\(\int \frac {1}{\sqrt {-2-2 x^2+3 x^4}} \, dx\) [49]

Optimal result

Integrand size = 16, antiderivative size = 148 \[ \int \frac {1}{\sqrt {-2-2 x^2+3 x^4}} \, dx=\frac {\sqrt {-2-\left (1-\sqrt {7}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {7}\right ) x^2}{2+\left (1-\sqrt {7}\right ) x^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{7} x}{\sqrt {-2-\left (1-\sqrt {7}\right ) x^2}}\right ),\frac {1}{14} \left (7-\sqrt {7}\right )\right )}{2 \sqrt [4]{7} \sqrt {\frac {1}{2+\left (1-\sqrt {7}\right ) x^2}} \sqrt {-2-2 x^2+3 x^4}} \]

[Out]

1/14*EllipticF(7^(1/4)*x*2^(1/2)/(-2-x^2*(1-7^(1/2)))^(1/2),1/14*(98-14*7^(1/2))^(1/2))*(-2-x^2*(1-7^(1/2)))^(
1/2)*((2+x^2*(1+7^(1/2)))/(2+x^2*(1-7^(1/2))))^(1/2)*7^(3/4)/(3*x^4-2*x^2-2)^(1/2)/(1/(2+x^2*(1-7^(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} \[ \int \frac {1}{\sqrt {-2-2 x^2+3 x^4}} \, dx=\frac {\sqrt {-\left (\left (1-\sqrt {7}\right ) x^2\right )-2} \sqrt {\frac {\left (1+\sqrt {7}\right ) x^2+2}{\left (1-\sqrt {7}\right ) x^2+2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{7} x}{\sqrt {-\left (\left (1-\sqrt {7}\right ) x^2\right )-2}}\right ),\frac {1}{14} \left (7-\sqrt {7}\right )\right )}{2 \sqrt [4]{7} \sqrt {\frac {1}{\left (1-\sqrt {7}\right ) x^2+2}} \sqrt {3 x^4-2 x^2-2}} \]

[In]

Int[1/Sqrt[-2 - 2*x^2 + 3*x^4],x]

[Out]

(Sqrt[-2 - (1 - Sqrt[7])*x^2]*Sqrt[(2 + (1 + Sqrt[7])*x^2)/(2 + (1 - Sqrt[7])*x^2)]*EllipticF[ArcSin[(Sqrt[2]*
7^(1/4)*x)/Sqrt[-2 - (1 - Sqrt[7])*x^2]], (7 - Sqrt[7])/14])/(2*7^(1/4)*Sqrt[(2 + (1 - Sqrt[7])*x^2)^(-1)]*Sqr
t[-2 - 2*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 (1-\sqrt {7}\right ) x^2} \sqrt {\frac {2+\left (1+\sqrt {7}\right ) x^2}{2+\left (1-\sqrt {7}\right ) x^2}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{7} x}{\sqrt {-2-\left (1-\sqrt {7}\right ) x^2}}\right )|\frac {1}{14} \left (7-\sqrt {7}\right )\right )}{2 \sqrt [4]{7} \sqrt {\frac {1}{2+\left (1-\sqrt {7}\right ) x^2}} \sqrt {-2-2 x^2+3 x^4}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 10.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\sqrt {-2-2 x^2+3 x^4}} \, dx=-\frac {i \sqrt {2+2 x^2-3 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{-1+\sqrt {7}}} x\right ),\frac {1}{3} \left (-4+\sqrt {7}\right )\right )}{\sqrt {1+\sqrt {7}} \sqrt {-2-2 x^2+3 x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {2 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {7}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}+\frac {\sqrt {7}}{2}\right ) x^{2}}\, F\left (\frac {\sqrt {-2-2 \sqrt {7}}\, x}{2}, \frac {i \sqrt {42}}{6}-\frac {i \sqrt {6}}{6}\right )}{\sqrt {-2-2 \sqrt {7}}\, \sqrt {3 x^{4}-2 x^{2}-2}}\)	\(84\)
elliptic	\(\frac {2 \sqrt {1-\left (-\frac {1}{2}-\frac {\sqrt {7}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}+\frac {\sqrt {7}}{2}\right ) x^{2}}\, F\left (\frac {\sqrt {-2-2 \sqrt {7}}\, x}{2}, \frac {i \sqrt {42}}{6}-\frac {i \sqrt {6}}{6}\right )}{\sqrt {-2-2 \sqrt {7}}\, \sqrt {3 x^{4}-2 x^{2}-2}}\)	\(84\)

[In]

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

[Out]

2/(-2-2*7^(1/2))^(1/2)*(1-(-1/2-1/2*7^(1/2))*x^2)^(1/2)*(1-(-1/2+1/2*7^(1/2))*x^2)^(1/2)/(3*x^4-2*x^2-2)^(1/2)
*EllipticF(1/2*(-2-2*7^(1/2))^(1/2)*x,1/6*I*42^(1/2)-1/6*I*6^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt {-2-2 x^2+3 x^4}} \, dx=-\frac {1}{12} \, {\left (\sqrt {7} \sqrt {2} \sqrt {-2} + \sqrt {2} \sqrt {-2}\right )} \sqrt {\sqrt {7} - 1} F(\arcsin \left (\frac {1}{2} \, \sqrt {2} x \sqrt {\sqrt {7} - 1}\right )\,|\,-\frac {1}{3} \, \sqrt {7} - \frac {4}{3}) \]

[In]

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

[Out]

-1/12*(sqrt(7)*sqrt(2)*sqrt(-2) + sqrt(2)*sqrt(-2))*sqrt(sqrt(7) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*x*sqrt(sqr
t(7) - 1)), -1/3*sqrt(7) - 4/3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(3*x^4 - 2*x^2 - 2)^(1/2),x)

[Out]

int(1/(3*x^4 - 2*x^2 - 2)^(1/2), x)