\(\int \frac {x^4}{(-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\) [1457]

Optimal result

Integrand size = 24, antiderivative size = 246 \[ \int \frac {x^4}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2}{45} x \left (-1+3 x^2\right )^{3/4}+\frac {8 x \sqrt [4]{-1+3 x^2}}{15 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {1}{9} \sqrt {\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}} \sqrt [4]{-1+3 x^2}}{x}\right )-\frac {8 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}+\frac {4 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{15 \sqrt {3} x} \] Output:

2/45*x*(3*x^2-1)^(3/4)+8*x*(3*x^2-1)^(1/4)/(15+15*(3*x^2-1)^(1/2))-1/27*ar 
ctan(1/2*6^(1/2)*x/(3*x^2-1)^(1/4))*6^(1/2)-1/27*arctanh(1/3*6^(1/2)/x*(3* 
x^2-1)^(1/4))*6^(1/2)-8/45*(x^2/(1+(3*x^2-1)^(1/2))^2)^(1/2)*(1+(3*x^2-1)^ 
(1/2))*EllipticE(sin(2*arctan((3*x^2-1)^(1/4))),1/2*2^(1/2))*3^(1/2)/x+4/4 
5*(x^2/(1+(3*x^2-1)^(1/2))^2)^(1/2)*(1+(3*x^2-1)^(1/2))*InverseJacobiAM(2* 
arctan((3*x^2-1)^(1/4)),1/2*2^(1/2))*3^(1/2)/x
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 5.57 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.72 \[ \int \frac {x^4}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {2 x \left (-1+3 x^2-3 x^2 \sqrt [4]{1-3 x^2} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )-\frac {4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )}{\left (-2+3 x^2\right ) \left (2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )\right )\right )}\right )}{45 \sqrt [4]{-1+3 x^2}} \] Input:

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

Output:

(2*x*(-1 + 3*x^2 - 3*x^2*(1 - 3*x^2)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, 3*x^ 
2, (3*x^2)/2] - (4*AppellF1[1/2, 1/4, 1, 3/2, 3*x^2, (3*x^2)/2])/((-2 + 3* 
x^2)*(2*AppellF1[1/2, 1/4, 1, 3/2, 3*x^2, (3*x^2)/2] + x^2*(2*AppellF1[3/2 
, 1/4, 2, 5/2, 3*x^2, (3*x^2)/2] + AppellF1[3/2, 5/4, 1, 5/2, 3*x^2, (3*x^ 
2)/2])))))/(45*(-1 + 3*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {349, 2009}

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.

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

\(\Big \downarrow \) 349

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{9} \sqrt {\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {4 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{15 \sqrt {3} x}-\frac {8 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) E\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}-\frac {1}{9} \sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {2}{45} \left (3 x^2-1\right )^{3/4} x+\frac {8 \sqrt [4]{3 x^2-1} x}{15 \left (\sqrt {3 x^2-1}+1\right )}\)

Input:

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

Output:

(2*x*(-1 + 3*x^2)^(3/4))/45 + (8*x*(-1 + 3*x^2)^(1/4))/(15*(1 + Sqrt[-1 + 
3*x^2])) - (Sqrt[2/3]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/9 - (Sqrt[ 
2/3]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/9 - (8*Sqrt[x^2/(1 + Sqrt[ 
-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticE[2*ArcTan[(-1 + 3*x^2)^(1/ 
4)], 1/2])/(15*Sqrt[3]*x) + (4*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqr 
t[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(15*Sqrt[3]*x 
)
 

Defintions of rubi rules used

rule 349

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol 
] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(1/4)*(c + d*x^2)), x], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In 
tegerQ[m/2])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((3*x^2 - 1)^(3/4)*x^4/(9*x^4 - 9*x^2 + 2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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