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

Optimal. Leaf size=83 \[ \frac{45 \sin (c+d x)}{512 d (5-3 \cos (c+d x))}+\frac{3 \sin (c+d x)}{32 d (5-3 \cos (c+d x))^2}+\frac{59 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{1024 d}+\frac{59 x}{2048} \]

[Out]

(59*x)/2048 + (59*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(1024*d) + (3*Sin[c + d*x])/(32*d*(5 - 3*Cos[c + d*
x])^2) + (45*Sin[c + d*x])/(512*d*(5 - 3*Cos[c + d*x]))

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Rubi [A]  time = 0.0626456, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2664, 2754, 12, 2657} \[ \frac{45 \sin (c+d x)}{512 d (5-3 \cos (c+d x))}+\frac{3 \sin (c+d x)}{32 d (5-3 \cos (c+d x))^2}+\frac{59 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{1024 d}+\frac{59 x}{2048} \]

Antiderivative was successfully verified.

[In]

Int[(5 - 3*Cos[c + d*x])^(-3),x]

[Out]

(59*x)/2048 + (59*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(1024*d) + (3*Sin[c + d*x])/(32*d*(5 - 3*Cos[c + d*
x])^2) + (45*Sin[c + d*x])/(512*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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 12

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

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 \cos (c+d x))^3} \, dx &=\frac{3 \sin (c+d x)}{32 d (5-3 \cos (c+d x))^2}-\frac{1}{32} \int \frac{-10-3 \cos (c+d x)}{(5-3 \cos (c+d x))^2} \, dx\\ &=\frac{3 \sin (c+d x)}{32 d (5-3 \cos (c+d x))^2}+\frac{45 \sin (c+d x)}{512 d (5-3 \cos (c+d x))}+\frac{1}{512} \int \frac{59}{5-3 \cos (c+d x)} \, dx\\ &=\frac{3 \sin (c+d x)}{32 d (5-3 \cos (c+d x))^2}+\frac{45 \sin (c+d x)}{512 d (5-3 \cos (c+d x))}+\frac{59}{512} \int \frac{1}{5-3 \cos (c+d x)} \, dx\\ &=\frac{59 x}{2048}+\frac{59 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{1024 d}+\frac{3 \sin (c+d x)}{32 d (5-3 \cos (c+d x))^2}+\frac{45 \sin (c+d x)}{512 d (5-3 \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.137174, size = 65, normalized size = 0.78 \[ \frac{546 \sin (c+d x)-135 \sin (2 (c+d x))+59 (5-3 \cos (c+d x))^2 \tan ^{-1}\left (2 \tan \left (\frac{1}{2} (c+d x)\right )\right )}{1024 d (5-3 \cos (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(59*ArcTan[2*Tan[(c + d*x)/2]]*(5 - 3*Cos[c + d*x])^2 + 546*Sin[c + d*x] - 135*Sin[2*(c + d*x)])/(1024*d*(5 -
3*Cos[c + d*x])^2)

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Maple [A]  time = 0.038, size = 83, normalized size = 1. \begin{align*}{\frac{51}{128\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{69}{512\,d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-2}}+{\frac{59}{1024\,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))^3,x)

[Out]

51/128/d/(4*tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1/2*c)^3+69/512/d/(4*tan(1/2*d*x+1/2*c)^2+1)^2*tan(1/2*d*x+1
/2*c)+59/1024/d*arctan(2*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.98967, size = 151, normalized size = 1.82 \begin{align*} \frac{\frac{6 \,{\left (\frac{23 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{68 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac{8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{16 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1} + 59 \, \arctan \left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{1024 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*(6*(23*sin(d*x + c)/(cos(d*x + c) + 1) + 68*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(8*sin(d*x + c)^2/(cos
(d*x + c) + 1)^2 + 16*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1) + 59*arctan(2*sin(d*x + c)/(cos(d*x + c) + 1)))
/d

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Fricas [A]  time = 1.64308, size = 258, normalized size = 3.11 \begin{align*} -\frac{59 \,{\left (9 \, \cos \left (d x + c\right )^{2} - 30 \, \cos \left (d x + c\right ) + 25\right )} \arctan \left (\frac{5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right ) + 12 \,{\left (45 \, \cos \left (d x + c\right ) - 91\right )} \sin \left (d x + c\right )}{2048 \,{\left (9 \, d \cos \left (d x + c\right )^{2} - 30 \, d \cos \left (d x + c\right ) + 25 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2048*(59*(9*cos(d*x + c)^2 - 30*cos(d*x + c) + 25)*arctan(1/4*(5*cos(d*x + c) - 3)/sin(d*x + c)) + 12*(45*c
os(d*x + c) - 91)*sin(d*x + c))/(9*d*cos(d*x + c)^2 - 30*d*cos(d*x + c) + 25*d)

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Sympy [A]  time = 5.79532, size = 364, normalized size = 4.39 \begin{align*} \begin{cases} \frac{x}{\left (5 - 3 \cosh{\left (2 \operatorname{atanh}{\left (\frac{1}{2} \right )} \right )}\right )^{3}} & \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 (5 - 3 \cos{\left (c \right )}\right )^{3}} & \text{for}\: d = 0 \\\frac{944 \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 ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{16384 d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1024 d} + \frac{472 \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 )}}{16384 d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1024 d} + \frac{59 \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 )}{16384 d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1024 d} + \frac{408 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{16384 d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1024 d} + \frac{138 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{16384 d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 8192 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 1024 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))**3,x)

[Out]

Piecewise((x/(5 - 3*cosh(2*atanh(1/2)))**3, Eq(c, -d*x - 2*I*atanh(1/2)) | Eq(c, -d*x + 2*I*atanh(1/2))), (x/(
5 - 3*cos(c))**3, Eq(d, 0)), (944*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x
/2)**4/(16384*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 1024*d) + 472*(atan(2*tan(c/2 + d*x/2)) + p
i*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**2/(16384*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**
2 + 1024*d) + 59*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(16384*d*tan(c/2 + d*x/2)**4 +
8192*d*tan(c/2 + d*x/2)**2 + 1024*d) + 408*tan(c/2 + d*x/2)**3/(16384*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2
+ d*x/2)**2 + 1024*d) + 138*tan(c/2 + d*x/2)/(16384*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 1024*
d), True))

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Giac [A]  time = 1.12363, size = 104, normalized size = 1.25 \begin{align*} \frac{59 \, d x + 59 \, c + \frac{12 \,{\left (68 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 23 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} - 118 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{2048 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(59*d*x + 59*c + 12*(68*tan(1/2*d*x + 1/2*c)^3 + 23*tan(1/2*d*x + 1/2*c))/(4*tan(1/2*d*x + 1/2*c)^2 + 1
)^2 - 118*arctan(sin(d*x + c)/(cos(d*x + c) - 3)))/d