Integrand size = 16, antiderivative size = 80 \[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=\frac {\sqrt {\frac {1}{6} \left (3+\sqrt {3}\right )} \sqrt {3-\sqrt {3}+2 x^2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{3} \left (3-\sqrt {3}\right )} x\right ),-1-\sqrt {3}\right )}{\sqrt {-3+\sqrt {3}-2 x^2}} \] Output:
1/6*(18+6*3^(1/2))^(1/2)*(3-3^(1/2)+2*x^2)^(1/2)*InverseJacobiAM(arctan(1/ 3*(9-3*3^(1/2))^(1/2)*x),(-1-3^(1/2))^(1/2))/(-3+3^(1/2)-2*x^2)^(1/2)
Result contains complex when optimal does not.
Time = 10.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=-\frac {i \sqrt {\frac {3}{2}-\frac {\sqrt {3}}{2}+x^2} \sqrt {3+\sqrt {3}+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {1}{\sqrt {3}}} x\right ),2+\sqrt {3}\right )}{\sqrt {\left (-3+\sqrt {3}\right ) \left (3+6 x^2+2 x^4\right )}} \] Input:
Integrate[1/Sqrt[-3 - 6*x^2 - 2*x^4],x]
Output:
((-I)*Sqrt[3/2 - Sqrt[3]/2 + x^2]*Sqrt[3 + Sqrt[3] + 2*x^2]*EllipticF[I*Ar cSinh[Sqrt[1 - 1/Sqrt[3]]*x], 2 + Sqrt[3]])/Sqrt[(-3 + Sqrt[3])*(3 + 6*x^2 + 2*x^4)]
Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1408, 27, 320}
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.
\(\displaystyle \int \frac {1}{\sqrt {-2 x^4-6 x^2-3}} \, dx\) |
\(\Big \downarrow \) 1408 |
\(\displaystyle 2 \sqrt {2} \int \frac {1}{2 \sqrt {-2 x^2+\sqrt {3}-3} \sqrt {2 x^2+\sqrt {3}+3}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {2} \int \frac {1}{\sqrt {-2 x^2+\sqrt {3}-3} \sqrt {2 x^2+\sqrt {3}+3}}dx\) |
\(\Big \downarrow \) 320 |
\(\displaystyle \frac {\sqrt {\frac {6}{3-\sqrt {3}}} \sqrt {2 x^2+\sqrt {3}+3} \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{3} \left (3+\sqrt {3}\right )} x\right ),-1+\sqrt {3}\right )}{\left (3+\sqrt {3}\right ) \sqrt {-2 x^2+\sqrt {3}-3} \sqrt {\frac {2 x^2+\sqrt {3}+3}{2 x^2-\sqrt {3}+3}}}\) |
Input:
Int[1/Sqrt[-3 - 6*x^2 - 2*x^4],x]
Output:
(Sqrt[6/(3 - Sqrt[3])]*Sqrt[3 + Sqrt[3] + 2*x^2]*EllipticF[ArcTan[Sqrt[(3 + Sqrt[3])/3]*x], -1 + Sqrt[3]])/((3 + Sqrt[3])*Sqrt[-3 + Sqrt[3] - 2*x^2] *Sqrt[(3 + Sqrt[3] + 2*x^2)/(3 - Sqrt[3] + 2*x^2)])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] && !SimplerSqrtQ[b/a, d/c]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b ^2 - 4*a*c, 2]}, Simp[2*Sqrt[-c] Int[1/(Sqrt[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[ c, 0]
Time = 0.55 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02
method | result | size |
default | \(\frac {3 \sqrt {1-\left (-1-\frac {\sqrt {3}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1+\frac {\sqrt {3}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-9-3 \sqrt {3}}\, x}{3}, \frac {\sqrt {6}}{2}-\frac {\sqrt {2}}{2}\right )}{\sqrt {-9-3 \sqrt {3}}\, \sqrt {-2 x^{4}-6 x^{2}-3}}\) | \(82\) |
elliptic | \(\frac {3 \sqrt {1-\left (-1-\frac {\sqrt {3}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1+\frac {\sqrt {3}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-9-3 \sqrt {3}}\, x}{3}, \frac {\sqrt {6}}{2}-\frac {\sqrt {2}}{2}\right )}{\sqrt {-9-3 \sqrt {3}}\, \sqrt {-2 x^{4}-6 x^{2}-3}}\) | \(82\) |
Input:
int(1/(-2*x^4-6*x^2-3)^(1/2),x,method=_RETURNVERBOSE)
Output:
3/(-9-3*3^(1/2))^(1/2)*(1-(-1-1/3*3^(1/2))*x^2)^(1/2)*(1-(-1+1/3*3^(1/2))* x^2)^(1/2)/(-2*x^4-6*x^2-3)^(1/2)*EllipticF(1/3*(-9-3*3^(1/2))^(1/2)*x,1/2 *6^(1/2)-1/2*2^(1/2))
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=\frac {1}{6} \, {\left (\sqrt {3} \sqrt {-3} + 3 \, \sqrt {-3}\right )} \sqrt {\frac {1}{3} \, \sqrt {3} - 1} F(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {3} - 1}\right )\,|\,\sqrt {3} + 2) \] Input:
integrate(1/(-2*x^4-6*x^2-3)^(1/2),x, algorithm="fricas")
Output:
1/6*(sqrt(3)*sqrt(-3) + 3*sqrt(-3))*sqrt(1/3*sqrt(3) - 1)*elliptic_f(arcsi n(x*sqrt(1/3*sqrt(3) - 1)), sqrt(3) + 2)
\[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=\int \frac {1}{\sqrt {- 2 x^{4} - 6 x^{2} - 3}}\, dx \] Input:
integrate(1/(-2*x**4-6*x**2-3)**(1/2),x)
Output:
Integral(1/sqrt(-2*x**4 - 6*x**2 - 3), x)
\[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=\int { \frac {1}{\sqrt {-2 \, x^{4} - 6 \, x^{2} - 3}} \,d x } \] Input:
integrate(1/(-2*x^4-6*x^2-3)^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(-2*x^4 - 6*x^2 - 3), x)
\[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=\int { \frac {1}{\sqrt {-2 \, x^{4} - 6 \, x^{2} - 3}} \,d x } \] Input:
integrate(1/(-2*x^4-6*x^2-3)^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(-2*x^4 - 6*x^2 - 3), x)
Timed out. \[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=\int \frac {1}{\sqrt {-2\,x^4-6\,x^2-3}} \,d x \] Input:
int(1/(- 6*x^2 - 2*x^4 - 3)^(1/2),x)
Output:
int(1/(- 6*x^2 - 2*x^4 - 3)^(1/2), x)
\[ \int \frac {1}{\sqrt {-3-6 x^2-2 x^4}} \, dx=-\left (\int \frac {\sqrt {-2 x^{4}-6 x^{2}-3}}{2 x^{4}+6 x^{2}+3}d x \right ) \] Input:
int(1/(-2*x^4-6*x^2-3)^(1/2),x)
Output:
- int(sqrt( - 2*x**4 - 6*x**2 - 3)/(2*x**4 + 6*x**2 + 3),x)