\(\int \frac {1}{-5-3 \sin (c+d x)} \, dx\) [30]

Optimal result

Integrand size = 12, antiderivative size = 31 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=-\frac {x}{4}-\frac {\arctan \left (\frac {\cos (c+d x)}{3+\sin (c+d x)}\right )}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, 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 = {2737} \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=-\frac {\arctan \left (\frac {\cos (c+d x)}{\sin (c+d x)+3}\right )}{2 d}-\frac {x}{4} \]

[In]

Int[(-5 - 3*Sin[c + d*x])^(-1),x]

[Out]

-1/4*x - ArcTan[Cos[c + d*x]/(3 + Sin[c + d*x])]/(2*d)

Rule 2737

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/(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] && NegQ[a]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=-\frac {\arctan \left (\frac {2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{2 d} \]

[In]

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

[Out]

-1/2*ArcTan[(2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])]/d

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(-\frac {\arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{2 d}\)	\(20\)
default	\(-\frac {\arctan \left (\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+\frac {3}{4}\right )}{2 d}\)	\(20\)
risch	\(-\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3 i\right )}{4 d}+\frac {i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i}{3}\right )}{4 d}\)	\(40\)
parallelrisch	\(\frac {i \left (\ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3-4 i\right )-\ln \left (5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3+4 i\right )\right )}{4 d}\)	\(42\)

[In]

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

[Out]

-1/2/d*arctan(5/4*tan(1/2*d*x+1/2*c)+3/4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=-\frac {\arctan \left (\frac {5 \, \sin \left (d x + c\right ) + 3}{4 \, \cos \left (d x + c\right )}\right )}{4 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).

Time = 0.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=\begin {cases} - \frac {\operatorname {atan}{\left (\frac {5 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4} + \frac {3}{4} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor }{2 d} & \text {for}\: d \neq 0 \\\frac {x}{- 3 \sin {\left (c \right )} - 5} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-(atan(5*tan(c/2 + d*x/2)/4 + 3/4) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(2*d), Ne(d, 0)), (x/(-3*si
n(c) - 5), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=-\frac {\arctan \left (\frac {5 \, \sin \left (d x + c\right )}{4 \, {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {3}{4}\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*arctan(5/4*sin(d*x + c)/(cos(d*x + c) + 1) + 3/4)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=-\frac {d x + c + 2 \, \arctan \left (-\frac {3 \, \cos \left (d x + c\right ) + \sin \left (d x + c\right ) + 3}{\cos \left (d x + c\right ) - 3 \, \sin \left (d x + c\right ) - 9}\right )}{4 \, d} \]

[In]

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

[Out]

-1/4*(d*x + c + 2*arctan(-(3*cos(d*x + c) + sin(d*x + c) + 3)/(cos(d*x + c) - 3*sin(d*x + c) - 9)))/d

Mupad [B] (verification not implemented)

Time = 5.94 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {1}{-5-3 \sin (c+d x)} \, dx=\frac {\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}}{2\,d}-\frac {\mathrm {atan}\left (\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {3}{4}\right )}{2\,d} \]

[In]

int(-1/(3*sin(c + d*x) + 5),x)

[Out]

(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)/(2*d) - atan((5*tan(c/2 + (d*x)/2))/4 + 3/4)/(2*d)