3.9.93 \(\int \frac {x^2}{\sqrt [4]{-2+3 x^2}} \, dx\) [893]

3.9.93.1 Optimal result

Integrand size = 15, antiderivative size = 222 \[ \int \frac {x^2}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2}{15} x \left (-2+3 x^2\right )^{3/4}+\frac {8 x \sqrt [4]{-2+3 x^2}}{15 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}-\frac {8 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}+\frac {4 \sqrt [4]{2} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{15 \sqrt {3} x} \]

output

2/15*x*(3*x^2-2)^(3/4)+8/15*x*(3*x^2-2)^(1/4)/(2^(1/2)+(3*x^2-2)^(1/2))-8/ 
45*2^(1/4)*(cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arct 
an(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)+4/45*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)
 

3.9.93.2 Mathematica [C] (verified)

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

Time = 5.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.26 \[ \int \frac {x^2}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2 x \left (-2+3 x^2+2^{3/4} \sqrt [4]{2-3 x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {3 x^2}{2}\right )\right )}{15 \sqrt [4]{-2+3 x^2}} \]

input

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

output

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

3.9.93.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {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.

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 228

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

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

input

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

output

(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])*EllipticF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^ 
(1/4)], 1/2])/(2^(3/4)*Sqrt[x^2])))/(15*Sqrt[3]*x)
 

3.9.93.3.1 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]
 

3.9.93.4 Maple [C] (warning: unable to verify)

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

Time = 2.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.19

method	result	size
meijerg	\(\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {3}{2};\frac {5}{2};\frac {3 x^{2}}{2}\right )}{6 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\)	\(42\)
risch	\(\frac {2 x \left (3 x^{2}-2\right )^{\frac {3}{4}}}{15}+\frac {2 \,2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{15 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\)	\(53\)

input

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

output

1/6*2^(3/4)/signum(-1+3/2*x^2)^(1/4)*(-signum(-1+3/2*x^2))^(1/4)*x^3*hyper 
geom([1/4,3/2],[5/2],3/2*x^2)
 

3.9.93.5 Fricas [F]

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

input

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

output

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

3.9.93.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.13 \[ \int \frac {x^2}{\sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} x^{3} e^{- \frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{6} \]

input

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

output

2**(3/4)*x**3*exp(-I*pi/4)*hyper((1/4, 3/2), (5/2,), 3*x**2/2)/6
 

3.9.93.7 Maxima [F]

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

input

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

output

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

3.9.93.8 Giac [F]

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

input

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

output

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

3.9.93.9 Mupad [F(-1)]

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

input

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

output

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