\(\int \frac {1}{\sqrt {-2-4 x^2+3 x^4}} \, dx\) [51]

Optimal result

Integrand size = 16, antiderivative size = 148 \[ \int \frac {1}{\sqrt {-2-4 x^2+3 x^4}} \, dx=\frac {\sqrt {-2-\left (2-\sqrt {10}\right ) x^2} \sqrt {\frac {2+\left (2+\sqrt {10}\right ) x^2}{2+\left (2-\sqrt {10}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-2-\left (2-\sqrt {10}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {\frac {1}{2+\left (2-\sqrt {10}\right ) x^2}} \sqrt {-2-4 x^2+3 x^4}} \]

[Out]

1/20*EllipticF(2^(3/4)*5^(1/4)*x/(-2-x^2*(2-10^(1/2)))^(1/2),1/10*(50-10*10^(1/2))^(1/2))*(-2-x^2*(2-10^(1/2))
)^(1/2)*((2+x^2*(2+10^(1/2)))/(2+x^2*(2-10^(1/2))))^(1/2)*10^(3/4)/(3*x^4-4*x^2-2)^(1/2)/(1/(2+x^2*(2-10^(1/2)
)))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 148, 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 = {1112} \[ \int \frac {1}{\sqrt {-2-4 x^2+3 x^4}} \, dx=\frac {\sqrt {-\left (\left (2-\sqrt {10}\right ) x^2\right )-2} \sqrt {\frac {\left (2+\sqrt {10}\right ) x^2+2}{\left (2-\sqrt {10}\right ) x^2+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-\left (\left (2-\sqrt {10}\right ) x^2\right )-2}}\right ),\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {\frac {1}{\left (2-\sqrt {10}\right ) x^2+2}} \sqrt {3 x^4-4 x^2-2}} \]

[In]

Int[1/Sqrt[-2 - 4*x^2 + 3*x^4],x]

[Out]

(Sqrt[-2 - (2 - Sqrt[10])*x^2]*Sqrt[(2 + (2 + Sqrt[10])*x^2)/(2 + (2 - Sqrt[10])*x^2)]*EllipticF[ArcSin[(2^(3/
4)*5^(1/4)*x)/Sqrt[-2 - (2 - Sqrt[10])*x^2]], (5 - Sqrt[10])/10])/(2*10^(1/4)*Sqrt[(2 + (2 - Sqrt[10])*x^2)^(-
1)]*Sqrt[-2 - 4*x^2 + 3*x^4])

Rule 1112

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-2-\left (2-\sqrt {10}\right ) x^2} \sqrt {\frac {2+\left (2+\sqrt {10}\right ) x^2}{2+\left (2-\sqrt {10}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-2-\left (2-\sqrt {10}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2 \sqrt [4]{10} \sqrt {\frac {1}{2+\left (2-\sqrt {10}\right ) x^2}} \sqrt {-2-4 x^2+3 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.06 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sqrt {-2-4 x^2+3 x^4}} \, dx=-\frac {i \sqrt {2+4 x^2-3 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1+\sqrt {\frac {5}{2}}} x\right ),\frac {1}{3} \left (-7+2 \sqrt {10}\right )\right )}{\sqrt {2+\sqrt {10}} \sqrt {-2-4 x^2+3 x^4}} \]

[In]

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

[Out]

((-I)*Sqrt[2 + 4*x^2 - 3*x^4]*EllipticF[I*ArcSinh[Sqrt[1 + Sqrt[5/2]]*x], (-7 + 2*Sqrt[10])/3])/(Sqrt[2 + Sqrt
[10]]*Sqrt[-2 - 4*x^2 + 3*x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {2 \sqrt {1-\left (-1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, F\left (\frac {\sqrt {-4-2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )}{\sqrt {-4-2 \sqrt {10}}\, \sqrt {3 x^{4}-4 x^{2}-2}}\)	\(84\)
elliptic	\(\frac {2 \sqrt {1-\left (-1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, F\left (\frac {\sqrt {-4-2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )}{\sqrt {-4-2 \sqrt {10}}\, \sqrt {3 x^{4}-4 x^{2}-2}}\)	\(84\)

[In]

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

[Out]

2/(-4-2*10^(1/2))^(1/2)*(1-(-1-1/2*10^(1/2))*x^2)^(1/2)*(1-(-1+1/2*10^(1/2))*x^2)^(1/2)/(3*x^4-4*x^2-2)^(1/2)*
EllipticF(1/2*(-4-2*10^(1/2))^(1/2)*x,1/3*I*15^(1/2)-1/3*I*6^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {-2-4 x^2+3 x^4}} \, dx=-\frac {1}{12} \, {\left (\sqrt {10} \sqrt {2} \sqrt {-2} + 2 \, \sqrt {2} \sqrt {-2}\right )} \sqrt {\sqrt {10} - 2} F(\arcsin \left (\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {10} - 2}\right )\,|\,-\frac {2}{3} \, \sqrt {10} - \frac {7}{3}) \]

[In]

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

[Out]

-1/12*(sqrt(10)*sqrt(2)*sqrt(-2) + 2*sqrt(2)*sqrt(-2))*sqrt(sqrt(10) - 2)*elliptic_f(arcsin(1/2*sqrt(2)*x*sqrt
(sqrt(10) - 2)), -2/3*sqrt(10) - 7/3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(3*x^4 - 4*x^2 - 2)^(1/2),x)

[Out]

int(1/(3*x^4 - 4*x^2 - 2)^(1/2), x)