\(\int \frac {1}{5-3 \cos (c+d x)} \, dx\) [22]

Optimal result

Integrand size = 12, antiderivative size = 33 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\frac {x}{4}+\frac {\arctan \left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d} \]

[Out]

1/4*x+1/2*arctan(sin(d*x+c)/(3-cos(d*x+c)))/d

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2736} \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\frac {\arctan \left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d}+\frac {x}{4} \]

[In]

Int[(5 - 3*Cos[c + d*x])^(-1),x]

[Out]

x/4 + ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])]/(2*d)

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}& = \frac {x}{4}+\frac {\arctan \left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.61 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\frac {\arctan \left (2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \]

[In]

Integrate[(5 - 3*Cos[c + d*x])^(-1),x]

[Out]

ArcTan[2*Tan[(c + d*x)/2]]/(2*d)

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.55

method	result	size
derivativedivides	\(\frac {\arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(18\)
default	\(\frac {\arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}\)	\(18\)
risch	\(-\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{3}\right )}{4 d}+\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-3\right )}{4 d}\)	\(38\)
parallelrisch	\(-\frac {i \left (\ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )-\ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )\right )}{4 d}\)	\(40\)

[In]

int(1/(5-3*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/2/d*arctan(2*tan(1/2*d*x+1/2*c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=-\frac {\arctan \left (\frac {5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right )}{4 \, d} \]

[In]

integrate(1/(5-3*cos(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*arctan(1/4*(5*cos(d*x + c) - 3)/sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\begin {cases} \frac {\operatorname {atan}{\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor }{2 d} & \text {for}\: d \neq 0 \\\frac {x}{5 - 3 \cos {\left (c \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(5-3*cos(d*x+c)),x)

[Out]

Piecewise(((atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(2*d), Ne(d, 0)), (x/(5 - 3*cos(c)),
True))

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\frac {\arctan \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{2 \, d} \]

[In]

integrate(1/(5-3*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/2*arctan(2*sin(d*x + c)/(cos(d*x + c) + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\frac {d x + c - 2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{4 \, d} \]

[In]

integrate(1/(5-3*cos(d*x+c)),x, algorithm="giac")

[Out]

1/4*(d*x + c - 2*arctan(sin(d*x + c)/(cos(d*x + c) - 3)))/d

Mupad [B] (verification not implemented)

Time = 15.49 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {1}{5-3 \cos (c+d x)} \, dx=\frac {\mathrm {atan}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}}{2\,d} \]

[In]

int(-1/(3*cos(c + d*x) - 5),x)

[Out]

atan(2*tan(c/2 + (d*x)/2))/(2*d) - (atan(tan(c/2 + (d*x)/2)) - (d*x)/2)/(2*d)