\(\int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx\) [65]

Optimal result

Integrand size = 16, antiderivative size = 63 \[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=\frac {\sqrt {-3+x^2} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {7} x}{\sqrt {-3+x^2}}\right ),\frac {1}{7}\right )}{\sqrt {7} \sqrt {-3-5 x^2+2 x^4}} \]

[Out]

1/7*EllipticF(x*7^(1/2)/(x^2-3)^(1/2),1/7*7^(1/2))*(x^2-3)^(1/2)*(2*x^2+1)^(1/2)*7^(1/2)/(2*x^4-5*x^2-3)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 63, 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 = {1111} \[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=\frac {\sqrt {x^2-3} \sqrt {2 x^2+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {7} x}{\sqrt {x^2-3}}\right ),\frac {1}{7}\right )}{\sqrt {7} \sqrt {2 x^4-5 x^2-3}} \]

[In]

Int[1/Sqrt[-3 - 5*x^2 + 2*x^4],x]

[Out]

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

Rule 1111

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]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*Sqrt[-a]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[ArcSin[x/Sqrt[(2*a +
(b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x] /; IntegerQ[q]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[
a, 0] && GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-3+x^2} \sqrt {1+2 x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {7} x}{\sqrt {-3+x^2}}\right )|\frac {1}{7}\right )}{\sqrt {7} \sqrt {-3-5 x^2+2 x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=-\frac {i \sqrt {1-\frac {x^2}{3}} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} x\right ),-\frac {1}{6}\right )}{\sqrt {2} \sqrt {-3-5 x^2+2 x^4}} \]

[In]

Integrate[1/Sqrt[-3 - 5*x^2 + 2*x^4],x]

[Out]

((-I)*Sqrt[1 - x^2/3]*Sqrt[1 + 2*x^2]*EllipticF[I*ArcSinh[Sqrt[2]*x], -1/6])/(Sqrt[2]*Sqrt[-3 - 5*x^2 + 2*x^4]
)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, F\left (i \sqrt {2}\, x , \frac {i \sqrt {6}}{6}\right )}{6 \sqrt {2 x^{4}-5 x^{2}-3}}\)	\(53\)
elliptic	\(-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+1}\, \sqrt {-3 x^{2}+9}\, F\left (i \sqrt {2}\, x , \frac {i \sqrt {6}}{6}\right )}{6 \sqrt {2 x^{4}-5 x^{2}-3}}\)	\(53\)

[In]

int(1/(2*x^4-5*x^2-3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*I*2^(1/2)*(2*x^2+1)^(1/2)*(-3*x^2+9)^(1/2)/(2*x^4-5*x^2-3)^(1/2)*EllipticF(I*2^(1/2)*x,1/6*I*6^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.27 \[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=-\frac {1}{3} \, \sqrt {3} \sqrt {-3} F(\arcsin \left (\frac {1}{3} \, \sqrt {3} x\right )\,|\,-6) \]

[In]

integrate(1/(2*x^4-5*x^2-3)^(1/2),x, algorithm="fricas")

[Out]

-1/3*sqrt(3)*sqrt(-3)*elliptic_f(arcsin(1/3*sqrt(3)*x), -6)

Sympy [F]

\[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2 x^{4} - 5 x^{2} - 3}}\, dx \]

[In]

integrate(1/(2*x**4-5*x**2-3)**(1/2),x)

[Out]

Integral(1/sqrt(2*x**4 - 5*x**2 - 3), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} - 5 \, x^{2} - 3}} \,d x } \]

[In]

integrate(1/(2*x^4-5*x^2-3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 - 5*x^2 - 3), x)

Giac [F]

\[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=\int { \frac {1}{\sqrt {2 \, x^{4} - 5 \, x^{2} - 3}} \,d x } \]

[In]

integrate(1/(2*x^4-5*x^2-3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(2*x^4 - 5*x^2 - 3), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {-3-5 x^2+2 x^4}} \, dx=\int \frac {1}{\sqrt {2\,x^4-5\,x^2-3}} \,d x \]

[In]

int(1/(2*x^4 - 5*x^2 - 3)^(1/2),x)

[Out]

int(1/(2*x^4 - 5*x^2 - 3)^(1/2), x)