\(\int \frac {2}{3-\cos (4+6 x)} \, dx\) [1]

Optimal result

Integrand size = 14, antiderivative size = 44 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {x}{\sqrt {2}}+\frac {\arctan \left (\frac {\sin (4+6 x)}{3+2 \sqrt {2}-\cos (4+6 x)}\right )}{3 \sqrt {2}} \]

[Out]

1/2*x*2^(1/2)+1/6*arctan(sin(4+6*x)/(3-cos(4+6*x)+2*2^(1/2)))*2^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 2736} \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {\arctan \left (\frac {\sin (6 x+4)}{-\cos (6 x+4)+2 \sqrt {2}+3}\right )}{3 \sqrt {2}}+\frac {x}{\sqrt {2}} \]

[In]

Int[2/(3 - Cos[4 + 6*x]),x]

[Out]

x/Sqrt[2] + ArcTan[Sin[4 + 6*x]/(3 + 2*Sqrt[2] - Cos[4 + 6*x])]/(3*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2736

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {1}{3-\cos (4+6 x)} \, dx \\ & = \frac {x}{\sqrt {2}}+\frac {\arctan \left (\frac {\sin (4+6 x)}{3+2 \sqrt {2}-\cos (4+6 x)}\right )}{3 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.50 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {\arctan \left (\sqrt {2} \tan (2+3 x)\right )}{3 \sqrt {2}} \]

[In]

Integrate[2/(3 - Cos[4 + 6*x]),x]

[Out]

ArcTan[Sqrt[2]*Tan[2 + 3*x]]/(3*Sqrt[2])

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.39

method	result	size
derivativedivides	\(\frac {\sqrt {2}\, \arctan \left (\tan \left (2+3 x \right ) \sqrt {2}\right )}{6}\)	\(17\)
default	\(\frac {\sqrt {2}\, \arctan \left (\tan \left (2+3 x \right ) \sqrt {2}\right )}{6}\)	\(17\)
risch	\(\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (2+3 x \right )}-3-2 \sqrt {2}\right )}{12}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i \left (2+3 x \right )}-3+2 \sqrt {2}\right )}{12}\)	\(48\)

[In]

int(2/(3-cos(4+6*x)),x,method=_RETURNVERBOSE)

[Out]

1/6*2^(1/2)*arctan(tan(2+3*x)*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=-\frac {1}{12} \, \sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (6 \, x + 4\right ) - \sqrt {2}}{4 \, \sin \left (6 \, x + 4\right )}\right ) \]

[In]

integrate(2/(3-cos(4+6*x)),x, algorithm="fricas")

[Out]

-1/12*sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(6*x + 4) - sqrt(2))/sin(6*x + 4))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {\sqrt {2} \left (\operatorname {atan}{\left (\sqrt {2} \tan {\left (3 x + 2 \right )} \right )} + \pi \left \lfloor {\frac {3 x - \frac {\pi }{2} + 2}{\pi }}\right \rfloor \right )}{6} \]

[In]

integrate(2/(3-cos(4+6*x)),x)

[Out]

sqrt(2)*(atan(sqrt(2)*tan(3*x + 2)) + pi*floor((3*x - pi/2 + 2)/pi))/6

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.59 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {1}{6} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sin \left (6 \, x + 4\right )}{\cos \left (6 \, x + 4\right ) + 1}\right ) \]

[In]

integrate(2/(3-cos(4+6*x)),x, algorithm="maxima")

[Out]

1/6*sqrt(2)*arctan(sqrt(2)*sin(6*x + 4)/(cos(6*x + 4) + 1))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.30 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {1}{6} \, \sqrt {2} {\left (3 \, x + \arctan \left (-\frac {\sqrt {2} \sin \left (6 \, x + 4\right ) - 2 \, \sin \left (6 \, x + 4\right )}{\sqrt {2} \cos \left (6 \, x + 4\right ) + \sqrt {2} - 2 \, \cos \left (6 \, x + 4\right ) + 2}\right ) + 2\right )} \]

[In]

integrate(2/(3-cos(4+6*x)),x, algorithm="giac")

[Out]

1/6*sqrt(2)*(3*x + arctan(-(sqrt(2)*sin(6*x + 4) - 2*sin(6*x + 4))/(sqrt(2)*cos(6*x + 4) + sqrt(2) - 2*cos(6*x
+ 4) + 2)) + 2)

Mupad [B] (verification not implemented)

Time = 27.40 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80 \[ \int \frac {2}{3-\cos (4+6 x)} \, dx=\frac {\sqrt {2}\,\left (3\,x-\mathrm {atan}\left (\mathrm {tan}\left (3\,x+2\right )\right )\right )}{6}+\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (3\,x+2\right )\right )}{6} \]

[In]

int(-2/(cos(6*x + 4) - 3),x)

[Out]

(2^(1/2)*(3*x - atan(tan(3*x + 2))))/6 + (2^(1/2)*atan(2^(1/2)*tan(3*x + 2)))/6