3.10.87 \(\int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [987]

3.10.87.1 Optimal result
3.10.87.2 Mathematica [A] (verified)
3.10.87.3 Rubi [A] (verified)
3.10.87.4 Maple [A] (verified)
3.10.87.5 Fricas [A] (verification not implemented)
3.10.87.6 Sympy [B] (verification not implemented)
3.10.87.7 Maxima [A] (verification not implemented)
3.10.87.8 Giac [B] (verification not implemented)
3.10.87.9 Mupad [B] (verification not implemented)

3.10.87.1 Optimal result

Integrand size = 31, antiderivative size = 105 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 (A-B) (a+a \sin (c+d x))^6}{3 a^3 d}-\frac {4 (A-2 B) (a+a \sin (c+d x))^7}{7 a^4 d}+\frac {(A-5 B) (a+a \sin (c+d x))^8}{8 a^5 d}+\frac {B (a+a \sin (c+d x))^9}{9 a^6 d} \]

output
2/3*(A-B)*(a+a*sin(d*x+c))^6/a^3/d-4/7*(A-2*B)*(a+a*sin(d*x+c))^7/a^4/d+1/ 
8*(A-5*B)*(a+a*sin(d*x+c))^8/a^5/d+1/9*B*(a+a*sin(d*x+c))^9/a^6/d
 
3.10.87.2 Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (1+\sin (c+d x))^6 \left (111 A-19 B-6 (27 A-19 B) \sin (c+d x)+21 (3 A-7 B) \sin ^2(c+d x)+56 B \sin ^3(c+d x)\right )}{504 d} \]

input
Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 
output
(a^3*(1 + Sin[c + d*x])^6*(111*A - 19*B - 6*(27*A - 19*B)*Sin[c + d*x] + 2 
1*(3*A - 7*B)*Sin[c + d*x]^2 + 56*B*Sin[c + d*x]^3))/(504*d)
 
3.10.87.3 Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {\int \left (B (\sin (c+d x) a+a)^8+a (A-5 B) (\sin (c+d x) a+a)^7-4 a^2 (A-2 B) (\sin (c+d x) a+a)^6+4 a^3 (A-B) (\sin (c+d x) a+a)^5\right )d(a \sin (c+d x))}{a^6 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {2}{3} a^3 (A-B) (a \sin (c+d x)+a)^6-\frac {4}{7} a^2 (A-2 B) (a \sin (c+d x)+a)^7+\frac {1}{8} a (A-5 B) (a \sin (c+d x)+a)^8+\frac {1}{9} B (a \sin (c+d x)+a)^9}{a^6 d}\)

input
Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 
output
((2*a^3*(A - B)*(a + a*Sin[c + d*x])^6)/3 - (4*a^2*(A - 2*B)*(a + a*Sin[c 
+ d*x])^7)/7 + (a*(A - 5*B)*(a + a*Sin[c + d*x])^8)/8 + (B*(a + a*Sin[c + 
d*x])^9)/9)/(a^6*d)
 

3.10.87.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 86
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3315
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, 
 x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege 
rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
 
