\(\int \frac {1}{x^2 (-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\) [1519]

Optimal result

Integrand size = 24, antiderivative size = 151 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{-1+3 x^2}}{2 x}+\frac {1}{4} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}} \sqrt [4]{-1+3 x^2}}{x}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{2 x} \] Output:

-1/2*(3*x^2-1)^(1/4)/x+1/8*arctan(1/2*6^(1/2)*x/(3*x^2-1)^(1/4))*6^(1/2)-1 
/8*arctanh(1/3*6^(1/2)/x*(3*x^2-1)^(1/4))*6^(1/2)-1/2*(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] (verified)

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

Time = 10.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.34 \[ \int \frac {1}{x^2 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\left (1-3 x^2\right )^{3/4} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {3}{4},1,\frac {1}{2},3 x^2,\frac {3 x^2}{2}\right )}{2 x \left (-1+3 x^2\right )^{3/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 149, 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 = {352, 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.

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

\(\Big \downarrow \) 352

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 352

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol 
] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(3/4)*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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