Integrand size = 28, antiderivative size = 83 \[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=\frac {2 \sqrt {-2 x^2+x^4} \arctan \left (\frac {1}{2} \sqrt {-2+x^2}\right )}{3 x \sqrt {-2+x^2}}-\frac {\sqrt {-2 x^2+x^4} \arctan \left (\sqrt {-2+x^2}\right )}{3 x \sqrt {-2+x^2}} \]
2/3*arctan(1/2*(x^2-2)^(1/2))*(x^4-2*x^2)^(1/2)/x/(x^2-2)^(1/2)-1/3*arctan ((x^2-2)^(1/2))*(x^4-2*x^2)^(1/2)/x/(x^2-2)^(1/2)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=\frac {x \sqrt {-2+x^2} \left (2 \arctan \left (\frac {1}{2} \sqrt {-2+x^2}\right )-\arctan \left (\sqrt {-2+x^2}\right )\right )}{3 \sqrt {x^2 \left (-2+x^2\right )}} \]
(x*Sqrt[-2 + x^2]*(2*ArcTan[Sqrt[-2 + x^2]/2] - ArcTan[Sqrt[-2 + x^2]]))/( 3*Sqrt[x^2*(-2 + x^2)])
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2467, 25, 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.
\(\displaystyle \int \frac {\sqrt {x^4-2 x^2}}{\left (x^2-1\right ) \left (x^2+2\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x^4-2 x^2} \int -\frac {x \sqrt {x^2-2}}{\left (1-x^2\right ) \left (x^2+2\right )}dx}{x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x^4-2 x^2} \int \frac {x \sqrt {x^2-2}}{\left (1-x^2\right ) \left (x^2+2\right )}dx}{x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 435 |
\(\displaystyle -\frac {\sqrt {x^4-2 x^2} \int \frac {\sqrt {x^2-2}}{\left (1-x^2\right ) \left (x^2+2\right )}dx^2}{2 x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 94 |
\(\displaystyle -\frac {\sqrt {x^4-2 x^2} \left (-\frac {1}{3} \int \frac {1}{\left (1-x^2\right ) \sqrt {x^2-2}}dx^2-\frac {4}{3} \int \frac {1}{\sqrt {x^2-2} \left (x^2+2\right )}dx^2\right )}{2 x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\sqrt {x^4-2 x^2} \left (-\frac {2}{3} \int \frac {1}{-x^4-1}d\sqrt {x^2-2}-\frac {8}{3} \int \frac {1}{x^4+4}d\sqrt {x^2-2}\right )}{2 x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\sqrt {x^4-2 x^2} \left (-\frac {2}{3} \int \frac {1}{-x^4-1}d\sqrt {x^2-2}-\frac {4}{3} \arctan \left (\frac {\sqrt {x^2-2}}{2}\right )\right )}{2 x \sqrt {x^2-2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {\sqrt {x^4-2 x^2} \left (\frac {2}{3} \arctan \left (\sqrt {x^2-2}\right )-\frac {4}{3} \arctan \left (\frac {\sqrt {x^2-2}}{2}\right )\right )}{2 x \sqrt {x^2-2}}\) |
-1/2*(Sqrt[-2*x^2 + x^4]*((-4*ArcTan[Sqrt[-2 + x^2]/2])/3 + (2*ArcTan[Sqrt [-2 + x^2]])/3))/(x*Sqrt[-2 + x^2])
3.9.92.3.1 Defintions of rubi rules used
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[((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]
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])
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[(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]
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]
Result contains complex when optimal does not.
Time = 1.70 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {i \left (-i \arctan \left (\frac {1}{\sqrt {x^{2}-2}}\right )+\operatorname {arctanh}\left (\frac {\left (i \sqrt {2}-x \right ) \sqrt {2}}{2 \sqrt {x^{2}-2}}\right )+\operatorname {arctanh}\left (\frac {\left (x +i \sqrt {2}\right ) \sqrt {2}}{2 \sqrt {x^{2}-2}}\right )\right )}{3}\) | \(60\) |
default | \(\frac {\sqrt {x^{4}-2 x^{2}}\, \left (\arctan \left (\frac {x +2}{\sqrt {x^{2}-2}}\right )-\arctan \left (\frac {x -2}{\sqrt {x^{2}-2}}\right )+4 \arctan \left (\frac {\sqrt {x^{2}-2}}{2}\right )\right )}{6 x \sqrt {x^{2}-2}}\) | \(63\) |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{7}-15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{5}+6 \sqrt {x^{4}-2 x^{2}}\, x^{4}+24 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}-16 \sqrt {x^{4}-2 x^{2}}\, x^{2}-12 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +8 \sqrt {x^{4}-2 x^{2}}}{\left (x^{2}+2\right )^{2} \left (x -1\right ) \left (x +1\right ) x}\right )}{6}\) | \(118\) |
1/3*I*(-I*arctan(1/(x^2-2)^(1/2))+arctanh(1/2*(I*2^(1/2)-x)*2^(1/2)/(x^2-2 )^(1/2))+arctanh(1/2*(x+I*2^(1/2))*2^(1/2)/(x^2-2)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=-\frac {1}{3} \, \arctan \left (\frac {\sqrt {x^{4} - 2 \, x^{2}}}{x}\right ) + \frac {2}{3} \, \arctan \left (\frac {\sqrt {x^{4} - 2 \, x^{2}}}{2 \, x}\right ) \]
\[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=\int \frac {\sqrt {x^{2} \left (x^{2} - 2\right )}}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 2\right )}\, dx \]
\[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=\int { \frac {\sqrt {x^{4} - 2 \, x^{2}}}{{\left (x^{2} + 2\right )} {\left (x^{2} - 1\right )}} \,d x } \]
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=\frac {1}{3} \, {\left (\arctan \left (i \, \sqrt {2}\right ) - 2 \, \arctan \left (\frac {1}{2} i \, \sqrt {2}\right )\right )} \mathrm {sgn}\left (x\right ) + \frac {2}{3} \, \arctan \left (\frac {1}{2} \, \sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{3} \, \arctan \left (\sqrt {x^{2} - 2}\right ) \mathrm {sgn}\left (x\right ) \]
1/3*(arctan(I*sqrt(2)) - 2*arctan(1/2*I*sqrt(2)))*sgn(x) + 2/3*arctan(1/2* sqrt(x^2 - 2))*sgn(x) - 1/3*arctan(sqrt(x^2 - 2))*sgn(x)
Timed out. \[ \int \frac {\sqrt {-2 x^2+x^4}}{\left (-1+x^2\right ) \left (2+x^2\right )} \, dx=\int \frac {\sqrt {x^4-2\,x^2}}{\left (x^2-1\right )\,\left (x^2+2\right )} \,d x \]