\(\int \frac {1}{x \sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\) [1442]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{2-3 x^2}}{\sqrt {2}+\sqrt {2-3 x^2}}\right )}{4\ 2^{3/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {\sqrt {2} \arctan \left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+\arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-\sqrt {2} \text {arctanh}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+\text {arctanh}\left (\frac {2 \sqrt [4]{4-6 x^2}}{2+\sqrt {4-6 x^2}}\right )}{4\ 2^{3/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 145, 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.

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

\(\Big \downarrow \) 349

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Time = 6.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-\left (-3 x^{2}+2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}}{-\left (-3 x^{2}+2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}}\right ) \sqrt {2}-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}}{2}\right ) \sqrt {2}+\ln \left (\frac {\sqrt {-3 x^{2}+2}-2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}}{\sqrt {-3 x^{2}+2}+2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}}\right )+2 \arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}-1\right )+2 \arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+1\right )\right )}{16}\)	\(160\)

Input:

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

Output:

-1/16*2^(1/4)*(ln((-(-3*x^2+2)^(1/4)-2^(1/4))/(-(-3*x^2+2)^(1/4)+2^(1/4))) 
*2^(1/2)-2*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(1/2)+ln(((-3*x^2+2)^(1/ 
2)-2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2))/((-3*x^2+2)^(1/2)+2^(3/4)*(-3*x^2+2)^ 
(1/4)+2^(1/2)))+2*arctan(2^(1/4)*(-3*x^2+2)^(1/4)-1)+2*arctan(2^(1/4)*(-3* 
x^2+2)^(1/4)+1))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {1}{32} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{4} \cdot 8^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{32} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{4} \cdot 8^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} - 1\right ) + \frac {1}{8} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (2 \, \sqrt {2} + 2 \cdot 8^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 2 \, \sqrt {-3 \, x^{2} + 2}\right ) - \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (2 \, \sqrt {2} - 2 \cdot 8^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 2 \, \sqrt {-3 \, x^{2} + 2}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) \] Input:

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

Output:

-1/32*8^(3/4)*arctan(1/4*8^(3/4)*(-3*x^2 + 2)^(1/4) + 1) - 1/32*8^(3/4)*ar 
ctan(1/4*8^(3/4)*(-3*x^2 + 2)^(1/4) - 1) + 1/8*2^(3/4)*arctan(1/2*2^(3/4)* 
(-3*x^2 + 2)^(1/4)) + 1/64*8^(3/4)*log(2*sqrt(2) + 2*8^(1/4)*(-3*x^2 + 2)^ 
(1/4) + 2*sqrt(-3*x^2 + 2)) - 1/64*8^(3/4)*log(2*sqrt(2) - 2*8^(1/4)*(-3*x 
^2 + 2)^(1/4) + 2*sqrt(-3*x^2 + 2)) - 1/16*2^(3/4)*log(2^(1/4) + (-3*x^2 + 
 2)^(1/4)) + 1/16*2^(3/4)*log(-2^(1/4) + (-3*x^2 + 2)^(1/4))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/x/(-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), x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {1}{16} \cdot 4^{\frac {3}{8}} \sqrt {2} \arctan \left (\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {3}{8}} \sqrt {2} \arctan \left (-\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{32} \cdot 4^{\frac {3}{8}} \sqrt {2} \log \left (4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) - \frac {1}{32} \cdot 4^{\frac {3}{8}} \sqrt {2} \log \left (-4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) + \frac {1}{8} \cdot 4^{\frac {1}{8}} \sqrt {2} \arctan \left (\frac {1}{4} \cdot 4^{\frac {7}{8}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} \cdot 4^{\frac {1}{8}} \sqrt {2} \log \left (-{\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) - \frac {1}{16} \cdot 4^{\frac {3}{8}} \log \left ({\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) \] Input:

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

Output:

-1/16*4^(3/8)*sqrt(2)*arctan(1/8*4^(7/8)*sqrt(2)*(4^(1/8)*sqrt(2) + 2*(-3* 
x^2 + 2)^(1/4))) - 1/16*4^(3/8)*sqrt(2)*arctan(-1/8*4^(7/8)*sqrt(2)*(4^(1/ 
8)*sqrt(2) - 2*(-3*x^2 + 2)^(1/4))) + 1/32*4^(3/8)*sqrt(2)*log(4^(1/8)*sqr 
t(2)*(-3*x^2 + 2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) - 1/32*4^(3/8)*sqrt( 
2)*log(-4^(1/8)*sqrt(2)*(-3*x^2 + 2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) + 
 1/8*4^(1/8)*sqrt(2)*arctan(1/4*4^(7/8)*(-3*x^2 + 2)^(1/4)) + 1/16*4^(1/8) 
*sqrt(2)*log(-(-3*x^2 + 2)^(1/4) + 4^(1/8)) - 1/16*4^(3/8)*log((-3*x^2 + 2 
)^(1/4) + 4^(1/8))
 

Mupad [B] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{8}+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\frac {2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{8} \] Input:

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

Output:

(2^(3/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4))/2))/8 - 2^(1/4)*atan(2^(1/4)*(2 
- 3*x^2)^(1/4)*(1/2 - 1i/2))*(1/8 - 1i/8) - 2^(1/4)*atan(2^(1/4)*(2 - 3*x^ 
2)^(1/4)*(1/2 + 1i/2))*(1/8 + 1i/8) + (2^(3/4)*atan((2^(3/4)*(2 - 3*x^2)^( 
1/4)*1i)/2)*1i)/8
 

Reduce [F]

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

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

Output:

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