\(\int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx\) [9]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=-\frac {1}{3} (4-\sin (2 x))^{3/2} \]

[Out]

-1/3*(4-sin(2*x))^(3/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2747, 32} \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=-\frac {1}{3} (4-\sin (2 x))^{3/2} \]

[In]

Int[Cos[2*x]*Sqrt[4 - Sin[2*x]],x]

[Out]

-1/3*(4 - Sin[2*x])^(3/2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=-\frac {1}{3} (4-\sin (2 x))^{3/2} \]

[In]

Integrate[Cos[2*x]*Sqrt[4 - Sin[2*x]],x]

[Out]

-1/3*(4 - Sin[2*x])^(3/2)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {\left (4-\sin \left (2 x \right )\right )^{\frac {3}{2}}}{3}\)	\(13\)
default	\(-\frac {\left (4-\sin \left (2 x \right )\right )^{\frac {3}{2}}}{3}\)	\(13\)

[In]

int(cos(2*x)*(4-sin(2*x))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*(4-sin(2*x))^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=\frac {1}{3} \, {\left (\sin \left (2 \, x\right ) - 4\right )} \sqrt {-\sin \left (2 \, x\right ) + 4} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (12) = 24\).

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=\frac {\sqrt {4 - \sin {\left (2 x \right )}} \sin {\left (2 x \right )}}{3} - \frac {4 \sqrt {4 - \sin {\left (2 x \right )}}}{3} \]

[In]

integrate(cos(2*x)*(4-sin(2*x))**(1/2),x)

[Out]

sqrt(4 - sin(2*x))*sin(2*x)/3 - 4*sqrt(4 - sin(2*x))/3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=-\frac {1}{3} \, {\left (-\sin \left (2 \, x\right ) + 4\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

-1/3*(-sin(2*x) + 4)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=-\frac {1}{3} \, {\left (-\sin \left (2 \, x\right ) + 4\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

-1/3*(-sin(2*x) + 4)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx=-\frac {{\left (4-\sin \left (2\,x\right )\right )}^{3/2}}{3} \]

[In]

int(cos(2*x)*(4 - sin(2*x))^(1/2),x)

[Out]

-(4 - sin(2*x))^(3/2)/3