Integrand size = 33, antiderivative size = 68 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=-\frac {x \left (1-x^2\right ) \sqrt {-2+x^2} \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \arctan \left (\sqrt {-2+x^2}\right )}{2 x^2-x^4} \] Output:
-x*(-x^2+1)*(x^2-2)^(1/2)*(-(-x^4+2*x^2)/(-x^2+1)^2)^(1/2)*arctan((x^2-2)^ (1/2))/(-x^4+2*x^2)
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=-\frac {\sqrt {\frac {x^2 \left (-2+x^2\right )}{\left (-1+x^2\right )^2}} \left (-1+x^2\right ) \arctan \left (\sqrt {-2+x^2}\right )}{x \sqrt {-2+x^2}} \] 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)*ArcTan[Sqrt[-2 + x^2]])/ (x*Sqrt[-2 + x^2]))
Time = 0.77 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {7270, 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.
\(\displaystyle \int \frac {\sqrt {\frac {x^4-2 x^2}{\left (1-x^2\right )^2}}}{2-x^2} \, dx\) |
\(\Big \downarrow \) 7270 |
\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \int \frac {\sqrt {x^4-2 x^2}}{\left (1-x^2\right ) \left (2-x^2\right )}dx}{\sqrt {x^4-2 x^2}}\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \int \frac {x \sqrt {x^2-2}}{\left (1-x^2\right ) \left (2-x^2\right )}dx}{x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 281 |
\(\displaystyle -\frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \int \frac {x}{\left (1-x^2\right ) \sqrt {x^2-2}}dx}{x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle -\frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \int \frac {1}{\left (1-x^2\right ) \sqrt {x^2-2}}dx^2}{2 x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \int \frac {1}{-x^4-1}d\sqrt {x^2-2}}{x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\left (1-x^2\right ) \sqrt {-\frac {2 x^2-x^4}{\left (1-x^2\right )^2}} \arctan \left (\sqrt {x^2-2}\right )}{x \sqrt {x^2-2}}\) |
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)]*ArcTan[Sqrt[-2 + x^2]])/(x*S qrt[-2 + x^2])
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]
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])
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])
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]
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]
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]
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93
method | result | size |
default | \(-\frac {\sqrt {\frac {x^{2} \left (x^{2}-2\right )}{\left (x^{2}-1\right )^{2}}}\, \left (x^{2}-1\right ) \left (\arctan \left (\frac {x -2}{\sqrt {x^{2}-2}}\right )-\arctan \left (\frac {2+x}{\sqrt {x^{2}-2}}\right )\right )}{2 x \sqrt {x^{2}-2}}\) | \(63\) |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}-2 x^{2} \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-\frac {-x^{4}+2 x^{2}}{x^{4}-2 x^{2}+1}}}{\left (1+x \right ) \left (-1+x \right ) x}\right )}{2}\) | \(106\) |
Input:
int(((x^4-2*x^2)/(-x^2+1)^2)^(1/2)/(-x^2+2),x,method=_RETURNVERBOSE)
Output:
-1/2*(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)))/x/(x^2-2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=-\arctan \left (\frac {{\left (x^{2} - 1\right )} \sqrt {\frac {x^{4} - 2 \, x^{2}}{x^{4} - 2 \, x^{2} + 1}}}{x}\right ) \] Input:
integrate(((x^4-2*x^2)/(-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)
\[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=- \int \frac {\sqrt {\frac {x^{4}}{x^{4} - 2 x^{2} + 1} - \frac {2 x^{2}}{x^{4} - 2 x^{2} + 1}}}{x^{2} - 2}\, dx \] Input:
integrate(((x**4-2*x**2)/(-x**2+1)**2)**(1/2)/(-x**2+2),x)
Output:
-Integral(sqrt(x**4/(x**4 - 2*x**2 + 1) - 2*x**2/(x**4 - 2*x**2 + 1))/(x** 2 - 2), x)
\[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=\int { -\frac {\sqrt {\frac {x^{4} - 2 \, x^{2}}{{\left (x^{2} - 1\right )}^{2}}}}{x^{2} - 2} \,d x } \] 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)
Time = 0.11 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.26 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=-\arctan \left (\sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x^{3} - x\right ) \] Input:
integrate(((x^4-2*x^2)/(-x^2+1)^2)^(1/2)/(-x^2+2),x, algorithm="giac")
Output:
-arctan(sqrt(x^2 - 2))*sgn(x^3 - x)
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)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.40 \[ \int \frac {\sqrt {\frac {-2 x^2+x^4}{\left (1-x^2\right )^2}}}{2-x^2} \, dx=-\mathit {atan} \left (\frac {\sqrt {x^{2}-2}\, x +x^{2}-2}{\sqrt {x^{2}-2}+x}\right ) \] 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))