\(\int \frac {1}{x^4 \sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\) [1448]

Optimal result

Integrand size = 24, antiderivative size = 184 \[ \int \frac {1}{x^4 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {\left (2-3 x^2\right )^{3/4}}{24 x^3}-\frac {3 \left (2-3 x^2\right )^{3/4}}{16 x}+\frac {3 \sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{32\ 2^{3/4}}+\frac {3 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} x \sqrt [4]{2-3 x^2}}{2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}\right )}{32\ 2^{3/4}}-\frac {3 \sqrt {3} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{8\ 2^{3/4}} \] Output:

-1/24*(-3*x^2+2)^(3/4)/x^3-3/16*(-3*x^2+2)^(3/4)/x+3/64*2^(1/4)*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)+3/ 
64*2^(1/4)*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)-3/16*2^(1/4)*EllipticE(sin(1/2*arcsin(1/2*x*6^(1/2))),2^(1 
/2))*3^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 11.12 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^4 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {1}{8} \left (2-3 x^2\right )^{3/4} \left (-\frac {2+9 x^2}{6 x^3}+\frac {9 x \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{\left (-4+3 x^2\right ) \left (4 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},1,\frac {3}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{4},2,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )-3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )\right )}\right ) \] Input:

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

Output:

((2 - 3*x^2)^(3/4)*(-1/6*(2 + 9*x^2)/x^3 + (9*x*AppellF1[1/2, -3/4, 1, 3/2 
, (3*x^2)/2, (3*x^2)/4])/((-4 + 3*x^2)*(4*AppellF1[1/2, -3/4, 1, 3/2, (3*x 
^2)/2, (3*x^2)/4] + x^2*(2*AppellF1[3/2, -3/4, 2, 5/2, (3*x^2)/2, (3*x^2)/ 
4] - 3*AppellF1[3/2, 1/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4])))))/8
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 184, 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 = {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.

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

\(\Big \downarrow \) 349

\(\displaystyle \int \left (-\frac {9}{16 \left (3 x^2-4\right ) \sqrt [4]{2-3 x^2}}+\frac {3}{16 x^2 \sqrt [4]{2-3 x^2}}+\frac {1}{4 x^4 \sqrt [4]{2-3 x^2}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \sqrt {3} E\left (\left .\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{8\ 2^{3/4}}+\frac {3 \sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{32\ 2^{3/4}}+\frac {3 \sqrt {3} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{32\ 2^{3/4}}-\frac {3 \left (2-3 x^2\right )^{3/4}}{16 x}-\frac {\left (2-3 x^2\right )^{3/4}}{24 x^3}\)

Input:

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

Output:

-1/24*(2 - 3*x^2)^(3/4)/x^3 - (3*(2 - 3*x^2)^(3/4))/(16*x) + (3*Sqrt[3]*Ar 
cTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/( 
32*2^(3/4)) + (3*Sqrt[3]*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt 
[3]*x*(2 - 3*x^2)^(1/4))])/(32*2^(3/4)) - (3*Sqrt[3]*EllipticE[ArcSin[Sqrt 
[3/2]*x]/2, 2])/(8*2^(3/4))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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