\(\int \frac {1}{x^4 (-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\) [1461]

Optimal result

Integrand size = 24, antiderivative size = 266 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=-\frac {\left (-1+3 x^2\right )^{3/4}}{6 x^3}-\frac {3 \left (-1+3 x^2\right )^{3/4}}{2 x}+\frac {9 x \sqrt [4]{-1+3 x^2}}{2 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {3}{8} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}} \sqrt [4]{-1+3 x^2}}{x}\right )-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{2 x}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{4 x} \] Output:

-1/6*(3*x^2-1)^(3/4)/x^3-3/2*(3*x^2-1)^(3/4)/x+9*x*(3*x^2-1)^(1/4)/(2+2*(3 
*x^2-1)^(1/2))-3/16*arctan(1/2*6^(1/2)*x/(3*x^2-1)^(1/4))*6^(1/2)-3/16*arc 
tanh(1/3*6^(1/2)/x*(3*x^2-1)^(1/4))*6^(1/2)-3/2*(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+3/4*(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 = 10.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\frac {1}{2} \left (-1+3 x^2\right )^{3/4} \left (-\frac {1+9 x^2}{3 x^3}+\frac {9 x \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )}{\left (-2+3 x^2\right ) \left (2 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{4},2,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )-3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )\right )\right )}\right ) \] Input:

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

Output:

((-1 + 3*x^2)^(3/4)*(-1/3*(1 + 9*x^2)/x^3 + (9*x*AppellF1[1/2, -3/4, 1, 3/ 
2, 3*x^2, (3*x^2)/2])/((-2 + 3*x^2)*(2*AppellF1[1/2, -3/4, 1, 3/2, 3*x^2, 
(3*x^2)/2] + x^2*(2*AppellF1[3/2, -3/4, 2, 5/2, 3*x^2, (3*x^2)/2] - 3*Appe 
llF1[3/2, 1/4, 1, 5/2, 3*x^2, (3*x^2)/2])))))/2
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 264, 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.

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

\(\Big \downarrow \) 349

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3}{8} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{4 x}-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) E\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{2 x}-\frac {3}{8} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {9 \sqrt [4]{3 x^2-1} x}{2 \left (\sqrt {3 x^2-1}+1\right )}-\frac {3 \left (3 x^2-1\right )^{3/4}}{2 x}-\frac {\left (3 x^2-1\right )^{3/4}}{6 x^3}\)

Input:

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

Output:

-1/6*(-1 + 3*x^2)^(3/4)/x^3 - (3*(-1 + 3*x^2)^(3/4))/(2*x) + (9*x*(-1 + 3* 
x^2)^(1/4))/(2*(1 + Sqrt[-1 + 3*x^2])) - (3*Sqrt[3/2]*ArcTan[(Sqrt[3/2]*x) 
/(-1 + 3*x^2)^(1/4)])/8 - (3*Sqrt[3/2]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^ 
(1/4)])/8 - (3*Sqrt[3]*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])/(2*x) + (3*Sqrt[3]*Sq 
rt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*ArcTan 
[(-1 + 3*x^2)^(1/4)], 1/2])/(4*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((3*x^2 - 1)^(3/4)/(9*x^8 - 9*x^6 + 2*x^4), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx=\int \frac {1}{3 \left (3 x^{2}-1\right )^{\frac {1}{4}} x^{6}-2 \left (3 x^{2}-1\right )^{\frac {1}{4}} x^{4}}d x \] Input:

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

Output:

int(1/(3*(3*x**2 - 1)**(1/4)*x**6 - 2*(3*x**2 - 1)**(1/4)*x**4),x)