3.10.87.4 Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.29

method result size
derivativedivides \(\frac {a^{3} \left (\frac {\left (\sin ^{9}\left (d x +c \right )\right ) B}{9}+\frac {\left (A +3 B \right ) \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (3 A +B \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (A -5 B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (-5 A -5 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (B -5 A \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A +3 B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )\right )}{d}\) \(135\)
default \(\frac {a^{3} \left (\frac {\left (\sin ^{9}\left (d x +c \right )\right ) B}{9}+\frac {\left (A +3 B \right ) \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (3 A +B \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (A -5 B \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (-5 A -5 B \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (B -5 A \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (A +3 B \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (3 A +B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}+A \sin \left (d x +c \right )\right )}{d}\) \(135\)
parallelrisch \(\frac {a^{3} \left (8 \left (-33 A -19 B \right ) \cos \left (2 d x +2 c \right )+4 \left (-25 A -11 B \right ) \cos \left (4 d x +4 c \right )+\frac {8 \left (B -5 A \right ) \cos \left (6 d x +6 c \right )}{3}+\left (A +3 B \right ) \cos \left (8 d x +8 c \right )+\frac {16 \left (17 A -4 B \right ) \sin \left (3 d x +3 c \right )}{3}+16 \left (-A -2 B \right ) \sin \left (5 d x +5 c \right )+\frac {4 \left (-12 A -11 B \right ) \sin \left (7 d x +7 c \right )}{7}+\frac {4 B \sin \left (9 d x +9 c \right )}{9}+88 \left (10 A +3 B \right ) \sin \left (d x +c \right )+\frac {1129 A}{3}+\frac {571 B}{3}\right )}{1024 d}\) \(164\)
risch \(\frac {55 \sin \left (d x +c \right ) A \,a^{3}}{64 d}+\frac {33 a^{3} B \sin \left (d x +c \right )}{128 d}+\frac {\sin \left (9 d x +9 c \right ) B \,a^{3}}{2304 d}+\frac {a^{3} \cos \left (8 d x +8 c \right ) A}{1024 d}+\frac {3 a^{3} \cos \left (8 d x +8 c \right ) B}{1024 d}-\frac {3 \sin \left (7 d x +7 c \right ) A \,a^{3}}{448 d}-\frac {11 \sin \left (7 d x +7 c \right ) B \,a^{3}}{1792 d}-\frac {5 a^{3} \cos \left (6 d x +6 c \right ) A}{384 d}+\frac {a^{3} \cos \left (6 d x +6 c \right ) B}{384 d}-\frac {\sin \left (5 d x +5 c \right ) A \,a^{3}}{64 d}-\frac {\sin \left (5 d x +5 c \right ) B \,a^{3}}{32 d}-\frac {25 a^{3} \cos \left (4 d x +4 c \right ) A}{256 d}-\frac {11 a^{3} \cos \left (4 d x +4 c \right ) B}{256 d}+\frac {17 \sin \left (3 d x +3 c \right ) A \,a^{3}}{192 d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{3}}{48 d}-\frac {33 a^{3} \cos \left (2 d x +2 c \right ) A}{128 d}-\frac {19 a^{3} \cos \left (2 d x +2 c \right ) B}{128 d}\) \(302\)
norman \(\frac {\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (6 A \,a^{3}+2 B \,a^{3}\right ) \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{3}+9 B \,a^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (11 A \,a^{3}+9 B \,a^{3}\right ) \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (37 A \,a^{3}+23 B \,a^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (37 A \,a^{3}+23 B \,a^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (55 A \,a^{3}+13 B \,a^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (55 A \,a^{3}+13 B \,a^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 A \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{3} \left (\tan ^{17}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (5 A +2 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (5 A +2 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (7 A +3 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a^{3} \left (7 A +3 B \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {24 a^{3} \left (23 A +3 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 d}+\frac {24 a^{3} \left (23 A +3 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7 d}+\frac {44 a^{3} \left (159 A +88 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}\) \(468\)

input
int(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x,method=_RETURNVERBO 
SE)
 
output
a^3/d*(1/9*sin(d*x+c)^9*B+1/8*(A+3*B)*sin(d*x+c)^8+1/7*(3*A+B)*sin(d*x+c)^ 
7+1/6*(A-5*B)*sin(d*x+c)^6+1/5*(-5*A-5*B)*sin(d*x+c)^5+1/4*(B-5*A)*sin(d*x 
+c)^4+1/3*(A+3*B)*sin(d*x+c)^3+1/2*(3*A+B)*sin(d*x+c)^2+A*sin(d*x+c))
 
3.10.87.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.23 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {63 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} - 336 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{6} + 8 \, {\left (7 \, B a^{3} \cos \left (d x + c\right )^{8} - {\left (27 \, A + 37 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 6 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \, {\left (3 \, A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{504 \, d} \]

input
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 
output
1/504*(63*(A + 3*B)*a^3*cos(d*x + c)^8 - 336*(A + B)*a^3*cos(d*x + c)^6 + 
8*(7*B*a^3*cos(d*x + c)^8 - (27*A + 37*B)*a^3*cos(d*x + c)^6 + 6*(3*A + B) 
*a^3*cos(d*x + c)^4 + 8*(3*A + B)*a^3*cos(d*x + c)^2 + 16*(3*A + B)*a^3)*s 
in(d*x + c))/d
 
3.10.87.6 Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (99) = 198\).

Time = 0.95 (sec) , antiderivative size = 418, normalized size of antiderivative = 3.98 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {8 A a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {4 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {4 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{3} \cos ^{8}{\left (c + d x \right )}}{24 d} - \frac {A a^{3} \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {8 B a^{3} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{2 d} - \frac {B a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac {B a^{3} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

