\(\int \frac {1}{\sqrt {3+4 x^2+2 x^4}} \, dx\) [83]

Optimal result

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

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

[In]

Int[1/Sqrt[3 + 4*x^2 + 2*x^4],x]

[Out]

((3 + Sqrt[6]*x^2)*Sqrt[(3 + 4*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(2/3)^(1/4)*x], 1/2 - 1/Sq
rt[6]])/(2*6^(1/4)*Sqrt[3 + 4*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+4 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}{2}-\frac {1}{\sqrt {6}}\right )}{2 \sqrt [4]{6} \sqrt {3+4 x^2+2 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 \[ \int \frac {1}{\sqrt {3+4 x^2+2 x^4}} \, dx=-\frac {i \sqrt {1-\frac {2 x^2}{-2-i \sqrt {2}}} \sqrt {1-\frac {2 x^2}{-2+i \sqrt {2}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {2}{-2-i \sqrt {2}}} x\right ),\frac {-2-i \sqrt {2}}{-2+i \sqrt {2}}\right )}{\sqrt {2} \sqrt {-\frac {1}{-2-i \sqrt {2}}} \sqrt {3+4 x^2+2 x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(4*x^2 + 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(4*x^2 + 2*x^4 + 3)^(1/2), x)