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

Optimal. Leaf size=16 \[ -\frac {1}{3} (4-\sin (2 x))^{3/2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2747, 32} \begin {gather*} -\frac {1}{3} (4-\sin (2 x))^{3/2} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \cos (2 x) \sqrt {4-\sin (2 x)} \, dx &=-\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*}

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Mathematica [A]
time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} -\frac {1}{3} (4-\sin (2 x))^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(4 - Sin[2*x])^(3/2)

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Maple [A]
time = 0.07, size = 13, normalized size = 0.81

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 4.35, size = 12, normalized size = 0.75 \begin {gather*} -\frac {1}{3} \, {\left (-\sin \left (2 \, x\right ) + 4\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-sin(2*x) + 4)^(3/2)

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Fricas [A]
time = 0.70, size = 18, normalized size = 1.12 \begin {gather*} \frac {1}{3} \, {\left (\sin \left (2 \, x\right ) - 4\right )} \sqrt {-\sin \left (2 \, x\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (12) = 24\).
time = 0.09, size = 29, normalized size = 1.81 \begin {gather*} \frac {\sqrt {4 - \sin {\left (2 x \right )}} \sin {\left (2 x \right )}}{3} - \frac {4 \sqrt {4 - \sin {\left (2 x \right )}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]
time = 0.83, size = 12, normalized size = 0.75 \begin {gather*} -\frac {1}{3} \, {\left (-\sin \left (2 \, x\right ) + 4\right )}^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-sin(2*x) + 4)^(3/2)

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Mupad [B]
time = 0.17, size = 12, normalized size = 0.75 \begin {gather*} -\frac {{\left (4-\sin \left (2\,x\right )\right )}^{3/2}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)*(4 - sin(2*x))^(1/2),x)

[Out]

-(4 - sin(2*x))^(3/2)/3

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