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

Optimal. Leaf size=155 \[ \frac {\sqrt {-\left (\left (2-\sqrt {10}\right ) x^2\right )-3} \sqrt {\frac {\left (2+\sqrt {10}\right ) x^2+3}{\left (2-\sqrt {10}\right ) x^2+3}} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-\left (\left (2-\sqrt {10}\right ) x^2\right )-3}}\right )|\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{\left (2-\sqrt {10}\right ) x^2+3}} \sqrt {2 x^4-4 x^2-3}} \]

[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]  time = 0.02, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1098} \[ \frac {\sqrt {-\left (2-\sqrt {10}\right ) x^2-3} \sqrt {\frac {\left (2+\sqrt {10}\right ) x^2+3}{\left (2-\sqrt {10}\right ) x^2+3}} F\left (\sin ^{-1}\left (\frac {2^{3/4} \sqrt [4]{5} x}{\sqrt {-\left (2-\sqrt {10}\right ) x^2-3}}\right )|\frac {1}{10} \left (5-\sqrt {10}\right )\right )}{2^{3/4} \sqrt {3} \sqrt [4]{5} \sqrt {\frac {1}{\left (2-\sqrt {10}\right ) x^2+3}} \sqrt {2 x^4-4 x^2-3}} \]

Antiderivative was successfully verified.

[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 1098

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

Rubi steps

\begin {align*} \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}} 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]  time = 0.06, size = 83, normalized size = 0.54 \[ -\frac {i \sqrt {-2 x^4+4 x^2+3} F\left (i \sinh ^{-1}\left (\sqrt {\frac {2}{-2+\sqrt {10}}} x\right )|-\frac {7}{3}+\frac {2 \sqrt {10}}{3}\right )}{\sqrt {2+\sqrt {10}} \sqrt {2 x^4-4 x^2-3}} \]

Warning: Unable to verify antiderivative.

[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])

________________________________________________________________________________________

fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {2 \, x^{4} - 4 \, x^{2} - 3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} - 4 \, x^{2} - 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [C]  time = 0.03, size = 84, normalized size = 0.54 \[ \frac {3 \sqrt {-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}+1}\, \sqrt {-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {-6-3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )}{\sqrt {-6-3 \sqrt {10}}\, \sqrt {2 x^{4}-4 x^{2}-3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/(-6-3*10^(1/2))^(1/2)*(-(-2/3-1/3*10^(1/2))*x^2+1)^(1/2)*(-(-2/3+1/3*10^(1/2))*x^2+1)^(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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} - 4 \, x^{2} - 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {2\,x^4-4\,x^2-3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 x^{4} - 4 x^{2} - 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________