, dx=-\frac {2\,3^{3/4}\,{\left (\frac {2}{x^2}+3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -\frac {2}{3\,x^2}\right )}{9\,x\,{\left (-3\,x^2-2\right )}^{1/4}} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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