3.54 \(\int \frac {1}{\sqrt {-3+6 x^2+2 x^4}} \, dx\)

Optimal. Leaf size=148 \[ \frac {\sqrt {\frac {3-\left (3-\sqrt {15}\right ) x^2}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {\left (3+\sqrt {15}\right ) x^2-3} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{15} x}{\sqrt {\left (3+\sqrt {15}\right ) x^2-3}}\right )|\frac {1}{10} \left (5+\sqrt {15}\right )\right )}{\sqrt {2} 3^{3/4} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {2 x^4+6 x^2-3}} \]

[Out]

1/30*EllipticF(15^(1/4)*x*2^(1/2)/(-3+x^2*(3+15^(1/2)))^(1/2),1/10*(50+10*15^(1/2))^(1/2))*((3-x^2*(3-15^(1/2)
))/(3-x^2*(3+15^(1/2))))^(1/2)*(-3+x^2*(3+15^(1/2)))^(1/2)*3^(1/4)*5^(3/4)*2^(1/2)/(2*x^4+6*x^2-3)^(1/2)/(1/(3
-x^2*(3+15^(1/2))))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 148, 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} \[ \frac {\sqrt {\frac {3-\left (3-\sqrt {15}\right ) x^2}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {\left (3+\sqrt {15}\right ) x^2-3} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{15} x}{\sqrt {\left (3+\sqrt {15}\right ) x^2-3}}\right )|\frac {1}{10} \left (5+\sqrt {15}\right )\right )}{\sqrt {2} 3^{3/4} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {2 x^4+6 x^2-3}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-3 + 6*x^2 + 2*x^4],x]

[Out]

(Sqrt[(3 - (3 - Sqrt[15])*x^2)/(3 - (3 + Sqrt[15])*x^2)]*Sqrt[-3 + (3 + Sqrt[15])*x^2]*EllipticF[ArcSin[(Sqrt[
2]*15^(1/4)*x)/Sqrt[-3 + (3 + Sqrt[15])*x^2]], (5 + Sqrt[15])/10])/(Sqrt[2]*3^(3/4)*5^(1/4)*Sqrt[(3 - (3 + Sqr
t[15])*x^2)^(-1)]*Sqrt[-3 + 6*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+6 x^2+2 x^4}} \, dx &=\frac {\sqrt {\frac {3-\left (3-\sqrt {15}\right ) x^2}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {-3+\left (3+\sqrt {15}\right ) x^2} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{15} x}{\sqrt {-3+\left (3+\sqrt {15}\right ) x^2}}\right )|\frac {1}{10} \left (5+\sqrt {15}\right )\right )}{\sqrt {2} 3^{3/4} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {-3+6 x^2+2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 77, normalized size = 0.52 \[ -\frac {i \sqrt {-2 x^4-6 x^2+3} F\left (i \sinh ^{-1}\left (\sqrt {-1+\sqrt {\frac {5}{3}}} x\right )|-4-\sqrt {15}\right )}{\sqrt {\sqrt {15}-3} \sqrt {2 x^4+6 x^2-3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/Sqrt[-3 + 6*x^2 + 2*x^4],x]

[Out]

((-I)*Sqrt[3 - 6*x^2 - 2*x^4]*EllipticF[I*ArcSinh[Sqrt[-1 + Sqrt[5/3]]*x], -4 - Sqrt[15]])/(Sqrt[-3 + Sqrt[15]
]*Sqrt[-3 + 6*x^2 + 2*x^4])

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {2 \, x^{4} + 6 \, x^{2} - 3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^4+6*x^2-3)^(1/2),x, algorithm="fricas")

[Out]

integral(1/sqrt(2*x^4 + 6*x^2 - 3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} + 6 \, x^{2} - 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^4+6*x^2-3)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(2*x^4 + 6*x^2 - 3), x)

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maple [C]  time = 0.03, size = 84, normalized size = 0.57 \[ \frac {3 \sqrt {-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}+1}\, \sqrt {-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}+1}\, \EllipticF \left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{\sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x^4+6*x^2-3)^(1/2),x)

[Out]

3/(9-3*15^(1/2))^(1/2)*(-(1-1/3*15^(1/2))*x^2+1)^(1/2)*(-(1+1/3*15^(1/2))*x^2+1)^(1/2)/(2*x^4+6*x^2-3)^(1/2)*E
llipticF(1/3*(9-3*15^(1/2))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 \, x^{4} + 6 \, x^{2} - 3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x^4+6*x^2-3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(2*x^4 + 6*x^2 - 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {2\,x^4+6\,x^2-3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(6*x^2 + 2*x^4 - 3)^(1/2),x)

[Out]

int(1/(6*x^2 + 2*x^4 - 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {2 x^{4} + 6 x^{2} - 3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*x**4+6*x**2-3)**(1/2),x)

[Out]

Integral(1/sqrt(2*x**4 + 6*x**2 - 3), x)

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