3.763 \(\int \frac {\cos (2 x)}{8+\sin ^2(2 x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\tan ^{-1}\left (\frac {\sin (2 x)}{2 \sqrt {2}}\right )}{4 \sqrt {2}} \]

[Out]

1/8*arctan(1/4*sin(2*x)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3190, 203} \[ \frac {\tan ^{-1}\left (\frac {\sin (2 x)}{2 \sqrt {2}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[2*x]/(8 + Sin[2*x]^2),x]

[Out]

ArcTan[Sin[2*x]/(2*Sqrt[2])]/(4*Sqrt[2])

Rule 203

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

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\cos (2 x)}{8+\sin ^2(2 x)} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{8+x^2} \, dx,x,\sin (2 x)\right )\\ &=\frac {\tan ^{-1}\left (\frac {\sin (2 x)}{2 \sqrt {2}}\right )}{4 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.87 \[ \frac {\tan ^{-1}\left (\frac {\sin (x) \cos (x)}{\sqrt {2}}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[2*x]/(8 + Sin[2*x]^2),x]

[Out]

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

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fricas [A]  time = 1.58, size = 15, normalized size = 0.65 \[ \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} \sin \left (2 \, x\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)/(8+sin(2*x)^2),x, algorithm="fricas")

[Out]

1/8*sqrt(2)*arctan(1/4*sqrt(2)*sin(2*x))

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giac [A]  time = 0.15, size = 15, normalized size = 0.65 \[ \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} \sin \left (2 \, x\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)/(8+sin(2*x)^2),x, algorithm="giac")

[Out]

1/8*sqrt(2)*arctan(1/4*sqrt(2)*sin(2*x))

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maple [A]  time = 0.04, size = 16, normalized size = 0.70 \[ \frac {\arctan \left (\frac {\sin \left (2 x \right ) \sqrt {2}}{4}\right ) \sqrt {2}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)/(8+sin(2*x)^2),x)

[Out]

1/8*arctan(1/4*sin(2*x)*2^(1/2))*2^(1/2)

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maxima [A]  time = 0.43, size = 15, normalized size = 0.65 \[ \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} \sin \left (2 \, x\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)/(8+sin(2*x)^2),x, algorithm="maxima")

[Out]

1/8*sqrt(2)*arctan(1/4*sqrt(2)*sin(2*x))

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mupad [B]  time = 0.07, size = 15, normalized size = 0.65 \[ \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sin \left (2\,x\right )}{4}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)/(sin(2*x)^2 + 8),x)

[Out]

(2^(1/2)*atan((2^(1/2)*sin(2*x))/4))/8

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sympy [A]  time = 0.27, size = 19, normalized size = 0.83 \[ \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sin {\left (2 x \right )}}{4} \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)/(8+sin(2*x)**2),x)

[Out]

sqrt(2)*atan(sqrt(2)*sin(2*x)/4)/8

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