\(\int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx\) [78]

Optimal result

Integrand size = 16, antiderivative size = 110 \[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=\frac {\sqrt {\frac {6+\left (9-\sqrt {57}\right ) x^2}{6+\left (9+\sqrt {57}\right ) x^2}} \left (6+\left (9+\sqrt {57}\right ) x^2\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (9+\sqrt {57}\right )} x\right ),\frac {1}{4} \left (-19+3 \sqrt {57}\right )\right )}{\sqrt {6 \left (9+\sqrt {57}\right )} \sqrt {3+9 x^2+2 x^4}} \]

[Out]

(1/(36+x^2*(54+6*57^(1/2))))^(1/2)*(36+x^2*(54+6*57^(1/2)))^(1/2)*EllipticF(x*(54+6*57^(1/2))^(1/2)/(36+x^2*(5
4+6*57^(1/2)))^(1/2),1/2*(-19+3*57^(1/2))^(1/2))*(6+x^2*(57^(1/2)+9))*((6+x^2*(9-57^(1/2)))/(6+x^2*(57^(1/2)+9
)))^(1/2)/(2*x^4+9*x^2+3)^(1/2)/(54+6*57^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 110, 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} \[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=\frac {\sqrt {\frac {\left (9-\sqrt {57}\right ) x^2+6}{\left (9+\sqrt {57}\right ) x^2+6}} \left (\left (9+\sqrt {57}\right ) x^2+6\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{6} \left (9+\sqrt {57}\right )} x\right ),\frac {1}{4} \left (-19+3 \sqrt {57}\right )\right )}{\sqrt {6 \left (9+\sqrt {57}\right )} \sqrt {2 x^4+9 x^2+3}} \]

[In]

Int[1/Sqrt[3 + 9*x^2 + 2*x^4],x]

[Out]

(Sqrt[(6 + (9 - Sqrt[57])*x^2)/(6 + (9 + Sqrt[57])*x^2)]*(6 + (9 + Sqrt[57])*x^2)*EllipticF[ArcTan[Sqrt[(9 + S
qrt[57])/6]*x], (-19 + 3*Sqrt[57])/4])/(Sqrt[6*(9 + Sqrt[57])]*Sqrt[3 + 9*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 {6+\left (9-\sqrt {57}\right ) x^2}{6+\left (9+\sqrt {57}\right ) x^2}} \left (6+\left (9+\sqrt {57}\right ) x^2\right ) F\left (\tan ^{-1}\left (\sqrt {\frac {1}{6} \left (9+\sqrt {57}\right )} x\right )|\frac {1}{4} \left (-19+3 \sqrt {57}\right )\right )}{\sqrt {6 \left (9+\sqrt {57}\right )} \sqrt {3+9 x^2+2 x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=-\frac {i \sqrt {\frac {-9+\sqrt {57}-4 x^2}{-9+\sqrt {57}}} \sqrt {9+\sqrt {57}+4 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {2 x}{\sqrt {9+\sqrt {57}}}\right ),\frac {23}{4}+\frac {3 \sqrt {57}}{4}\right )}{2 \sqrt {3+9 x^2+2 x^4}} \]

[In]

Integrate[1/Sqrt[3 + 9*x^2 + 2*x^4],x]

[Out]

((-1/2*I)*Sqrt[(-9 + Sqrt[57] - 4*x^2)/(-9 + Sqrt[57])]*Sqrt[9 + Sqrt[57] + 4*x^2]*EllipticF[I*ArcSinh[(2*x)/S
qrt[9 + Sqrt[57]]], 23/4 + (3*Sqrt[57])/4])/Sqrt[3 + 9*x^2 + 2*x^4]

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75

method	result	size
default	\(\frac {6 \sqrt {1-\left (-\frac {3}{2}+\frac {\sqrt {57}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{2}-\frac {\sqrt {57}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-54+6 \sqrt {57}}}{6}, \frac {3 \sqrt {6}}{4}+\frac {\sqrt {38}}{4}\right )}{\sqrt {-54+6 \sqrt {57}}\, \sqrt {2 x^{4}+9 x^{2}+3}}\)	\(82\)
elliptic	\(\frac {6 \sqrt {1-\left (-\frac {3}{2}+\frac {\sqrt {57}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{2}-\frac {\sqrt {57}}{6}\right ) x^{2}}\, F\left (\frac {x \sqrt {-54+6 \sqrt {57}}}{6}, \frac {3 \sqrt {6}}{4}+\frac {\sqrt {38}}{4}\right )}{\sqrt {-54+6 \sqrt {57}}\, \sqrt {2 x^{4}+9 x^{2}+3}}\)	\(82\)

[In]

int(1/(2*x^4+9*x^2+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

6/(-54+6*57^(1/2))^(1/2)*(1-(-3/2+1/6*57^(1/2))*x^2)^(1/2)*(1-(-3/2-1/6*57^(1/2))*x^2)^(1/2)/(2*x^4+9*x^2+3)^(
1/2)*EllipticF(1/6*x*(-54+6*57^(1/2))^(1/2),3/4*6^(1/2)+1/4*38^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=-\frac {1}{24} \, {\left (\sqrt {19} \sqrt {6} + 3 \, \sqrt {6} \sqrt {3}\right )} \sqrt {\sqrt {19} \sqrt {3} - 9} F(\arcsin \left (\frac {1}{6} \, \sqrt {6} \sqrt {\sqrt {19} \sqrt {3} - 9} x\right )\,|\,\frac {3}{4} \, \sqrt {19} \sqrt {3} + \frac {23}{4}) \]

[In]

integrate(1/(2*x^4+9*x^2+3)^(1/2),x, algorithm="fricas")

[Out]

-1/24*(sqrt(19)*sqrt(6) + 3*sqrt(6)*sqrt(3))*sqrt(sqrt(19)*sqrt(3) - 9)*elliptic_f(arcsin(1/6*sqrt(6)*sqrt(sqr
t(19)*sqrt(3) - 9)*x), 3/4*sqrt(19)*sqrt(3) + 23/4)

Sympy [F]

\[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2 x^{4} + 9 x^{2} + 3}}\, dx \]

[In]

integrate(1/(2*x**4+9*x**2+3)**(1/2),x)

[Out]

Integral(1/sqrt(2*x**4 + 9*x**2 + 3), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} + 9 \, x^{2} + 3}} \,d x } \]

[In]

integrate(1/(2*x^4+9*x^2+3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 + 9*x^2 + 3), x)

Giac [F]

\[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} + 9 \, x^{2} + 3}} \,d x } \]

[In]

integrate(1/(2*x^4+9*x^2+3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(2*x^4 + 9*x^2 + 3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {3+9 x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2\,x^4+9\,x^2+3}} \,d x \]

[In]

int(1/(9*x^2 + 2*x^4 + 3)^(1/2),x)

[Out]

int(1/(9*x^2 + 2*x^4 + 3)^(1/2), x)