\(\int \frac {1}{x^2 (2-3 x^2)^{3/4} (4-3 x^2)} \, dx\) [1489]

Optimal result

Integrand size = 24, antiderivative size = 166 \[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\sqrt [4]{2-3 x^2}}{8 x}+\frac {\sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} x \sqrt [4]{2-3 x^2}}{2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}\right )}{16 \sqrt [4]{2}}+\frac {\sqrt {3} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{4 \sqrt [4]{2}} \] Output:

-1/8*(-3*x^2+2)^(1/4)/x+1/32*2^(3/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2 
)^(1/2))*3^(1/2)/x/(-3*x^2+2)^(1/4))*3^(1/2)-1/32*2^(3/4)*arctanh(3^(1/2)* 
x*(-3*x^2+2)^(1/4)/(2^(3/4)+2^(1/4)*(-3*x^2+2)^(1/2)))*3^(1/2)+1/8*2^(3/4) 
*InverseJacobiAM(1/2*arcsin(1/2*x*6^(1/2)),2^(1/2))*3^(1/2)
 

Mathematica [C] (verified)

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

Time = 11.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {1}{2},\frac {3}{4},1,\frac {1}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{4\ 2^{3/4} x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 352

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {3} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{4 \sqrt [4]{2}}+\frac {\sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{16 \sqrt [4]{2}}-\frac {\sqrt [4]{2-3 x^2}}{8 x}\)

Input:

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

Output:

-1/8*(2 - 3*x^2)^(1/4)/x + (Sqrt[3]*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x 
^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(16*2^(1/4)) - (Sqrt[3]*ArcTanh[(2^(3 
/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(16*2^(1/4) 
) + (Sqrt[3]*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(4*2^(1/4))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((-3*x^2 + 2)^(1/4)/(9*x^6 - 18*x^4 + 8*x^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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