input
integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
 
output
Piecewise((8*A*a**3*sin(c + d*x)**7/(35*d) + 4*A*a**3*sin(c + d*x)**5*cos( 
c + d*x)**2/(5*d) + 8*A*a**3*sin(c + d*x)**5/(15*d) + A*a**3*sin(c + d*x)* 
*3*cos(c + d*x)**4/d + 4*A*a**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - A* 
a**3*sin(c + d*x)**2*cos(c + d*x)**6/(6*d) + A*a**3*sin(c + d*x)*cos(c + d 
*x)**4/d - A*a**3*cos(c + d*x)**8/(24*d) - A*a**3*cos(c + d*x)**6/(2*d) + 
8*B*a**3*sin(c + d*x)**9/(315*d) + 4*B*a**3*sin(c + d*x)**7*cos(c + d*x)** 
2/(35*d) + 8*B*a**3*sin(c + d*x)**7/(35*d) + B*a**3*sin(c + d*x)**5*cos(c 
+ d*x)**4/(5*d) + 4*B*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + B*a**3* 
sin(c + d*x)**3*cos(c + d*x)**4/d - B*a**3*sin(c + d*x)**2*cos(c + d*x)**6 
/(2*d) - B*a**3*cos(c + d*x)**8/(8*d) - B*a**3*cos(c + d*x)**6/(6*d), Ne(d 
, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c)**5, True))
 
3.10.87.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {56 \, B a^{3} \sin \left (d x + c\right )^{9} + 63 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{8} + 72 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{7} + 84 \, {\left (A - 5 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} - 504 \, {\left (A + B\right )} a^{3} \sin \left (d x + c\right )^{5} - 126 \, {\left (5 \, A - B\right )} a^{3} \sin \left (d x + c\right )^{4} + 168 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} + 252 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} + 504 \, A a^{3} \sin \left (d x + c\right )}{504 \, d} \]

input
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 
output
1/504*(56*B*a^3*sin(d*x + c)^9 + 63*(A + 3*B)*a^3*sin(d*x + c)^8 + 72*(3*A 
 + B)*a^3*sin(d*x + c)^7 + 84*(A - 5*B)*a^3*sin(d*x + c)^6 - 504*(A + B)*a 
^3*sin(d*x + c)^5 - 126*(5*A - B)*a^3*sin(d*x + c)^4 + 168*(A + 3*B)*a^3*s 
in(d*x + c)^3 + 252*(3*A + B)*a^3*sin(d*x + c)^2 + 504*A*a^3*sin(d*x + c)) 
/d
 
3.10.87.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (97) = 194\).

Time = 0.44 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.19 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B a^{3} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (5 \, A a^{3} - B a^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (25 \, A a^{3} + 11 \, B a^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {{\left (33 \, A a^{3} + 19 \, B a^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {{\left (12 \, A a^{3} + 11 \, B a^{3}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (A a^{3} + 2 \, B a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{64 \, d} + \frac {{\left (17 \, A a^{3} - 4 \, B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {11 \, {\left (10 \, A a^{3} + 3 \, B a^{3}\right )} \sin \left (d x + c\right )}{128 \, d} \]

input
integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 
output
1/2304*B*a^3*sin(9*d*x + 9*c)/d + 1/1024*(A*a^3 + 3*B*a^3)*cos(8*d*x + 8*c 
)/d - 1/384*(5*A*a^3 - B*a^3)*cos(6*d*x + 6*c)/d - 1/256*(25*A*a^3 + 11*B* 
a^3)*cos(4*d*x + 4*c)/d - 1/128*(33*A*a^3 + 19*B*a^3)*cos(2*d*x + 2*c)/d - 
 1/1792*(12*A*a^3 + 11*B*a^3)*sin(7*d*x + 7*c)/d - 1/64*(A*a^3 + 2*B*a^3)* 
sin(5*d*x + 5*c)/d + 1/192*(17*A*a^3 - 4*B*a^3)*sin(3*d*x + 3*c)/d + 11/12 
8*(10*A*a^3 + 3*B*a^3)*sin(d*x + c)/d
 
3.10.87.9 Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.49 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3\,\left (A+3\,B\right )}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^7\,\left (3\,A+B\right )}{7}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A-5\,B\right )}{6}+\frac {a^3\,{\sin \left (c+d\,x\right )}^8\,\left (A+3\,B\right )}{8}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (5\,A-B\right )}{4}+A\,a^3\,\sin \left (c+d\,x\right )-a^3\,{\sin \left (c+d\,x\right )}^5\,\left (A+B\right )}{d} \]

input
int(cos(c + d*x)^5*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)
 
output
((a^3*sin(c + d*x)^2*(3*A + B))/2 + (a^3*sin(c + d*x)^3*(A + 3*B))/3 + (a^ 
3*sin(c + d*x)^7*(3*A + B))/7 + (a^3*sin(c + d*x)^6*(A - 5*B))/6 + (a^3*si 
n(c + d*x)^8*(A + 3*B))/8 + (B*a^3*sin(c + d*x)^9)/9 - (a^3*sin(c + d*x)^4 
*(5*A - B))/4 + A*a^3*sin(c + d*x) - a^3*sin(c + d*x)^5*(A + B))/d