### 3.22 $$\int \frac{1}{5-3 \sin (c+d x)} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0125714, antiderivative size = 33, normalized size of antiderivative = 1., 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 = {2657} $\frac{x}{4}-\frac{\tan ^{-1}\left (\frac{\cos (c+d x)}{3-\sin (c+d x)}\right )}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(5 - 3*Sin[c + d*x])^(-1),x]

[Out]

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

Rule 2657

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(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] && PosQ[a]

Rubi steps

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

Mathematica [A]  time = 0.0251863, size = 56, normalized size = 1.7 $-\frac{\tan ^{-1}\left (\frac{2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}\right )}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 20, normalized size = 0.6 \begin{align*}{\frac{1}{2\,d}\arctan \left ({\frac{5}{4}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3}{4}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.85592, size = 35, normalized size = 1.06 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.968, size = 72, normalized size = 2.18 \begin{align*} \frac{\arctan \left (\frac{5 \, \sin \left (d x + c\right ) - 3}{4 \, \cos \left (d x + c\right )}\right )}{4 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.910273, size = 46, normalized size = 1.39 \begin{align*} \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}{5 - 3 \sin{\left (c \right )}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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/(5 - 3*
sin(c)), True))

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Giac [A]  time = 1.11797, size = 68, normalized size = 2.06 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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