\(\int \frac {1}{\sqrt {x^2 (3-3 x^2+x^4)}} \, dx\) [138]

Optimal result

Integrand size = 18, antiderivative size = 45 \[ \int \frac {1}{\sqrt {x^2 \left (3-3 x^2+x^4\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x \left (6-3 x^2\right )}{2 \sqrt {3} \sqrt {3 x^2-3 x^4+x^6}}\right )}{2 \sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, 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 = {2021, 1918, 212} \[ \int \frac {1}{\sqrt {x^2 \left (3-3 x^2+x^4\right )}} \, dx=-\frac {\text {arctanh}\left (\frac {x \left (6-3 x^2\right )}{2 \sqrt {3} \sqrt {x^6-3 x^4+3 x^2}}\right )}{2 \sqrt {3}} \]

[In]

Int[1/Sqrt[x^2*(3 - 3*x^2 + x^4)],x]

[Out]

-1/2*ArcTanh[(x*(6 - 3*x^2))/(2*Sqrt[3]*Sqrt[3*x^2 - 3*x^4 + x^6])]/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 1918

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - 2), Subst[Int[1/(4*a
- x^2), x], x, x*((2*a + b*x^(n - 2))/Sqrt[a*x^2 + b*x^n + c*x^r])], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r
, 2*n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]

Rule 2021

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && GeneralizedTrinomialQ[u, x] &&  !Gen
eralizedTrinomialMatchQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {3 x^2-3 x^4+x^6}} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {x \left (6-3 x^2\right )}{\sqrt {3 x^2-3 x^4+x^6}}\right ) \\ & = -\frac {\tanh ^{-1}\left (\frac {x \left (6-3 x^2\right )}{2 \sqrt {3} \sqrt {3 x^2-3 x^4+x^6}}\right )}{2 \sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\sqrt {x^2 \left (3-3 x^2+x^4\right )}} \, dx=\frac {x \sqrt {3-3 x^2+x^4} \text {arctanh}\left (\frac {x^2-\sqrt {3-3 x^2+x^4}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt {x^2 \left (3-3 x^2+x^4\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (x^{2}-2\right ) \sqrt {3}\, x}{2 \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}\right )}{6}\)	\(34\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +2 \sqrt {x^{6}-3 x^{4}+3 x^{2}}}{x^{3}}\right )}{6}\)	\(53\)
default	\(\frac {\sqrt {x^{4}-3 x^{2}+3}\, x \sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (x^{2}-2\right ) \sqrt {3}}{2 \sqrt {x^{4}-3 x^{2}+3}}\right )}{6 \sqrt {x^{2} \left (x^{4}-3 x^{2}+3\right )}}\)	\(58\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.22 \[ \int \frac {1}{\sqrt {x^2 \left (3-3 x^2+x^4\right )}} \, dx=\frac {1}{6} \, \sqrt {3} \log \left (-\frac {3 \, x^{3} + 2 \, \sqrt {3} {\left (x^{3} - 2 \, x\right )} + 2 \, \sqrt {x^{6} - 3 \, x^{4} + 3 \, x^{2}} {\left (\sqrt {3} + 2\right )} - 6 \, x}{x^{3}}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\sqrt {x^2 \left (3-3 x^2+x^4\right )}} \, dx=\frac {\sqrt {3} \log \left (x^{2} + \sqrt {3} - \sqrt {x^{4} - 3 \, x^{2} + 3}\right ) - \sqrt {3} \log \left (-x^{2} + \sqrt {3} + \sqrt {x^{4} - 3 \, x^{2} + 3}\right )}{6 \, \mathrm {sgn}\left (x\right )} \]

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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