3.27 \(\int \frac{1}{(-5+3 \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=58 \[ \frac{3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}+\frac{5 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac{5 x}{64} \]

[Out]

(5*x)/64 + (5*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(32*d) + (3*Sin[c + d*x])/(16*d*(5 - 3*Cos[c + d*x]))

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Rubi [A]  time = 0.0327876, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2664, 12, 2658} \[ \frac{3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}+\frac{5 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac{5 x}{64} \]

Antiderivative was successfully verified.

[In]

Int[(-5 + 3*Cos[c + d*x])^(-2),x]

[Out]

(5*x)/64 + (5*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(32*d) + (3*Sin[c + d*x])/(16*d*(5 - 3*Cos[c + d*x]))

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

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

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))^2} \, dx &=\frac{3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}-\frac{1}{16} \int \frac{5}{-5+3 \cos (c+d x)} \, dx\\ &=\frac{3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}-\frac{5}{16} \int \frac{1}{-5+3 \cos (c+d x)} \, dx\\ &=\frac{5 x}{64}+\frac{5 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac{3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.0234029, size = 43, normalized size = 0.74 \[ \frac{5 \tan ^{-1}\left (2 \tan \left (\frac{1}{2} (c+d x)\right )\right )-\frac{6 \sin (c+d x)}{3 \cos (c+d x)-5}}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 3*Cos[c + d*x])^(-2),x]

[Out]

(5*ArcTan[2*Tan[(c + d*x)/2]] - (6*Sin[c + d*x])/(-5 + 3*Cos[c + d*x]))/(32*d)

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Maple [A]  time = 0.034, size = 48, normalized size = 0.8 \begin{align*}{\frac{3}{64\,d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+{\frac{1}{4}} \right ) ^{-1}}+{\frac{5}{32\,d}\arctan \left ( 2\,\tan \left ( 1/2\,dx+c/2 \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-5+3*cos(d*x+c))^2,x)

[Out]

3/64/d*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1/4)+5/32/d*arctan(2*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.95581, size = 93, normalized size = 1.6 \begin{align*} \frac{\frac{6 \, \sin \left (d x + c\right )}{{\left (\frac{4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + 5 \, \arctan \left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

1/32*(6*sin(d*x + c)/((4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)*(cos(d*x + c) + 1)) + 5*arctan(2*sin(d*x + c
)/(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.65515, size = 163, normalized size = 2.81 \begin{align*} -\frac{5 \,{\left (3 \, \cos \left (d x + c\right ) - 5\right )} \arctan \left (\frac{5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right ) + 12 \, \sin \left (d x + c\right )}{64 \,{\left (3 \, d \cos \left (d x + c\right ) - 5 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/64*(5*(3*cos(d*x + c) - 5)*arctan(1/4*(5*cos(d*x + c) - 3)/sin(d*x + c)) + 12*sin(d*x + c))/(3*d*cos(d*x +
c) - 5*d)

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Sympy [A]  time = 2.21745, size = 192, normalized size = 3.31 \begin{align*} \begin{cases} \frac{x}{\left (-5 + 3 \cosh{\left (2 \operatorname{atanh}{\left (\frac{1}{2} \right )} \right )}\right )^{2}} & \text{for}\: c = - d x - 2 i \operatorname{atanh}{\left (\frac{1}{2} \right )} \vee c = - d x + 2 i \operatorname{atanh}{\left (\frac{1}{2} \right )} \\\frac{x}{\left (3 \cos{\left (c \right )} - 5\right )^{2}} & \text{for}\: d = 0 \\\frac{20 \left (\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 \right ) \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{128 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 32 d} + \frac{5 \left (\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 \right )}{128 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 32 d} + \frac{6 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{128 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 32 d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))**2,x)

[Out]

Piecewise((x/(-5 + 3*cosh(2*atanh(1/2)))**2, Eq(c, -d*x - 2*I*atanh(1/2)) | Eq(c, -d*x + 2*I*atanh(1/2))), (x/
(3*cos(c) - 5)**2, Eq(d, 0)), (20*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x
/2)**2/(128*d*tan(c/2 + d*x/2)**2 + 32*d) + 5*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(
128*d*tan(c/2 + d*x/2)**2 + 32*d) + 6*tan(c/2 + d*x/2)/(128*d*tan(c/2 + d*x/2)**2 + 32*d), True))

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Giac [A]  time = 1.19969, size = 82, normalized size = 1.41 \begin{align*} \frac{5 \, d x + 5 \, c + \frac{12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} - 10 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/64*(5*d*x + 5*c + 12*tan(1/2*d*x + 1/2*c)/(4*tan(1/2*d*x + 1/2*c)^2 + 1) - 10*arctan(sin(d*x + c)/(cos(d*x +
c) - 3)))/d