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

Optimal. Leaf size=81 \[ \frac {45 \sin (c+d x)}{512 d (3 \cos (c+d x)+5)}+\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}+\frac {59 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{1024 d}-\frac {59 x}{2048} \]

[Out]

-59/2048*x+59/1024*arctan(sin(d*x+c)/(3+cos(d*x+c)))/d+3/32*sin(d*x+c)/d/(5+3*cos(d*x+c))^2+45/512*sin(d*x+c)/
d/(5+3*cos(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 81, normalized size of antiderivative = 1.00, 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, 2658} \[ \frac {45 \sin (c+d x)}{512 d (3 \cos (c+d x)+5)}+\frac {3 \sin (c+d x)}{32 d (3 \cos (c+d x)+5)^2}+\frac {59 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{1024 d}-\frac {59 x}{2048} \]

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 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]

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]

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*}

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Mathematica [A]  time = 0.10, size = 65, normalized size = 0.80 \[ \frac {546 \sin (c+d x)+135 \sin (2 (c+d x))+59 (3 \cos (c+d x)+5)^2 \tan ^{-1}\left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right )}{1024 d (3 \cos (c+d x)+5)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(59*ArcTan[2*Cot[(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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fricas [A]  time = 0.93, size = 90, normalized size = 1.11 \[ \frac {59 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 25\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) + 12 \, {\left (45 \, \cos \left (d x + c\right ) + 91\right )} \sin \left (d x + c\right )}{2048 \, {\left (9 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 25 \, d\right )}} \]

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*co
s(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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giac [A]  time = 0.42, size = 75, normalized size = 0.93 \[ -\frac {59 \, d x + 59 \, c - \frac {12 \, {\left (23 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 68 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4\right )}^{2}} - 118 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{2048 \, d} \]

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

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maple [A]  time = 0.04, size = 79, normalized size = 0.98 \[ \frac {69 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{512 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )^{2}}+\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )^{2}}-\frac {59 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1024 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.26, size = 111, normalized size = 1.37 \[ \frac {\frac {6 \, {\left (\frac {68 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {23 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 16} - 59 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{1024 \, d} \]

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

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mupad [B]  time = 0.36, size = 95, normalized size = 1.17 \[ \frac {59\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{1024\,d}-\frac {59\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{1024\,d}+\frac {\frac {69\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{512}+\frac {51\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(3*cos(c + d*x) + 5)^3,x)

[Out]

(59*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(1024*d) - (59*atan(tan(c/2 + (d*x)/2)/2))/(1024*d) + ((51*tan(c/2 +
(d*x)/2))/128 + (69*tan(c/2 + (d*x)/2)^3)/512)/(d*(8*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 16))

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sympy [A]  time = 3.05, size = 362, normalized size = 4.47 \[ \begin {cases} \frac {x}{\left (-5 - 3 \cosh {\left (2 \operatorname {atanh}{\relax (2 )} \right )}\right )^{3}} & \text {for}\: c = - d x - 2 i \operatorname {atanh}{\relax (2 )} \vee c = - d x + 2 i \operatorname {atanh}{\relax (2 )} \\\frac {x}{\left (- 3 \cos {\relax (c )} - 5\right )^{3}} & \text {for}\: d = 0 \\- \frac {59 \left (\operatorname {atan}{\left (\frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \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 )}}{1024 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 )} + 16384 d} - \frac {472 \left (\operatorname {atan}{\left (\frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \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 )}}{1024 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 )} + 16384 d} - \frac {944 \left (\operatorname {atan}{\left (\frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{1024 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 )} + 16384 d} + \frac {138 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 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 )} + 16384 d} + \frac {408 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 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 )} + 16384 d} & \text {otherwise} \end {cases} \]

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(2)))**3, Eq(c, -d*x - 2*I*atanh(2)) | Eq(c, -d*x + 2*I*atanh(2))), (x/(-3*co
s(c) - 5)**3, Eq(d, 0)), (-59*(atan(tan(c/2 + d*x/2)/2) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)*
*4/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 16384*d) - 472*(atan(tan(c/2 + d*x/2)/2) + pi*fl
oor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**2/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 1
6384*d) - 944*(atan(tan(c/2 + d*x/2)/2) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(1024*d*tan(c/2 + d*x/2)**4 + 819
2*d*tan(c/2 + d*x/2)**2 + 16384*d) + 138*tan(c/2 + d*x/2)**3/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*
x/2)**2 + 16384*d) + 408*tan(c/2 + d*x/2)/(1024*d*tan(c/2 + d*x/2)**4 + 8192*d*tan(c/2 + d*x/2)**2 + 16384*d),
True))

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