\(\int \frac {1}{\sqrt {2+5 x^2+2 x^4}} \, dx\) [123]

Optimal result

Integrand size = 16, antiderivative size = 58 \[ \int \frac {1}{\sqrt {2+5 x^2+2 x^4}} \, dx=\frac {\sqrt {\frac {2+x^2}{1+2 x^2}} \left (1+2 x^2\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {2} x\right ),\frac {3}{4}\right )}{2 \sqrt {2+5 x^2+2 x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, 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 = {1113} \[ \int \frac {1}{\sqrt {2+5 x^2+2 x^4}} \, dx=\frac {\sqrt {\frac {x^2+2}{2 x^2+1}} \left (2 x^2+1\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {2} x\right ),\frac {3}{4}\right )}{2 \sqrt {2 x^4+5 x^2+2}} \]

[In]

Int[1/Sqrt[2 + 5*x^2 + 2*x^4],x]

[Out]

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

Rule 1113

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] &&  !(PosQ[(b - q)/a] && SimplerSq
rtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {2+5 x^2+2 x^4}} \, dx=-\frac {i \sqrt {2+x^2} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} x\right ),\frac {1}{4}\right )}{2 \sqrt {2+5 x^2+2 x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.50 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, 2\right )}{2 \sqrt {2 x^{4}+5 x^{2}+2}}\)	\(48\)
elliptic	\(-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, F\left (\frac {i \sqrt {2}\, x}{2}, 2\right )}{2 \sqrt {2 x^{4}+5 x^{2}+2}}\)	\(48\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.19 \[ \int \frac {1}{\sqrt {2+5 x^2+2 x^4}} \, dx=-i \, F(\arcsin \left (\frac {1}{2} i \, \sqrt {2} x\right )\,|\,4) \]

[In]

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

[Out]

-I*elliptic_f(arcsin(1/2*I*sqrt(2)*x), 4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(5*x^2 + 2*x^4 + 2)^(1/2),x)

[Out]

int(1/(5*x^2 + 2*x^4 + 2)^(1/2), x)