\(\int \frac {1}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\) [1518]

Optimal result

Integrand size = 21, antiderivative size = 129 \[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )}{2 \sqrt {6}}-\frac {\text {arctanh}\left (\frac {\sqrt {\frac {2}{3}} \sqrt [4]{-1+3 x^2}}{x}\right )}{2 \sqrt {6}}-\frac {\sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{2 \sqrt {3} x} \] Output:

1/12*arctan(1/2*6^(1/2)*x/(3*x^2-1)^(1/4))*6^(1/2)-1/12*arctanh(1/3*6^(1/2 
)/x*(3*x^2-1)^(1/4))*6^(1/2)-1/6*(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 complex when optimal does not.

Time = 5.62 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{-1} \sqrt {x^2} \left (\operatorname {EllipticPi}\left (-i,\arcsin \left ((-1)^{3/4} \sqrt [4]{-1+3 x^2}\right ),-1\right )+\operatorname {EllipticPi}\left (i,\arcsin \left ((-1)^{3/4} \sqrt [4]{-1+3 x^2}\right ),-1\right )\right )}{\sqrt {3} x} \] Input:

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

Output:

((-1)^(1/4)*Sqrt[x^2]*(EllipticPi[-I, ArcSin[(-1)^(3/4)*(-1 + 3*x^2)^(1/4) 
], -1] + EllipticPi[I, ArcSin[(-1)^(3/4)*(-1 + 3*x^2)^(1/4)], -1]))/(Sqrt[ 
3]*x)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {311, 25, 232, 351, 761}

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.

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

\(\Big \downarrow \) 311

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 232

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

\(\Big \downarrow \) 351

\(\displaystyle -\frac {\sqrt {x^2} \int \frac {1}{\sqrt {3} \sqrt {x^2}}d\sqrt [4]{3 x^2-1}}{\sqrt {3} x}-\frac {3}{2} \left (\frac {\text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{3 \sqrt {6}}-\frac {\arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )}{3 \sqrt {6}}\right )\)

\(\Big \downarrow \) 761

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

Input:

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

Output:

(-3*(-1/3*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]/Sqrt[6] + ArcTanh[(Sqrt 
[3/2]*x)/(-1 + 3*x^2)^(1/4)]/(3*Sqrt[6])))/2 - (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*Sqrt[3]*x)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[2*(Sqrt[(-b)*(x^2/a)]/( 
b*x))   Subst[Int[1/Sqrt[1 - x^4/a], x], x, (a + b*x^2)^(1/4)], x] /; FreeQ 
[{a, b}, x] && NegQ[a]
 

rule 311

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Sim 
p[1/c   Int[1/(a + b*x^2)^(3/4), x], x] - Simp[d/c   Int[x^2/((a + b*x^2)^( 
3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0]
 

rule 351

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] : 
> Simp[(-b/(Sqrt[2]*a*d*Rt[-b^2/a, 4]^3))*ArcTan[(Rt[-b^2/a, 4]*x)/(Sqrt[2] 
*(a + b*x^2)^(1/4))], x] + Simp[(b/(Sqrt[2]*a*d*Rt[-b^2/a, 4]^3))*ArcTanh[( 
Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c, d}, x] 
 && EqQ[b*c - 2*a*d, 0] && NegQ[b^2/a]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/(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)
 

Giac [F]

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

integrate(1/(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)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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