\(\int \frac {1}{\sqrt {3-5 x^2+2 x^4}} \, dx\) [92]

Optimal result

Integrand size = 16, antiderivative size = 92 \[ \int \frac {1}{\sqrt {3-5 x^2+2 x^4}} \, dx=\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-5 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{24} \left (12+5 \sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-5 x^2+2 x^4}} \]

[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)),1/12*(72+30*6^(1/2))^(1/2))*(3+x^2*6^(1/2))*((2*x^4-5*x^2+3)/(3+x^2*6^(1/2))^2)^(1/2
)*6^(3/4)/(2*x^4-5*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 = {1110} \[ \int \frac {1}{\sqrt {3-5 x^2+2 x^4}} \, dx=\frac {\left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-5 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{24} \left (12+5 \sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {2 x^4-5 x^2+3}} \]

[In]

Int[1/Sqrt[3 - 5*x^2 + 2*x^4],x]

[Out]

((3 + Sqrt[6]*x^2)*Sqrt[(3 - 5*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan[(2/3)^(1/4)*x], (12 + 5*Sq
rt[6])/24])/(2*6^(1/4)*Sqrt[3 - 5*x^2 + 2*x^4])

Rule 1110

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] && GtQ[b^2 - 4*a*c, 0] && GtQ[c/a, 0] && LtQ[b/a, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-5 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}{24} \left (12+5 \sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3-5 x^2+2 x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\sqrt {3-5 x^2+2 x^4}} \, dx=\frac {\sqrt {3-2 x^2} \sqrt {1-x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{3}} x\right ),\frac {3}{2}\right )}{\sqrt {6-10 x^2+4 x^4}} \]

[In]

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

[Out]

(Sqrt[3 - 2*x^2]*Sqrt[1 - x^2]*EllipticF[ArcSin[Sqrt[2/3]*x], 3/2])/Sqrt[6 - 10*x^2 + 4*x^4]

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46

method	result	size
default	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, F\left (x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}\)	\(42\)
elliptic	\(\frac {\sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+9}\, F\left (x , \frac {\sqrt {6}}{3}\right )}{3 \sqrt {2 x^{4}-5 x^{2}+3}}\)	\(42\)

[In]

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

[Out]

1/3*(-x^2+1)^(1/2)*(-6*x^2+9)^(1/2)/(2*x^4-5*x^2+3)^(1/2)*EllipticF(x,1/3*6^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.10 \[ \int \frac {1}{\sqrt {3-5 x^2+2 x^4}} \, dx=\frac {1}{3} \, \sqrt {3} F(\arcsin \left (x\right )\,|\,\frac {2}{3}) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(2*x^4 - 5*x^2 + 3)^(1/2),x)

[Out]

int(1/(2*x^4 - 5*x^2 + 3)^(1/2), x)