\(\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx\) [66]

Optimal result

Integrand size = 16, antiderivative size = 52 \[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=\frac {\left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+5 x^2+3 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, 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 = {1114} \[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=\frac {\left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^4+5 x^2+2}} \]

[In]

Int[1/Sqrt[2 + 5*x^2 + 3*x^4],x]

[Out]

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

Rule 1114

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (1+x^2\right ) \sqrt {\frac {2+3 x^2}{1+x^2}} F\left (\tan ^{-1}(x)|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+5 x^2+3 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=-\frac {i \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{\sqrt {6+15 x^2+9 x^4}} \]

[In]

Integrate[1/Sqrt[2 + 5*x^2 + 3*x^4],x]

[Out]

((-I)*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/Sqrt[6 + 15*x^2 + 9*x^4]

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, F\left (i x , \frac {\sqrt {6}}{2}\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(44\)
elliptic	\(-\frac {i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, F\left (i x , \frac {\sqrt {6}}{2}\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(44\)

[In]

int(1/(3*x^4+5*x^2+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.21 \[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=-\frac {1}{2} i \, \sqrt {2} F(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) \]

[In]

integrate(1/(3*x^4+5*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int \frac {1}{\sqrt {3 x^{4} + 5 x^{2} + 2}}\, dx \]

[In]

integrate(1/(3*x**4+5*x**2+2)**(1/2),x)

[Out]

Integral(1/sqrt(3*x**4 + 5*x**2 + 2), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}} \,d x } \]

[In]

integrate(1/(3*x^4+5*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(3*x^4 + 5*x^2 + 2), x)

Giac [F]

\[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{4} + 5 \, x^{2} + 2}} \,d x } \]

[In]

integrate(1/(3*x^4+5*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(3*x^4 + 5*x^2 + 2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx=\int \frac {1}{\sqrt {3\,x^4+5\,x^2+2}} \,d x \]

[In]

int(1/(5*x^2 + 3*x^4 + 2)^(1/2),x)

[Out]

int(1/(5*x^2 + 3*x^4 + 2)^(1/2), x)