\(\int \frac {1}{\sqrt {2-2 x^2+3 x^4}} \, dx\) [73]

Optimal result

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

[Out]

1/12*(cos(2*arctan(1/2*3^(1/4)*2^(3/4)*x))^2)^(1/2)/cos(2*arctan(1/2*3^(1/4)*2^(3/4)*x))*EllipticF(sin(2*arcta
n(1/2*3^(1/4)*2^(3/4)*x)),1/6*(18+3*6^(1/2))^(1/2))*(2+x^2*6^(1/2))*((3*x^4-2*x^2+2)/(2+x^2*6^(1/2))^2)^(1/2)*
6^(3/4)/(3*x^4-2*x^2+2)^(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} \[ \int \frac {1}{\sqrt {2-2 x^2+3 x^4}} \, dx=\frac {\left (\sqrt {6} x^2+2\right ) \sqrt {\frac {3 x^4-2 x^2+2}{\left (\sqrt {6} x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right ),\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3 x^4-2 x^2+2}} \]

[In]

Int[1/Sqrt[2 - 2*x^2 + 3*x^4],x]

[Out]

((2 + Sqrt[6]*x^2)*Sqrt[(2 - 2*x^2 + 3*x^4)/(2 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(3/2)^(1/4)*x], (6 + Sqrt[
6])/12])/(2*6^(1/4)*Sqrt[2 - 2*x^2 + 3*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 (2+\sqrt {6} x^2\right ) \sqrt {\frac {2-2 x^2+3 x^4}{\left (2+\sqrt {6} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\sqrt [4]{\frac {3}{2}} x\right )|\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2-2 x^2+3 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.07 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\sqrt {2-2 x^2+3 x^4}} \, dx=-\frac {i \sqrt {1-\frac {3 x^2}{1-i \sqrt {5}}} \sqrt {1-\frac {3 x^2}{1+i \sqrt {5}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {3}{1-i \sqrt {5}}} x\right ),\frac {1-i \sqrt {5}}{1+i \sqrt {5}}\right )}{\sqrt {3} \sqrt {-\frac {1}{1-i \sqrt {5}}} \sqrt {2-2 x^2+3 x^4}} \]

[In]

Integrate[1/Sqrt[2 - 2*x^2 + 3*x^4],x]

[Out]

((-I)*Sqrt[1 - (3*x^2)/(1 - I*Sqrt[5])]*Sqrt[1 - (3*x^2)/(1 + I*Sqrt[5])]*EllipticF[I*ArcSinh[Sqrt[-3/(1 - I*S
qrt[5])]*x], (1 - I*Sqrt[5])/(1 + I*Sqrt[5])])/(Sqrt[3]*Sqrt[-(1 - I*Sqrt[5])^(-1)]*Sqrt[2 - 2*x^2 + 3*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {2 \sqrt {1-\left (\frac {i \sqrt {5}}{2}+\frac {1}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {5}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {2+2 i \sqrt {5}}}{2}, \frac {\sqrt {-6-3 i \sqrt {5}}}{3}\right )}{\sqrt {2+2 i \sqrt {5}}\, \sqrt {3 x^{4}-2 x^{2}+2}}\)	\(87\)
elliptic	\(\frac {2 \sqrt {1-\left (\frac {i \sqrt {5}}{2}+\frac {1}{2}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{2}-\frac {i \sqrt {5}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {2+2 i \sqrt {5}}}{2}, \frac {\sqrt {-6-3 i \sqrt {5}}}{3}\right )}{\sqrt {2+2 i \sqrt {5}}\, \sqrt {3 x^{4}-2 x^{2}+2}}\)	\(87\)

[In]

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

[Out]

2/(2+2*I*5^(1/2))^(1/2)*(1-(1/2*I*5^(1/2)+1/2)*x^2)^(1/2)*(1-(1/2-1/2*I*5^(1/2))*x^2)^(1/2)/(3*x^4-2*x^2+2)^(1
/2)*EllipticF(1/2*x*(2+2*I*5^(1/2))^(1/2),1/3*(-6-3*I*5^(1/2))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\sqrt {2-2 x^2+3 x^4}} \, dx=-\frac {1}{6} \, \sqrt {\sqrt {-5} + 1} {\left (\sqrt {-5} - 1\right )} F(\arcsin \left (\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {-5} + 1}\right )\,|\,-\frac {1}{3} \, \sqrt {-5} - \frac {2}{3}) \]

[In]

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

[Out]

-1/6*sqrt(sqrt(-5) + 1)*(sqrt(-5) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*x*sqrt(sqrt(-5) + 1)), -1/3*sqrt(-5) - 2/
3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(3*x^4 - 2*x^2 + 2)^(1/2),x)

[Out]

int(1/(3*x^4 - 2*x^2 + 2)^(1/2), x)