\(\int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx\) [179]

Optimal result

Integrand size = 23, antiderivative size = 38 \[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\sqrt {\frac {2}{3}} x \sqrt {1-4 x^4}+\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {2} x\right ),-1\right )}{\sqrt {3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {254, 201, 227} \[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {2} x\right ),-1\right )}{\sqrt {3}}+\sqrt {\frac {2}{3}} \sqrt {1-4 x^4} x \]

[In]

Int[Sqrt[3 - 6*x^2]*Sqrt[2 + 4*x^2],x]

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 254

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_.)*((a2_.) + (b2_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[(a1*a2 + b1*b2*x^(2*
n))^p, x] /; FreeQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a
2, 0]))

Rubi steps \begin{align*} \text {integral}& = \int \sqrt {6-24 x^4} \, dx \\ & = \sqrt {\frac {2}{3}} x \sqrt {1-4 x^4}+4 \int \frac {1}{\sqrt {6-24 x^4}} \, dx \\ & = \sqrt {\frac {2}{3}} x \sqrt {1-4 x^4}+\frac {2 F\left (\left .\sin ^{-1}\left (\sqrt {2} x\right )\right |-1\right )}{\sqrt {3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.58 \[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\sqrt {6} x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},4 x^4\right ) \]

[In]

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

[Out]

Sqrt[6]*x*Hypergeometric2F1[-1/2, 1/4, 5/4, 4*x^4]

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (29 ) = 58\).

Time = 2.75 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.97

method	result	size
default	\(-\frac {\sqrt {-6 x^{2}+3}\, \sqrt {2}\, \sqrt {2 x^{2}+1}\, \left (\sqrt {2}\, \sqrt {3}\, \sqrt {-6 x^{2}+3}\, \sqrt {2 x^{2}+1}\, F\left (\sqrt {2}\, x , i\right )-12 x^{5}+3 x \right )}{9 \left (4 x^{4}-1\right )}\)	\(75\)
elliptic	\(-\frac {\sqrt {-6 x^{2}+3}\, \sqrt {4 x^{2}+2}\, \sqrt {-24 x^{4}+6}\, \left (\frac {x \sqrt {-24 x^{4}+6}}{3}+\frac {2 \sqrt {2}\, \sqrt {-2 x^{2}+1}\, \sqrt {2 x^{2}+1}\, F\left (\sqrt {2}\, x , i\right )}{\sqrt {-24 x^{4}+6}}\right )}{6 \left (4 x^{4}-1\right )}\)	\(92\)
risch	\(-\frac {x \left (2 x^{2}-1\right ) \left (2 x^{2}+1\right ) \sqrt {\left (-6 x^{2}+3\right ) \left (4 x^{2}+2\right )}\, \sqrt {6}}{3 \sqrt {-\left (2 x^{2}-1\right ) \left (2 x^{2}+1\right )}\, \sqrt {-6 x^{2}+3}\, \sqrt {4 x^{2}+2}}+\frac {\sqrt {2}\, \sqrt {-2 x^{2}+1}\, \sqrt {2 x^{2}+1}\, F\left (\sqrt {2}\, x , i\right ) \sqrt {\left (-6 x^{2}+3\right ) \left (4 x^{2}+2\right )}\, \sqrt {6}}{3 \sqrt {-4 x^{4}+1}\, \sqrt {-6 x^{2}+3}\, \sqrt {4 x^{2}+2}}\)	\(153\)

[In]

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

[Out]

-1/9*(-6*x^2+3)^(1/2)*2^(1/2)*(2*x^2+1)^(1/2)*(2^(1/2)*3^(1/2)*(-6*x^2+3)^(1/2)*(2*x^2+1)^(1/2)*EllipticF(2^(1
/2)*x,I)-12*x^5+3*x)/(4*x^4-1)

Fricas [A] (verification not implemented)

none

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\frac {1}{3} \, \sqrt {4 \, x^{2} + 2} \sqrt {-6 \, x^{2} + 3} x + \frac {1}{3} \, \sqrt {2} \sqrt {-6} F(\arcsin \left (\frac {\sqrt {2}}{2 \, x}\right )\,|\,-1) \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\sqrt {6} \int \sqrt {1 - 2 x^{2}} \sqrt {2 x^{2} + 1}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\int { \sqrt {4 \, x^{2} + 2} \sqrt {-6 \, x^{2} + 3} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\int { \sqrt {4 \, x^{2} + 2} \sqrt {-6 \, x^{2} + 3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {3-6 x^2} \sqrt {2+4 x^2} \, dx=\int \sqrt {4\,x^2+2}\,\sqrt {3-6\,x^2} \,d x \]

[In]

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

[Out]

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