\(\int \frac {\sqrt {\frac {-2 x^2+x^4}{(1-x^2)^2}}}{2+x^2} \, dx\) [394]

Optimal result

Integrand size = 31, antiderivative size = 123 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2+x^2} \, dx=-\frac {2 \left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \arctan \left (\frac {1}{2} \sqrt {-2+x^2}\right )}{3 x \sqrt {-2+x^2}}+\frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \arctan \left (\sqrt {-2+x^2}\right )}{3 x \sqrt {-2+x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.57 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2+x^2} \, dx=\frac {\sqrt {\frac {x^2 \left (-2+x^2\right )}{\left (-1+x^2\right )^2}} \left (-1+x^2\right ) \left (2 \arctan \left (\frac {1}{2} \sqrt {-2+x^2}\right )-\arctan \left (\sqrt {-2+x^2}\right )\right )}{3 x \sqrt {-2+x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.67, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {7270, 2467, 435, 94, 73, 216, 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.

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

\(\Big \downarrow \) 7270

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 435

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

\(\Big \downarrow \) 94

\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \left (-\frac {1}{3} \int \frac {1}{\left (1-x^2\right ) \sqrt {x^2-2}}dx^2-\frac {4}{3} \int \frac {1}{\sqrt {x^2-2} \left (x^2+2\right )}dx^2\right )}{2 x \sqrt {x^2-2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \left (-\frac {2}{3} \int \frac {1}{-x^4-1}d\sqrt {x^2-2}-\frac {8}{3} \int \frac {1}{x^4+4}d\sqrt {x^2-2}\right )}{2 x \sqrt {x^2-2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \left (-\frac {2}{3} \int \frac {1}{-x^4-1}d\sqrt {x^2-2}-\frac {4}{3} \arctan \left (\frac {\sqrt {x^2-2}}{2}\right )\right )}{2 x \sqrt {x^2-2}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \left (\frac {2}{3} \arctan \left (\sqrt {x^2-2}\right )-\frac {4}{3} \arctan \left (\frac {\sqrt {x^2-2}}{2}\right )\right )}{2 x \sqrt {x^2-2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

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 94

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[(e + f*x)^(p - 1)/(a + b*x), x], 
x] - Simp[(d*e - c*f)/(b*c - a*d)   Int[(e + f*x)^(p - 1)/(c + d*x), x], x] 
 /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]
 

rule 216

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

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 435

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^2)^(p_.)*((c_.) + (d_.)*(x_)^2)^(q_.)*(( 
e_.) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2) 
*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^2], x] /; FreeQ[{a, b, c, d, 
 e, f, p, q, r}, x] && IntegerQ[(m - 1)/2]
 

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 7270

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p 
]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p])))   Int[u*v 
^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !Free 
Q[v, x] &&  !FreeQ[w, x]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {\sqrt {\frac {x^{2} \left (x^{2}-2\right )}{\left (x^{2}-1\right )^{2}}}\, \left (x^{2}-1\right ) \left (-\arctan \left (\frac {x -2}{\sqrt {x^{2}-2}}\right )+\arctan \left (\frac {2+x}{\sqrt {x^{2}-2}}\right )+4 \arctan \left (\frac {\sqrt {x^{2}-2}}{2}\right )\right )}{6 x \sqrt {x^{2}-2}}\)	\(75\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{7}+6 \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}\, x^{6}-15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{5}-22 x^{4} \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}+24 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+24 x^{2} \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}-12 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -8 \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}}{x \left (x^{2}+2\right )^{2} \left (1+x \right ) \left (-1+x \right )}\right )}{6}\)	\(198\)

Input:

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

Output:

1/6*(x^2*(x^2-2)/(x^2-1)^2)^(1/2)*(x^2-1)*(-arctan((x-2)/(x^2-2)^(1/2))+ar 
ctan((2+x)/(x^2-2)^(1/2))+4*arctan(1/2*(x^2-2)^(1/2)))/x/(x^2-2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2+x^2} \, dx=-\frac {1}{3} \, \arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {\frac {x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{x}\right ) + \frac {2}{3} \, \arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {\frac {x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{2 \, x}\right ) \] Input:

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

Output:

-1/3*arctan((x^2 - 1)*sqrt((x^4 - 2*x^2)/(x^4 - 2*x^2 + 1))/x) + 2/3*arcta 
n(1/2*(x^2 - 1)*sqrt((x^4 - 2*x^2)/(x^4 - 2*x^2 + 1))/x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.32 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2+x^2} \, dx=\frac {2}{3} \, \arctan \left (\frac {1}{2} \, \sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x^{3} - x\right ) - \frac {1}{3} \, \arctan \left (\sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x^{3} - x\right ) \] Input:

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

Output:

2/3*arctan(1/2*sqrt(x^2 - 2))*sgn(x^3 - x) - 1/3*arctan(sqrt(x^2 - 2))*sgn 
(x^3 - x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2+x^2} \, dx=-\frac {\mathit {atan} \left (\frac {\sqrt {x^{2}-2}\, x +x^{2}-2}{\sqrt {x^{2}-2}+x}\right )}{3}+\frac {2 \mathit {atan} \left (\frac {\sqrt {x^{2}-2}\, x +x^{2}-2}{2 \sqrt {x^{2}-2}+2 x}\right )}{3} \] Input:

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

Output:

( - atan((sqrt(x**2 - 2)*x + x**2 - 2)/(sqrt(x**2 - 2) + x)) + 2*atan((sqr 
t(x**2 - 2)*x + x**2 - 2)/(2*sqrt(x**2 - 2) + 2*x)))/3