3.9.93 \(\int \frac {\sqrt {1-\frac {1}{(-1+x^2)^2}}}{2-x^2} \, dx\) [893]

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

[Out]

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

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Rubi [A]
time = 0.29, antiderivative size = 73, normalized size of antiderivative = 1.55, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6854, 6857, 2015, 1160, 21, 267, 455, 52, 65, 209} \begin {gather*} \frac {\left (1-x^2\right ) \sqrt {x^4-2 x^2} \sqrt {1-\frac {1}{\left (1-x^2\right )^2}} \text {ArcTan}\left (\sqrt {x^2-2}\right )}{x \sqrt {x^2-2} \sqrt {\left (x^2-1\right )^2-1}} \end {gather*}

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 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 52

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 65

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 209

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

Rule 267

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 455

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 1160

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 2015

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 6854

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 6857

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]

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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 5.52, size = 44, normalized size = 0.94 \begin {gather*} -\frac {\left (-1+x^2\right ) \sqrt {1-\frac {1}{\left (-1+x^2\right )^2}} \tan ^{-1}\left (\sqrt {-2+x^2}\right )}{x \sqrt {-2+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.23, size = 63, normalized size = 1.34

method	result	size
default	\(-\frac {\sqrt {\frac {x^{2} \left (x^{2}-2\right )}{\left (x^{2}-1\right )^{2}}}\, \left (x^{2}-1\right ) \left (\arctan \left (\frac {x -2}{\sqrt {x^{2}-2}}\right )-\arctan \left (\frac {x +2}{\sqrt {x^{2}-2}}\right )\right )}{2 x \sqrt {x^{2}-2}}\)	\(63\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 x^{2} \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}+3 x \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}}{x \left (1+x \right ) \left (-1+x \right )}\right )}{2}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.37, size = 36, normalized size = 0.77 \begin {gather*} -\arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {\frac {x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{x}\right ) \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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 = 5.51, size = 18, normalized size = 0.38 \begin {gather*} -\arctan \left (\sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x^{3} - x\right ) \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int -\frac {\sqrt {1-\frac {1}{{\left (x^2-1\right )}^2}}}{x^2-2} \,d x \end {gather*}

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]

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

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