3.893 \(\int \frac{\sqrt{1-\frac{1}{(-1+x^2)^2}}}{2-x^2} \, dx\)

Optimal. Leaf size=47 \[ \frac{\left (1-x^2\right ) \sqrt{1-\frac{1}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\sqrt{x^2-2}\right )}{x \sqrt{x^2-2}} \]

[Out]

((1 - x^2)*Sqrt[1 - (1 - x^2)^(-2)]*ArcTan[Sqrt[-2 + x^2]])/(x*Sqrt[-2 + x^2])

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Rubi [A]  time = 0.460336, antiderivative size = 73, normalized size of antiderivative = 1.55, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6722, 6725, 1990, 1146, 21, 261, 444, 50, 63, 203} \[ \frac{\left (1-x^2\right ) \sqrt{x^4-2 x^2} \sqrt{1-\frac{1}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\sqrt{x^2-2}\right )}{x \sqrt{x^2-2} \sqrt{\left (x^2-1\right )^2-1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - (-1 + x^2)^(-2)]/(2 - x^2),x]

[Out]

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

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 1990

Int[(u_)^(q_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*ExpandToSum[v, x]^p, x] /; FreeQ[{p, q}, x] &&
BinomialQ[u, x] && TrinomialQ[v, x] &&  !(BinomialMatchQ[u, x] && TrinomialMatchQ[v, x])

Rule 1146

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(b*x^2 + c*x^4)^FracPart
[p]/(x^(2*FracPart[p])*(b + c*x^2)^FracPart[p]), Int[x^(2*p)*(d + e*x^2)^q*(b + c*x^2)^p, x], x] /; FreeQ[{b,
c, d, e, p, q}, x] &&  !IntegerQ[p]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}}{2-x^2} \, dx &=\frac{\left (\left (-1+x^2\right ) \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{\sqrt{-1+\left (-1+x^2\right )^2}}{\left (2-x^2\right ) \left (-1+x^2\right )} \, dx}{\sqrt{-1+\left (-1+x^2\right )^2}}\\ &=\frac{\left (\left (-1+x^2\right ) \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \left (\frac{\sqrt{-1+\left (-1+x^2\right )^2}}{2-x^2}+\frac{\sqrt{-1+\left (-1+x^2\right )^2}}{-1+x^2}\right ) \, dx}{\sqrt{-1+\left (-1+x^2\right )^2}}\\ &=\frac{\left (\left (-1+x^2\right ) \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{\sqrt{-1+\left (-1+x^2\right )^2}}{2-x^2} \, dx}{\sqrt{-1+\left (-1+x^2\right )^2}}+\frac{\left (\left (-1+x^2\right ) \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{\sqrt{-1+\left (-1+x^2\right )^2}}{-1+x^2} \, dx}{\sqrt{-1+\left (-1+x^2\right )^2}}\\ &=\frac{\left (\left (-1+x^2\right ) \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{\sqrt{-2 x^2+x^4}}{2-x^2} \, dx}{\sqrt{-1+\left (-1+x^2\right )^2}}+\frac{\left (\left (-1+x^2\right ) \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{\sqrt{-2 x^2+x^4}}{-1+x^2} \, dx}{\sqrt{-1+\left (-1+x^2\right )^2}}\\ &=\frac{\left (\left (-1+x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{x \sqrt{-2+x^2}}{2-x^2} \, dx}{x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}+\frac{\left (\left (-1+x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{x \sqrt{-2+x^2}}{-1+x^2} \, dx}{x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}\\ &=\frac{\left (\left (-1+x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{-2+x}}{-1+x} \, dx,x,x^2\right )}{2 x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}-\frac{\left (\left (-1+x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \int \frac{x}{\sqrt{-2+x^2}} \, dx}{x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}\\ &=-\frac{\left (\left (-1+x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-2+x} (-1+x)} \, dx,x,x^2\right )}{2 x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}\\ &=-\frac{\left (\left (-1+x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (-1+x^2\right )^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-2+x^2}\right )}{x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}\\ &=\frac{\left (1-x^2\right ) \sqrt{-2 x^2+x^4} \sqrt{1-\frac{1}{\left (1-x^2\right )^2}} \tan ^{-1}\left (\sqrt{-2+x^2}\right )}{x \sqrt{-2+x^2} \sqrt{-1+\left (-1+x^2\right )^2}}\\ \end{align*}

Mathematica [A]  time = 0.11678, size = 91, normalized size = 1.94 \[ \frac{1}{2} \tan ^{-1}\left (\frac{(x-1) (x+1) (x+2) \sqrt{\frac{x^2 \left (x^2-2\right )}{\left (x^2-1\right )^2}}}{x \left (x^2-2\right )}\right )-\frac{1}{2} \tan ^{-1}\left (\frac{(x-2) (x-1) (x+1) \sqrt{\frac{x^2 \left (x^2-2\right )}{\left (x^2-1\right )^2}}}{x \left (x^2-2\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[((-2 + x)*(-1 + x)*(1 + x)*Sqrt[(x^2*(-2 + x^2))/(-1 + x^2)^2])/(x*(-2 + x^2))]/2 + ArcTan[((-1 + x)*(
1 + x)*(2 + x)*Sqrt[(x^2*(-2 + x^2))/(-1 + x^2)^2])/(x*(-2 + x^2))]/2

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Maple [A]  time = 0.016, size = 63, normalized size = 1.3 \begin{align*} -{\frac{{x}^{2}-1}{2\,x}\sqrt{{\frac{{x}^{2} \left ({x}^{2}-2 \right ) }{ \left ({x}^{2}-1 \right ) ^{2}}}} \left ( \arctan \left ({(-2+x){\frac{1}{\sqrt{{x}^{2}-2}}}} \right ) -\arctan \left ({(2+x){\frac{1}{\sqrt{{x}^{2}-2}}}} \right ) \right ){\frac{1}{\sqrt{{x}^{2}-2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-1/(x^2-1)^2)^(1/2)/(-x^2+2),x)

[Out]

-1/2*(x^2*(x^2-2)/(x^2-1)^2)^(1/2)*(x^2-1)*(arctan((-2+x)/(x^2-2)^(1/2))-arctan((2+x)/(x^2-2)^(1/2)))/x/(x^2-2
)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sqrt{-\frac{1}{{\left (x^{2} - 1\right )}^{2}} + 1}}{x^{2} - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46195, size = 81, normalized size = 1.72 \begin{align*} -\arctan \left (\frac{{\left (x^{2} - 1\right )} \sqrt{\frac{x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan((x^2 - 1)*sqrt((x^4 - 2*x^2)/(x^4 - 2*x^2 + 1))/x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{\sqrt{\frac{x^{4}}{x^{4} - 2 x^{2} + 1} - \frac{2 x^{2}}{x^{4} - 2 x^{2} + 1}}}{x^{2} - 2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1152, size = 24, normalized size = 0.51 \begin{align*} -\arctan \left (\sqrt{x^{2} - 2}\right ) \mathrm{sgn}\left (x^{3} - x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan(sqrt(x^2 - 2))*sgn(x^3 - x)