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

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

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [A]  time = 0.02, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos (x) \sqrt {\cos (2 x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 1.32, size = 77, normalized size = 2.33 \[ -\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {{\left (32 \, \sqrt {2} \cos \relax (x)^{4} - 48 \, \sqrt {2} \cos \relax (x)^{2} + 17 \, \sqrt {2}\right )} \sqrt {2 \, \cos \relax (x)^{2} - 1}}{8 \, {\left (8 \, \cos \relax (x)^{4} - 10 \, \cos \relax (x)^{2} + 3\right )} \sin \relax (x)}\right ) + \frac {1}{2} \, \sqrt {2 \, \cos \relax (x)^{2} - 1} \sin \relax (x) \]

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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giac [A]  time = 0.69, size = 27, normalized size = 0.82 \[ \frac {1}{4} \, \sqrt {2} \arcsin \left (\sqrt {2} \sin \relax (x)\right ) + \frac {1}{2} \, \sqrt {-2 \, \sin \relax (x)^{2} + 1} \sin \relax (x) \]

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)

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maple [B]  time = 0.16, size = 62, normalized size = 1.88

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\relax (x )\right )-1\right ) \left (\sin ^{2}\relax (x )\right )}\, \left (-\sqrt {2}\, \arcsin \left (4 \left (\sin ^{2}\relax (x )\right )-1\right )-4 \sqrt {-2 \left (\sin ^{4}\relax (x )\right )+\sin ^{2}\relax (x )}\right )}{8 \sin \relax (x ) \sqrt {2 \left (\cos ^{2}\relax (x )\right )-1}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*
cos(x)^2-1)^(1/2)

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maxima [B]  time = 1.15, size = 488, normalized size = 14.79 \[ \text {result too large to display} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \sqrt {\cos \left (2\,x\right )}\,\cos \relax (x) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(2*x)^(1/2)*cos(x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos {\relax (x )} \sqrt {\cos {\left (2 x \right )}}\, dx \]

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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