\(\int \frac {1}{\sqrt {-3-4 x^2+2 x^4}} \, dx\) [64]

Optimal result

Integrand size = 16, antiderivative size = 155 \[ \int \frac {1}{\sqrt {-3-4 x^2+2 x^4}} \, dx=\frac {\sqrt {-3-\left (2-\sqrt {10}\right ) x^2} \sqrt {\frac {3+\left (2+\sqrt {10}\right ) x^2}{3+\left (2-\sqrt {10}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-3-\left (2-\sqrt {10}\right ) x^2}}\right ),\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{3+\left (2-\sqrt {10}\right ) x^2}} \sqrt {-3-4 x^2+2 x^4}} \]

[Out]

1/30*EllipticF(2^(3/4)*5^(1/4)*x/(-3-x^2*(2-10^(1/2)))^(1/2),1/10*(50-10*10^(1/2))^(1/2))*(-3-x^2*(2-10^(1/2))
)^(1/2)*((3+x^2*(2+10^(1/2)))/(3+x^2*(2-10^(1/2))))^(1/2)*2^(1/4)*5^(3/4)*3^(1/2)/(2*x^4-4*x^2-3)^(1/2)/(1/(3+
x^2*(2-10^(1/2))))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 155, 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} \[ \int \frac {1}{\sqrt {-3-4 x^2+2 x^4}} \, dx=\frac {\sqrt {-\left (\left (2-\sqrt {10}\right ) x^2\right )-3} \sqrt {\frac {\left (2+\sqrt {10}\right ) x^2+3}{\left (2-\sqrt {10}\right ) x^2+3}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-\left (\left (2-\sqrt {10}\right ) x^2\right )-3}}\right ),\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{\left (2-\sqrt {10}\right ) x^2+3}} \sqrt {2 x^4-4 x^2-3}} \]

[In]

Int[1/Sqrt[-3 - 4*x^2 + 2*x^4],x]

[Out]

(Sqrt[-3 - (2 - Sqrt[10])*x^2]*Sqrt[(3 + (2 + Sqrt[10])*x^2)/(3 + (2 - Sqrt[10])*x^2)]*EllipticF[ArcSin[(2^(3/
4)*5^(1/4)*x)/Sqrt[-3 - (2 - Sqrt[10])*x^2]], (5 - Sqrt[10])/10])/(2^(3/4)*Sqrt[3]*5^(1/4)*Sqrt[(3 + (2 - Sqrt
[10])*x^2)^(-1)]*Sqrt[-3 - 4*x^2 + 2*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 {-3-\left (2-\sqrt {10}\right ) x^2} \sqrt {\frac {3+\left (2+\sqrt {10}\right ) x^2}{3+\left (2-\sqrt {10}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-3-\left (2-\sqrt {10}\right ) x^2}}\right )|\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{3+\left (2-\sqrt {10}\right ) x^2}} \sqrt {-3-4 x^2+2 x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {-3-4 x^2+2 x^4}} \, dx=-\frac {i \sqrt {3+4 x^2-2 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{-2+\sqrt {10}}} x\right ),-\frac {7}{3}+\frac {2 \sqrt {10}}{3}\right )}{\sqrt {2+\sqrt {10}} \sqrt {-3-4 x^2+2 x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.54

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

[In]

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

[Out]

3/(-6-3*10^(1/2))^(1/2)*(1-(-2/3-1/3*10^(1/2))*x^2)^(1/2)*(1-(-2/3+1/3*10^(1/2))*x^2)^(1/2)/(2*x^4-4*x^2-3)^(1
/2)*EllipticF(1/3*(-6-3*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 = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.32 \[ \int \frac {1}{\sqrt {-3-4 x^2+2 x^4}} \, dx=-\frac {1}{18} \, {\left (\sqrt {10} \sqrt {3} \sqrt {-3} + 2 \, \sqrt {3} \sqrt {-3}\right )} \sqrt {\sqrt {10} - 2} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} x \sqrt {\sqrt {10} - 2}\right )\,|\,-\frac {2}{3} \, \sqrt {10} - \frac {7}{3}) \]

[In]

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

[Out]

-1/18*(sqrt(10)*sqrt(3)*sqrt(-3) + 2*sqrt(3)*sqrt(-3))*sqrt(sqrt(10) - 2)*elliptic_f(arcsin(1/3*sqrt(3)*x*sqrt
(sqrt(10) - 2)), -2/3*sqrt(10) - 7/3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(2*x^4 - 4*x^2 - 3)^(1/2),x)

[Out]

int(1/(2*x^4 - 4*x^2 - 3)^(1/2), x)