\(\int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx\) [956]

Optimal result

Integrand size = 15, antiderivative size = 240 \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {8}{135} x \left (-2+3 x^2\right )^{3/4}+\frac {2}{27} x^3 \left (-2+3 x^2\right )^{3/4}+\frac {32 x \sqrt [4]{-2+3 x^2}}{135 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}-\frac {32 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{135 \sqrt {3} x}+\frac {16 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{135 \sqrt {3} x} \] Output:

8/135*x*(3*x^2-2)^(3/4)+2/27*x^3*(3*x^2-2)^(3/4)+32*x*(3*x^2-2)^(1/4)/(135 
*2^(1/2)+135*(3*x^2-2)^(1/2))-32/405*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+16/405*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 = 63, normalized size of antiderivative = 0.26 \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2 x \left (-8+2 x^2+15 x^4+4\ 2^{3/4} \sqrt [4]{2-3 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {3 x^2}{2}\right )\right )}{135 \sqrt [4]{-2+3 x^2}} \] Input:

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

Output:

(2*x*(-8 + 2*x^2 + 15*x^4 + 4*2^(3/4)*(2 - 3*x^2)^(1/4)*Hypergeometric2F1[ 
1/4, 1/2, 3/2, (3*x^2)/2]))/(135*(-2 + 3*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {262, 262, 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.

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {4}{9} \int \frac {x^2}{\sqrt [4]{3 x^2-2}}dx+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {4}{9} \left (\frac {4}{15} \int \frac {1}{\sqrt [4]{3 x^2-2}}dx+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 228

\(\displaystyle \frac {4}{9} \left (\frac {4 \sqrt {\frac {2}{3}} \sqrt {x^2} \int \frac {\sqrt {\frac {2}{3}} \sqrt {3 x^2-2}}{\sqrt {x^2}}d\sqrt [4]{3 x^2-2}}{15 x}+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4}{9} \left (\frac {8 \sqrt {x^2} \int \frac {\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}}{15 \sqrt {3} x}+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {4}{9} \left (\frac {8 \sqrt {x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}-\sqrt {2} \int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {6} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{15 \sqrt {3} x}+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4}{9} \left (\frac {8 \sqrt {x^2} \left (\sqrt {2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}-\int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{15 \sqrt {3} x}+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {4}{9} \left (\frac {8 \sqrt {x^2} \left (\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {x^2}}-\int \frac {\sqrt {2}-\sqrt {3 x^2-2}}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-2}\right )}{15 \sqrt {3} x}+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {4}{9} \left (\frac {8 \sqrt {x^2} \left (\frac {\sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{2^{3/4} \sqrt {x^2}}-\frac {\sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{\sqrt {x^2}}+\frac {\sqrt {3} \sqrt {x^2} \sqrt [4]{3 x^2-2}}{\sqrt {3 x^2-2}+\sqrt {2}}\right )}{15 \sqrt {3} x}+\frac {2}{15} \left (3 x^2-2\right )^{3/4} x\right )+\frac {2}{27} \left (3 x^2-2\right )^{3/4} x^3\)

Input:

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

Output:

(2*x^3*(-2 + 3*x^2)^(3/4))/27 + (4*((2*x*(-2 + 3*x^2)^(3/4))/15 + (8*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])*Elli 
pticF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(2^(3/4)*Sqrt[x^2])))/(1 
5*Sqrt[3]*x)))/9
 

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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.18

method	result	size
meijerg	\(\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {7}{2}\right ], \frac {3 x^{2}}{2}\right )}{10 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\)	\(42\)
risch	\(\frac {2 x \left (5 x^{2}+4\right ) \left (3 x^{2}-2\right )^{\frac {3}{4}}}{135}+\frac {8 \,2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{135 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\)	\(60\)

Input:

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

Output:

1/10*2^(3/4)/signum(-1+3/2*x^2)^(1/4)*(-signum(-1+3/2*x^2))^(1/4)*x^5*hype 
rgeom([1/4,5/2],[7/2],3/2*x^2)
 

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.12 \[ \int \frac {x^4}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} x^{5} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{10} \] Input:

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

Output:

2**(3/4)*x**5*exp(-I*pi/4)*hyper((1/4, 5/2), (7/2,), 3*x**2/2)/10
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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