\(\int \frac {1}{\sqrt {3+2 x^2+2 x^4}} \, dx\) [85]

Optimal result

Integrand size = 16, antiderivative size = 92 \[ \int \frac {1}{\sqrt {3+2 x^2+2 x^4}} \, dx=\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3+2 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3+2 x^2+2 x^4}} \]

[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)),1/6*(18-3*6^(1/2))^(1/2))*(3+x^2*6^(1/2))*((2*x^4+2*x^2+3)/(3+x^2*6^(1/2))^2)^(1/2)*
6^(3/4)/(2*x^4+2*x^2+3)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 92, 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} \[ \int \frac {1}{\sqrt {3+2 x^2+2 x^4}} \, dx=\frac {\left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2 x^4+2 x^2+3}} \]

[In]

Int[1/Sqrt[3 + 2*x^2 + 2*x^4],x]

[Out]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.06 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\sqrt {3+2 x^2+2 x^4}} \, dx=-\frac {i \sqrt {1-\frac {2 x^2}{-1-i \sqrt {5}}} \sqrt {1-\frac {2 x^2}{-1+i \sqrt {5}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {2}{-1-i \sqrt {5}}} x\right ),\frac {-1-i \sqrt {5}}{-1+i \sqrt {5}}\right )}{\sqrt {2} \sqrt {-\frac {1}{-1-i \sqrt {5}}} \sqrt {3+2 x^2+2 x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.95

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\sqrt {3+2 x^2+2 x^4}} \, dx=-\frac {1}{6} \, {\left (\sqrt {-5} + 1\right )} \sqrt {\sqrt {-5} - 1} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} x \sqrt {\sqrt {-5} - 1}\right )\,|\,\frac {1}{3} \, \sqrt {-5} - \frac {2}{3}) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(2*x^2 + 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(2*x^2 + 2*x^4 + 3)^(1/2), x)