\(\int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx\) [137]

Optimal result

Integrand size = 18, antiderivative size = 40 \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (2-x^2\right )}{2 \sqrt {3-3 x^2+x^4}}\right )}{2 \sqrt {3}} \]

[Out]

-1/6*arctanh(1/2*(-x^2+2)*3^(1/2)/(x^4-3*x^2+3)^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1128, 738, 212} \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {3} \left (2-x^2\right )}{2 \sqrt {x^4-3 x^2+3}}\right )}{2 \sqrt {3}} \]

[In]

Int[1/(x*Sqrt[3 - 3*x^2 + x^4]),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 1128

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {3-3 x+x^2}} \, dx,x,x^2\right ) \\ & = -\text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {3 \left (2-x^2\right )}{\sqrt {3-3 x^2+x^4}}\right ) \\ & = -\frac {\tanh ^{-1}\left (\frac {\sqrt {3} \left (2-x^2\right )}{2 \sqrt {3-3 x^2+x^4}}\right )}{2 \sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=\frac {\text {arctanh}\left (\frac {x^2-\sqrt {3-3 x^2+x^4}}{\sqrt {3}}\right )}{\sqrt {3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.72

method	result	size
pseudoelliptic	\(\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (x^{2}-2\right ) \sqrt {3}}{2 \sqrt {x^{4}-3 x^{2}+3}}\right )}{6}\)	\(29\)
default	\(-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-3 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}-3 x^{2}+3}}\right )}{6}\)	\(31\)
elliptic	\(-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (-3 x^{2}+6\right ) \sqrt {3}}{6 \sqrt {x^{4}-3 x^{2}+3}}\right )}{6}\)	\(31\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+2 \sqrt {x^{4}-3 x^{2}+3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{x^{2}}\right )}{6}\)	\(47\)

[In]

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

[Out]

1/6*3^(1/2)*arctanh(1/2*(x^2-2)*3^(1/2)/(x^4-3*x^2+3)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (-\frac {3 \, x^{2} + 2 \, \sqrt {3} {\left (x^{2} - 2\right )} + 2 \, \sqrt {x^{4} - 3 \, x^{2} + 3} {\left (\sqrt {3} + 2\right )} - 6}{x^{2}}\right ) \]

[In]

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

[Out]

1/6*sqrt(3)*log(-(3*x^2 + 2*sqrt(3)*(x^2 - 2) + 2*sqrt(x^4 - 3*x^2 + 3)*(sqrt(3) + 2) - 6)/x^2)

Sympy [F]

\[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=\int \frac {1}{x \sqrt {x^{4} - 3 x^{2} + 3}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=-\frac {1}{6} \, \sqrt {3} \operatorname {arsinh}\left (-\sqrt {3} + \frac {2 \, \sqrt {3}}{x^{2}}\right ) \]

[In]

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

[Out]

-1/6*sqrt(3)*arcsinh(-sqrt(3) + 2*sqrt(3)/x^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38 \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} - \sqrt {x^{4} - 3 \, x^{2} + 3}\right ) - \frac {1}{6} \, \sqrt {3} \log \left (-x^{2} + \sqrt {3} + \sqrt {x^{4} - 3 \, x^{2} + 3}\right ) \]

[In]

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

[Out]

1/6*sqrt(3)*log(x^2 + sqrt(3) - sqrt(x^4 - 3*x^2 + 3)) - 1/6*sqrt(3)*log(-x^2 + sqrt(3) + sqrt(x^4 - 3*x^2 + 3
))

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x \sqrt {3-3 x^2+x^4}} \, dx=-\frac {\sqrt {3}\,\left (\ln \left (x^2-\frac {2\,\sqrt {3}\,\sqrt {x^4-3\,x^2+3}}{3}-2\right )+\ln \left (\frac {1}{x^2}\right )\right )}{6} \]

[In]

int(1/(x*(x^4 - 3*x^2 + 3)^(1/2)),x)

[Out]

-(3^(1/2)*(log(x^2 - (2*3^(1/2)*(x^4 - 3*x^2 + 3)^(1/2))/3 - 2) + log(1/x^2)))/6