\(\int \sqrt {12-3 x^2} \, dx\) [95]

Optimal result

Integrand size = 11, antiderivative size = 35 \[ \int \sqrt {12-3 x^2} \, dx=\frac {1}{2} \sqrt {3} x \sqrt {4-x^2}+2 \sqrt {3} \arcsin \left (\frac {x}{2}\right ) \]

[Out]

1/2*3^(1/2)*x*(-x^2+4)^(1/2)+2*3^(1/2)*arcsin(1/2*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {201, 222} \[ \int \sqrt {12-3 x^2} \, dx=2 \sqrt {3} \arcsin \left (\frac {x}{2}\right )+\frac {1}{2} \sqrt {3} \sqrt {4-x^2} x \]

[In]

Int[Sqrt[12 - 3*x^2],x]

[Out]

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

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \sqrt {3} x \sqrt {4-x^2}+6 \int \frac {1}{\sqrt {12-3 x^2}} \, dx \\ & = \frac {1}{2} \sqrt {3} x \sqrt {4-x^2}+2 \sqrt {3} \arcsin \left (\frac {x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \sqrt {12-3 x^2} \, dx=\frac {1}{2} \sqrt {3} \left (x \sqrt {4-x^2}-8 \arctan \left (\frac {\sqrt {4-x^2}}{2+x}\right )\right ) \]

[In]

Integrate[Sqrt[12 - 3*x^2],x]

[Out]

(Sqrt[3]*(x*Sqrt[4 - x^2] - 8*ArcTan[Sqrt[4 - x^2]/(2 + x)]))/2

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {x \sqrt {-3 x^{2}+12}}{2}+2 \sqrt {3}\, \arcsin \left (\frac {x}{2}\right )\)	\(23\)
risch	\(-\frac {3 x \left (x^{2}-4\right )}{2 \sqrt {-3 x^{2}+12}}+2 \sqrt {3}\, \arcsin \left (\frac {x}{2}\right )\)	\(28\)
meijerg	\(\frac {i \sqrt {3}\, \left (-i \sqrt {\pi }\, x \sqrt {-\frac {x^{2}}{4}+1}-2 i \sqrt {\pi }\, \arcsin \left (\frac {x}{2}\right )\right )}{\sqrt {\pi }}\)	\(37\)
pseudoelliptic	\(\frac {x \sqrt {-3 x^{2}+12}}{2}-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {-3 x^{2}+12}\, \sqrt {3}}{3 x}\right )\)	\(37\)
trager	\(\frac {x \sqrt {-3 x^{2}+12}}{2}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+3\right ) \sqrt {-3 x^{2}+12}+3 x \right )\)	\(43\)

[In]

int((-3*x^2+12)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*x*(-3*x^2+12)^(1/2)+2*3^(1/2)*arcsin(1/2*x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \sqrt {12-3 x^2} \, dx=\frac {1}{2} \, \sqrt {-3 \, x^{2} + 12} x - 2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt {-3 \, x^{2} + 12}}{3 \, x}\right ) \]

[In]

integrate((-3*x^2+12)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(-3*x^2 + 12)*x - 2*sqrt(3)*arctan(1/3*sqrt(3)*sqrt(-3*x^2 + 12)/x)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \sqrt {12-3 x^2} \, dx=\frac {\sqrt {3} x \sqrt {4 - x^{2}}}{2} + 2 \sqrt {3} \operatorname {asin}{\left (\frac {x}{2} \right )} \]

[In]

integrate((-3*x**2+12)**(1/2),x)

[Out]

sqrt(3)*x*sqrt(4 - x**2)/2 + 2*sqrt(3)*asin(x/2)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.63 \[ \int \sqrt {12-3 x^2} \, dx=\frac {1}{2} \, \sqrt {-3 \, x^{2} + 12} x + 2 \, \sqrt {3} \arcsin \left (\frac {1}{2} \, x\right ) \]

[In]

integrate((-3*x^2+12)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(-3*x^2 + 12)*x + 2*sqrt(3)*arcsin(1/2*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.66 \[ \int \sqrt {12-3 x^2} \, dx=\frac {1}{2} \, \sqrt {3} {\left (\sqrt {-x^{2} + 4} x + 4 \, \arcsin \left (\frac {1}{2} \, x\right )\right )} \]

[In]

integrate((-3*x^2+12)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int \sqrt {12-3 x^2} \, dx=2\,\sqrt {3}\,\mathrm {asin}\left (\frac {x}{2}\right )+\frac {\sqrt {3}\,x\,\sqrt {4-x^2}}{2} \]

[In]

int((12 - 3*x^2)^(1/2),x)

[Out]

2*3^(1/2)*asin(x/2) + (3^(1/2)*x*(4 - x^2)^(1/2))/2