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

Optimal. Leaf size=33 \[ -\frac {\tan ^{-1}\left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d}-\frac {x}{4} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2658} \[ -\frac {\tan ^{-1}\left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d}-\frac {x}{4} \]

Antiderivative was successfully verified.

[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 2658

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

Rubi steps

\begin {align*} \int \frac {1}{-5+3 \cos (c+d x)} \, dx &=-\frac {x}{4}-\frac {\tan ^{-1}\left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.61 \[ -\frac {\tan ^{-1}\left (2 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTan[2*Tan[(c + d*x)/2]]/d

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fricas [A]  time = 0.92, size = 26, normalized size = 0.79 \[ \frac {\arctan \left (\frac {5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.38, size = 30, normalized size = 0.91 \[ -\frac {d x + c - 2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.03, size = 18, normalized size = 0.55 \[ -\frac {\arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.39, size = 24, normalized size = 0.73 \[ -\frac {\arctan \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.25, size = 38, normalized size = 1.15 \[ \frac {\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}}{2\,d}-\frac {\mathrm {atan}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*cos(c + d*x) - 5),x)

[Out]

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

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sympy [A]  time = 0.59, size = 42, normalized size = 1.27 \[ \begin {cases} - \frac {\operatorname {atan}{\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor }{2 d} & \text {for}\: d \neq 0 \\\frac {x}{3 \cos {\relax (c )} - 5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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/(3*cos(c) - 5)
, True))

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