\(\int \frac {1}{\sqrt {-3+2 x^2+2 x^4}} \, dx\) [58]

Optimal result

Integrand size = 16, antiderivative size = 143 \[ \int \frac {1}{\sqrt {-3+2 x^2+2 x^4}} \, dx=\frac {\sqrt {\frac {3-\left (1-\sqrt {7}\right ) x^2}{3-\left (1+\sqrt {7}\right ) x^2}} \sqrt {-3+\left (1+\sqrt {7}\right ) x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{7} x}{\sqrt {-3+\left (1+\sqrt {7}\right ) x^2}}\right ),\frac {1}{14} \left (7+\sqrt {7}\right )\right )}{\sqrt {6} \sqrt [4]{7} \sqrt {\frac {1}{3-\left (1+\sqrt {7}\right ) x^2}} \sqrt {-3+2 x^2+2 x^4}} \]

[Out]

1/42*EllipticF(7^(1/4)*x*2^(1/2)/(-3+x^2*(1+7^(1/2)))^(1/2),1/14*(98+14*7^(1/2))^(1/2))*((3-x^2*(1-7^(1/2)))/(
3-x^2*(1+7^(1/2))))^(1/2)*(-3+x^2*(1+7^(1/2)))^(1/2)*7^(3/4)*6^(1/2)/(2*x^4+2*x^2-3)^(1/2)/(1/(3-x^2*(1+7^(1/2
))))^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 143, 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} \[ \int \frac {1}{\sqrt {-3+2 x^2+2 x^4}} \, dx=\frac {\sqrt {\frac {3-\left (1-\sqrt {7}\right ) x^2}{3-\left (1+\sqrt {7}\right ) x^2}} \sqrt {\left (1+\sqrt {7}\right ) x^2-3} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{7} x}{\sqrt {\left (1+\sqrt {7}\right ) x^2-3}}\right ),\frac {1}{14} \left (7+\sqrt {7}\right )\right )}{\sqrt {6} \sqrt [4]{7} \sqrt {\frac {1}{3-\left (1+\sqrt {7}\right ) x^2}} \sqrt {2 x^4+2 x^2-3}} \]

[In]

Int[1/Sqrt[-3 + 2*x^2 + 2*x^4],x]

[Out]

(Sqrt[(3 - (1 - Sqrt[7])*x^2)/(3 - (1 + Sqrt[7])*x^2)]*Sqrt[-3 + (1 + Sqrt[7])*x^2]*EllipticF[ArcSin[(Sqrt[2]*
7^(1/4)*x)/Sqrt[-3 + (1 + Sqrt[7])*x^2]], (7 + Sqrt[7])/14])/(Sqrt[6]*7^(1/4)*Sqrt[(3 - (1 + Sqrt[7])*x^2)^(-1
)]*Sqrt[-3 + 2*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 {\frac {3-\left (1-\sqrt {7}\right ) x^2}{3-\left (1+\sqrt {7}\right ) x^2}} \sqrt {-3+\left (1+\sqrt {7}\right ) x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{7} x}{\sqrt {-3+\left (1+\sqrt {7}\right ) x^2}}\right )|\frac {1}{14} \left (7+\sqrt {7}\right )\right )}{\sqrt {6} \sqrt [4]{7} \sqrt {\frac {1}{3-\left (1+\sqrt {7}\right ) x^2}} \sqrt {-3+2 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 = 0.58 \[ \int \frac {1}{\sqrt {-3+2 x^2+2 x^4}} \, dx=-\frac {i \sqrt {3-2 x^2-2 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{1+\sqrt {7}}} x\right ),-\frac {4}{3}-\frac {\sqrt {7}}{3}\right )}{\sqrt {-1+\sqrt {7}} \sqrt {-3+2 x^2+2 x^4}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.59

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.35 \[ \int \frac {1}{\sqrt {-3+2 x^2+2 x^4}} \, dx=-\frac {1}{18} \, {\left (\sqrt {7} \sqrt {3} \sqrt {-3} - \sqrt {3} \sqrt {-3}\right )} \sqrt {\sqrt {7} + 1} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} x \sqrt {\sqrt {7} + 1}\right )\,|\,\frac {1}{3} \, \sqrt {7} - \frac {4}{3}) \]

[In]

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

[Out]

-1/18*(sqrt(7)*sqrt(3)*sqrt(-3) - sqrt(3)*sqrt(-3))*sqrt(sqrt(7) + 1)*elliptic_f(arcsin(1/3*sqrt(3)*x*sqrt(sqr
t(7) + 1)), 1/3*sqrt(7) - 4/3)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(2*x^2 + 2*x^4 - 3)^(1/2),x)

[Out]

int(1/(2*x^2 + 2*x^4 - 3)^(1/2), x)