\(\int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx\) [88]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 90 \[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{24} \left (12+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-x^2+2 x^4}} \]

[Out]

1/12*(cos(2*arctan(1/3*2^(1/4)*3^(3/4)*x))^2)^(1/2)/cos(2*arctan(1/3*2^(1/4)*3^(3/4)*x))*EllipticF(sin(2*arcta
n(1/3*2^(1/4)*3^(3/4)*x)),1/12*(72+6*6^(1/2))^(1/2))*(3+x^2*6^(1/2))*((2*x^4-x^2+3)/(3+x^2*6^(1/2))^2)^(1/2)*6
^(3/4)/(2*x^4-x^2+3)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1117} \[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=\frac {\left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{24} \left (12+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2 x^4-x^2+3}} \]

[In]

Int[1/Sqrt[3 - x^2 + 2*x^4],x]

[Out]

((3 + Sqrt[6]*x^2)*Sqrt[(3 - x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(2/3)^(1/4)*x], (12 + Sqrt[6
])/24])/(2*6^(1/4)*Sqrt[3 - x^2 + 2*x^4])

Rule 1117

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{24} \left (12+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-x^2+2 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.06 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=-\frac {i \sqrt {1-\frac {4 x^2}{1-i \sqrt {23}}} \sqrt {1-\frac {4 x^2}{1+i \sqrt {23}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {-\frac {1}{1-i \sqrt {23}}} x\right ),\frac {1-i \sqrt {23}}{1+i \sqrt {23}}\right )}{2 \sqrt {-\frac {1}{1-i \sqrt {23}}} \sqrt {3-x^2+2 x^4}} \]

[In]

Integrate[1/Sqrt[3 - x^2 + 2*x^4],x]

[Out]

((-1/2*I)*Sqrt[1 - (4*x^2)/(1 - I*Sqrt[23])]*Sqrt[1 - (4*x^2)/(1 + I*Sqrt[23])]*EllipticF[I*ArcSinh[2*Sqrt[-(1
 - I*Sqrt[23])^(-1)]*x], (1 - I*Sqrt[23])/(1 + I*Sqrt[23])])/(Sqrt[-(1 - I*Sqrt[23])^(-1)]*Sqrt[3 - x^2 + 2*x^
4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97

method result size
default \(\frac {6 \sqrt {1-\left (\frac {1}{6}+\frac {i \sqrt {23}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{6}-\frac {i \sqrt {23}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {6+6 i \sqrt {23}}}{6}, \frac {\sqrt {-33-3 i \sqrt {23}}}{6}\right )}{\sqrt {6+6 i \sqrt {23}}\, \sqrt {2 x^{4}-x^{2}+3}}\) \(87\)
elliptic \(\frac {6 \sqrt {1-\left (\frac {1}{6}+\frac {i \sqrt {23}}{6}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{6}-\frac {i \sqrt {23}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {6+6 i \sqrt {23}}}{6}, \frac {\sqrt {-33-3 i \sqrt {23}}}{6}\right )}{\sqrt {6+6 i \sqrt {23}}\, \sqrt {2 x^{4}-x^{2}+3}}\) \(87\)

[In]

int(1/(2*x^4-x^2+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

6/(6+6*I*23^(1/2))^(1/2)*(1-(1/6+1/6*I*23^(1/2))*x^2)^(1/2)*(1-(1/6-1/6*I*23^(1/2))*x^2)^(1/2)/(2*x^4-x^2+3)^(
1/2)*EllipticF(1/6*x*(6+6*I*23^(1/2))^(1/2),1/6*(-33-3*I*23^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=-\frac {1}{72} \, \sqrt {6} \sqrt {3} \sqrt {\sqrt {-23} + 1} {\left (\sqrt {-23} - 1\right )} F(\arcsin \left (\frac {1}{6} \, \sqrt {6} x \sqrt {\sqrt {-23} + 1}\right )\,|\,-\frac {1}{12} \, \sqrt {-23} - \frac {11}{12}) \]

[In]

integrate(1/(2*x^4-x^2+3)^(1/2),x, algorithm="fricas")

[Out]

-1/72*sqrt(6)*sqrt(3)*sqrt(sqrt(-23) + 1)*(sqrt(-23) - 1)*elliptic_f(arcsin(1/6*sqrt(6)*x*sqrt(sqrt(-23) + 1))
, -1/12*sqrt(-23) - 11/12)

Sympy [F]

\[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2 x^{4} - x^{2} + 3}}\, dx \]

[In]

integrate(1/(2*x**4-x**2+3)**(1/2),x)

[Out]

Integral(1/sqrt(2*x**4 - x**2 + 3), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} - x^{2} + 3}} \,d x } \]

[In]

integrate(1/(2*x^4-x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 - x^2 + 3), x)

Giac [F]

\[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} - x^{2} + 3}} \,d x } \]

[In]

integrate(1/(2*x^4-x^2+3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(2*x^4 - x^2 + 3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {3-x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2\,x^4-x^2+3}} \,d x \]

[In]

int(1/(2*x^4 - x^2 + 3)^(1/2),x)

[Out]

int(1/(2*x^4 - x^2 + 3)^(1/2), x)