3.429 \(\int \cos (x) \sqrt{\cos (2 x)} \, dx\)

Optimal. Leaf size=33 \[ \frac{\sin ^{-1}\left (\sqrt{2} \sin (x)\right )}{2 \sqrt{2}}+\frac{1}{2} \sin (x) \sqrt{\cos (2 x)} \]

[Out]

ArcSin[Sqrt[2]*Sin[x]]/(2*Sqrt[2]) + (Sqrt[Cos[2*x]]*Sin[x])/2

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Rubi [A]  time = 0.0220386, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4356, 195, 216} \[ \frac{\sin ^{-1}\left (\sqrt{2} \sin (x)\right )}{2 \sqrt{2}}+\frac{1}{2} \sin (x) \sqrt{\cos (2 x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]*Sqrt[Cos[2*x]],x]

[Out]

ArcSin[Sqrt[2]*Sin[x]]/(2*Sqrt[2]) + (Sqrt[Cos[2*x]]*Sin[x])/2

Rule 4356

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[d/(b
*c), Subst[Int[SubstFor[1, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a +
b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cos] || EqQ[F, cos])

Rule 195

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 216

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*} \int \cos (x) \sqrt{\cos (2 x)} \, dx &=\operatorname{Subst}\left (\int \sqrt{1-2 x^2} \, dx,x,\sin (x)\right )\\ &=\frac{1}{2} \sqrt{\cos (2 x)} \sin (x)+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-2 x^2}} \, dx,x,\sin (x)\right )\\ &=\frac{\sin ^{-1}\left (\sqrt{2} \sin (x)\right )}{2 \sqrt{2}}+\frac{1}{2} \sqrt{\cos (2 x)} \sin (x)\\ \end{align*}

Mathematica [A]  time = 0.0183587, size = 32, normalized size = 0.97 \[ \frac{1}{4} \left (\sqrt{2} \sin ^{-1}\left (\sqrt{2} \sin (x)\right )+2 \sin (x) \sqrt{\cos (2 x)}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]*Sqrt[Cos[2*x]],x]

[Out]

(Sqrt[2]*ArcSin[Sqrt[2]*Sin[x]] + 2*Sqrt[Cos[2*x]]*Sin[x])/4

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Maple [B]  time = 0.056, size = 61, normalized size = 1.9 \begin{align*}{\frac{1}{8\,\sin \left ( x \right ) }\sqrt{ \left ( 2\, \left ( \cos \left ( x \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( x \right ) \right ) ^{2}} \left ( \sqrt{2}\arcsin \left ( 4\, \left ( \sin \left ( x \right ) \right ) ^{2}-1 \right ) +4\,\sqrt{-2\, \left ( \sin \left ( x \right ) \right ) ^{4}+ \left ( \sin \left ( x \right ) \right ) ^{2}} \right ){\frac{1}{\sqrt{2\, \left ( \cos \left ( x \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)*cos(2*x)^(1/2),x)

[Out]

1/8*((2*cos(x)^2-1)*sin(x)^2)^(1/2)*(2^(1/2)*arcsin(4*sin(x)^2-1)+4*(-2*sin(x)^4+sin(x)^2)^(1/2))/sin(x)/(2*co
s(x)^2-1)^(1/2)

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Maxima [B]  time = 1.60023, size = 659, normalized size = 19.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(2)*(2*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))*sin
(2*x) - (cos(2*x) - 1)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))) + arctan2(-(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4
*x) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))*sin(2*x) - cos(2*x)*sin(1/2*arctan2(sin(4*x), cos(4*x
) + 1))), (cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*(cos(2*x)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))
+ sin(2*x)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))) + 1) - arctan2(-(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1
)^(1/4)*(cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))*sin(2*x) - cos(2*x)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))
), (cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*(cos(2*x)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1)) + sin(2
*x)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))) - 1) - arctan2((cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*
sin(1/2*arctan2(sin(4*x), cos(4*x) + 1)), (cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin
(4*x), cos(4*x) + 1)) + 1) + arctan2((cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*sin(1/2*arctan2(sin(4*x)
, cos(4*x) + 1)), (cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1)) -
1))

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Fricas [B]  time = 3.21268, size = 242, normalized size = 7.33 \begin{align*} -\frac{1}{16} \, \sqrt{2} \arctan \left (\frac{{\left (32 \, \sqrt{2} \cos \left (x\right )^{4} - 48 \, \sqrt{2} \cos \left (x\right )^{2} + 17 \, \sqrt{2}\right )} \sqrt{2 \, \cos \left (x\right )^{2} - 1}}{8 \,{\left (8 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 3\right )} \sin \left (x\right )}\right ) + \frac{1}{2} \, \sqrt{2 \, \cos \left (x\right )^{2} - 1} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sqrt(2)*arctan(1/8*(32*sqrt(2)*cos(x)^4 - 48*sqrt(2)*cos(x)^2 + 17*sqrt(2))*sqrt(2*cos(x)^2 - 1)/((8*cos
(x)^4 - 10*cos(x)^2 + 3)*sin(x))) + 1/2*sqrt(2*cos(x)^2 - 1)*sin(x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos{\left (x \right )} \sqrt{\cos{\left (2 x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)*cos(2*x)**(1/2),x)

[Out]

Integral(cos(x)*sqrt(cos(2*x)), x)

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Giac [A]  time = 1.11515, size = 36, normalized size = 1.09 \begin{align*} \frac{1}{4} \, \sqrt{2} \arcsin \left (\sqrt{2} \sin \left (x\right )\right ) + \frac{1}{2} \, \sqrt{-2 \, \sin \left (x\right )^{2} + 1} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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