\(\int \frac {2+x^2}{(-1+x^2) \sqrt {-1-x^2+x^4}} \, dx\) [323]

Optimal result

Integrand size = 27, antiderivative size = 28 \[ \int \frac {2+x^2}{\left (-1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx=-\arctan \left (\frac {\sqrt {-1-x^2+x^4}}{x \left (-2+x^2\right )}\right ) \]

[Out]

-arctan((x^4-x^2-1)^(1/2)/x/(x^2-2))

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.25 (sec) , antiderivative size = 520, normalized size of antiderivative = 18.57, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1724, 1112, 1228, 1470, 554, 432, 430, 552, 551} \[ \int \frac {2+x^2}{\left (-1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx=\frac {3 \left (1+\sqrt {5}\right ) \sqrt {2 x^2+\sqrt {5}-1} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {2} \left (1-\sqrt {5}\right ) \sqrt {x^4-x^2-1}}-\frac {3 \sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x^2+2}{\left (1-\sqrt {5}\right ) x^2+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (1-\sqrt {5}\right ) \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x^2+2}} \sqrt {x^4-x^2-1}}+\frac {\sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2} \sqrt {\frac {\left (1+\sqrt {5}\right ) x^2+2}{\left (1-\sqrt {5}\right ) x^2+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-\left (\left (1-\sqrt {5}\right ) x^2\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\frac {1}{\left (1-\sqrt {5}\right ) x^2+2}} \sqrt {x^4-x^2-1}}+\frac {3 \sqrt {2} \sqrt {2 x^2+\sqrt {5}-1} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\left (1-\sqrt {5}\right ) \sqrt {x^4-x^2-1}} \]

[In]

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

[Out]

(3*(1 + Sqrt[5])*Sqrt[-1 + Sqrt[5] + 2*x^2]*Sqrt[1 - (2*x^2)/(1 + Sqrt[5])]*EllipticF[ArcSin[Sqrt[2/(1 + Sqrt[
5])]*x], (-3 - Sqrt[5])/2])/(Sqrt[2]*(1 - Sqrt[5])*Sqrt[-1 - x^2 + x^4]) + (Sqrt[-2 - (1 - Sqrt[5])*x^2]*Sqrt[
(2 + (1 + Sqrt[5])*x^2)/(2 + (1 - Sqrt[5])*x^2)]*EllipticF[ArcSin[(Sqrt[2]*5^(1/4)*x)/Sqrt[-2 - (1 - Sqrt[5])*
x^2]], (5 - Sqrt[5])/10])/(2*5^(1/4)*Sqrt[(2 + (1 - Sqrt[5])*x^2)^(-1)]*Sqrt[-1 - x^2 + x^4]) - (3*Sqrt[-2 - (
1 - Sqrt[5])*x^2]*Sqrt[(2 + (1 + Sqrt[5])*x^2)/(2 + (1 - Sqrt[5])*x^2)]*EllipticF[ArcSin[(Sqrt[2]*5^(1/4)*x)/S
qrt[-2 - (1 - Sqrt[5])*x^2]], (5 - Sqrt[5])/10])/(5^(1/4)*(1 - Sqrt[5])*Sqrt[(2 + (1 - Sqrt[5])*x^2)^(-1)]*Sqr
t[-1 - x^2 + x^4]) + (3*Sqrt[2]*Sqrt[-1 + Sqrt[5] + 2*x^2]*Sqrt[1 - (2*x^2)/(1 + Sqrt[5])]*EllipticPi[(1 + Sqr
t[5])/2, ArcSin[Sqrt[2/(1 + Sqrt[5])]*x], (-3 - Sqrt[5])/2])/((1 - Sqrt[5])*Sqrt[-1 - x^2 + x^4])

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 552

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d/c)*x^2]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d/c)*x^2]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 0]

Rule 554

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[d/b, Int[1/
(Sqrt[c + d*x^2]*Sqrt[e + f*x^2]), x], x] + Dist[(b*c - a*d)/b, Int[1/((a + b*x^2)*Sqrt[c + d*x^2]*Sqrt[e + f*
x^2]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NegQ[d/c]

Rule 1112

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[Sqrt[(2*a + (
b - q)*x^2)/(2*a + (b + q)*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)
*x^2)]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x]] /; FreeQ[{a, b, c}, x] && Gt
Q[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]

Rule 1228

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[2*(c/(2*c*d - e*(b - q))), Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[e/(2*c*d - e*(b - q)), Int[
(b - q + 2*c*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a
*c, 0] &&  !LtQ[c, 0]

Rule 1470

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((f_) + (g_.)*(x_)^(n_))^(r_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^
(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + (c*x^n)/e)^FracPar
t[p]), Int[(d + e*x^n)^(p + q)*(f + g*x^n)^r*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p,
q, r}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p]

Rule 1724

Int[((A_.) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist
[B/e, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[(e*A - d*B)/e, Int[1/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]),
x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^
2 - a*e^2, 0] && NegQ[c/a]

