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

3.9.93.1 Optimal result

Integrand size = 25, antiderivative size = 47 \[ \int \frac {\sqrt {1-\frac {1}{\left (-1+x^2\right )^2}}}{2-x^2} \, dx=\frac {\left (1-x^2\right ) \sqrt {1-\frac {1}{\left (1-x^2\right )^2}} \arctan \left (\sqrt {-2+x^2}\right )}{x \sqrt {-2+x^2}} \]

output

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

3.9.93.2 Mathematica [A] (verified)

Time = 5.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {1-\frac {1}{\left (-1+x^2\right )^2}}}{2-x^2} \, dx=-\frac {\left (-1+x^2\right ) \sqrt {1-\frac {1}{\left (-1+x^2\right )^2}} \arctan \left (\sqrt {-2+x^2}\right )}{x \sqrt {-2+x^2}} \]

input

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

output

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

3.9.93.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {7273, 25, 2467, 281, 353, 73, 217}

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.

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

\(\Big \downarrow \) 7273

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 281

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\left (1-x^2\right ) \sqrt {1-\frac {1}{\left (1-x^2\right )^2}} \int \frac {1}{-x^4-1}d\sqrt {x^2-2}}{x \sqrt {x^2-2}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {1-\frac {1}{\left (1-x^2\right )^2}} \arctan \left (\sqrt {x^2-2}\right )}{x \sqrt {x^2-2}}\)

input

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

output

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

3.9.93.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

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

rule 281

Int[(u_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_ 
Symbol] :> Simp[(b/d)^p   Int[u*(c + d*x^n)^(p + q), x], x] /; FreeQ[{a, b, 
 c, d, n, p, q}, x] && EqQ[b*c - a*d, 0] && IntegerQ[p] &&  !(IntegerQ[q] & 
& SimplerQ[a + b*x^n, c + d*x^n])
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7273

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*v^n)^Fra 
cPart[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] && Bi 
nomialQ[v, x] &&  !LinearQ[v, x]
 

3.9.93.4 Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.34

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

input

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

output

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

3.9.93.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {1-\frac {1}{\left (-1+x^2\right )^2}}}{2-x^2} \, dx=-\arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {\frac {x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{x}\right ) \]

input

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

output

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

3.9.93.6 Sympy [F]

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

input

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

output

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

3.9.93.7 Maxima [F]

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

input

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

output

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

3.9.93.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {1-\frac {1}{\left (-1+x^2\right )^2}}}{2-x^2} \, dx=-\arctan \left (\sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x^{3} - x\right ) \]

input

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

output

-arctan(sqrt(x^2 - 2))*sgn(x^3 - x)
 

3.9.93.9 Mupad [F(-1)]

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

input

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

output

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