Rubi steps \begin{align*} \text {integral}& = 3 \int \frac {1}{\left (-1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx+\int \frac {1}{\sqrt {-1-x^2+x^4}} \, dx \\ & = \frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \int \frac {-1-\sqrt {5}+2 x^2}{\left (-1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx}{-1+\sqrt {5}}+\frac {6 \int \frac {1}{\sqrt {-1-x^2+x^4}} \, dx}{-1+\sqrt {5}} \\ & = \frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (1-\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {\left (3 \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}\right ) \int \frac {\sqrt {-1-\sqrt {5}+2 x^2}}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \left (-1+x^2\right )} \, dx}{\left (-1+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}} \\ & = \frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (1-\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {\left (6 \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}} \, dx}{\left (-1+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}-\frac {\left (3 \left (1-\sqrt {5}\right ) \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {-1-\sqrt {5}+2 x^2}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \left (-1+x^2\right ) \sqrt {-1-\sqrt {5}+2 x^2}} \, dx}{\left (-1+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}} \\ & = \frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (1-\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {\left (6 \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}} \, dx}{\left (-1+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}-\frac {\left (3 \left (1-\sqrt {5}\right ) \sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}\right ) \int \frac {1}{\sqrt {-\frac {1}{-1-\sqrt {5}}+\frac {x^2}{2}} \left (-1+x^2\right ) \sqrt {1+\frac {2 x^2}{-1-\sqrt {5}}}} \, dx}{\left (-1+\sqrt {5}\right ) \sqrt {-1-x^2+x^4}} \\ & = \frac {3 \left (1+\sqrt {5}\right ) \sqrt {-1+\sqrt {5}+2 x^2} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{\sqrt {2} \left (1-\sqrt {5}\right ) \sqrt {-1-x^2+x^4}}+\frac {\sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{2 \sqrt [4]{5} \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \sqrt {-2-\left (1-\sqrt {5}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {5}\right ) x^2}{2+\left (1-\sqrt {5}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{5} x}{\sqrt {-2-\left (1-\sqrt {5}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {5}\right )\right )}{\sqrt [4]{5} \left (1-\sqrt {5}\right ) \sqrt {\frac {1}{2+\left (1-\sqrt {5}\right ) x^2}} \sqrt {-1-x^2+x^4}}-\frac {3 \left (1+\sqrt {5}\right ) \sqrt {-1+\sqrt {5}+2 x^2} \sqrt {1-\frac {2 x^2}{1+\sqrt {5}}} \operatorname {EllipticPi}\left (\frac {1}{2} \left (1+\sqrt {5}\right ),\arcsin \left (\sqrt {\frac {2}{1+\sqrt {5}}} x\right ),\frac {1}{2} \left (-3-\sqrt {5}\right )\right )}{2 \sqrt {2} \sqrt {-1-x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {2+x^2}{\left (-1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx=-\arctan \left (\frac {\sqrt {-1-x^2+x^4}}{x \left (-2+x^2\right )}\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 3.43

method	result	size
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{6}+5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \sqrt {x^{4}-x^{2}-1}\, x^{3}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-4 x \sqrt {x^{4}-x^{2}-1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{\left (x -1\right )^{3} \left (1+x \right )^{3}}\right )}{2}\)	\(96\)
default	\(\frac {2 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-2-2 \sqrt {5}}}{2}, \frac {i \sqrt {5}}{2}-\frac {i}{2}\right )}{\sqrt {-2-2 \sqrt {5}}\, \sqrt {x^{4}-x^{2}-1}}-\frac {3 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {5}}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right ) x^{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, x , \frac {1}{-\frac {1}{2}-\frac {\sqrt {5}}{2}}, \frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}}\right )}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, \sqrt {x^{4}-x^{2}-1}}\)	\(174\)
elliptic	\(\frac {2 \sqrt {1+\frac {x^{2}}{2}+\frac {\sqrt {5}\, x^{2}}{2}}\, \sqrt {1-\frac {\sqrt {5}\, x^{2}}{2}+\frac {x^{2}}{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-2-2 \sqrt {5}}}{2}, \frac {i \sqrt {5}}{2}-\frac {i}{2}\right )}{\sqrt {-2-2 \sqrt {5}}\, \sqrt {x^{4}-x^{2}-1}}-\frac {3 \sqrt {1+\frac {x^{2}}{2}+\frac {\sqrt {5}\, x^{2}}{2}}\, \sqrt {1-\frac {\sqrt {5}\, x^{2}}{2}+\frac {x^{2}}{2}}\, \operatorname {EllipticPi}\left (\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, x , \frac {1}{-\frac {1}{2}-\frac {\sqrt {5}}{2}}, \frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}}\right )}{\sqrt {-\frac {1}{2}-\frac {\sqrt {5}}{2}}\, \sqrt {x^{4}-x^{2}-1}}\)	\(178\)

[In]

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

[Out]

1/2*RootOf(_Z^2+1)*ln(-(-RootOf(_Z^2+1)*x^6+5*RootOf(_Z^2+1)*x^4+2*(x^4-x^2-1)^(1/2)*x^3-5*RootOf(_Z^2+1)*x^2-
4*x*(x^4-x^2-1)^(1/2)-RootOf(_Z^2+1))/(x-1)^3/(1+x)^3)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {2+x^2}{\left (-1+x^2\right ) \sqrt {-1-x^2+x^4}} \, dx=-\frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {x^{4} - x^{2} - 1} {\left (x^{3} - 2 \, x\right )}}{x^{6} - 5 \, x^{4} + 5 \, x^{2} + 1}\right ) \]

[In]

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

[Out]

-1/2*arctan(2*sqrt(x^4 - x^2 - 1)*(x^3 - 2*x)/(x^6 - 5*x^4 + 5*x^2 + 1))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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