\(\int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\) [232]

Optimal result

Integrand size = 29, antiderivative size = 243 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {1}{8} \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) x+\frac {\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \sin (c+d x)}{30 b d}+\frac {\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{120 d}+\frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {B (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d} \] Output:

1/8*(12*A*a^2*b+3*A*b^3+4*B*a^3+9*B*a*b^2)*x+1/30*(15*A*a^3*b+60*A*a*b^3-3 
*B*a^4+52*B*a^2*b^2+16*B*b^4)*sin(d*x+c)/b/d+1/120*(30*A*a^2*b+45*A*b^3-6* 
B*a^3+71*B*a*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/60*(15*A*a*b-3*B*a^2+16*B*b^2) 
*(a+b*cos(d*x+c))^2*sin(d*x+c)/b/d+1/20*(5*A*b-B*a)*(a+b*cos(d*x+c))^3*sin 
(d*x+c)/b/d+1/5*B*(a+b*cos(d*x+c))^4*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 3.25 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.72 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {60 \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) (c+d x)+60 \left (8 a^3 A+18 a A b^2+18 a^2 b B+5 b^3 B\right ) \sin (c+d x)+120 \left (3 a^2 A b+A b^3+a^3 B+3 a b^2 B\right ) \sin (2 (c+d x))+10 b \left (12 a A b+12 a^2 B+5 b^2 B\right ) \sin (3 (c+d x))+15 b^2 (A b+3 a B) \sin (4 (c+d x))+6 b^3 B \sin (5 (c+d x))}{480 d} \] Input:

Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]),x]
 

Output:

(60*(12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*(c + d*x) + 60*(8*a^3*A + 
 18*a*A*b^2 + 18*a^2*b*B + 5*b^3*B)*Sin[c + d*x] + 120*(3*a^2*A*b + A*b^3 
+ a^3*B + 3*a*b^2*B)*Sin[2*(c + d*x)] + 10*b*(12*a*A*b + 12*a^2*B + 5*b^2* 
B)*Sin[3*(c + d*x)] + 15*b^2*(A*b + 3*a*B)*Sin[4*(c + d*x)] + 6*b^3*B*Sin[ 
5*(c + d*x)])/(480*d)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3447, 3042, 3502, 3042, 3232, 3042, 3232, 3042, 3213}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3447

\(\displaystyle \int (a+b \cos (c+d x))^3 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\int (a+b \cos (c+d x))^3 (4 b B+(5 A b-a B) \cos (c+d x))dx}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (4 b B+(5 A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {1}{4} \int (a+b \cos (c+d x))^2 \left (b (15 A b+13 a B)+\left (-3 B a^2+15 A b a+16 b^2 B\right ) \cos (c+d x)\right )dx+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b (15 A b+13 a B)+\left (-3 B a^2+15 A b a+16 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (a+b \cos (c+d x)) \left (b \left (33 B a^2+75 A b a+32 b^2 B\right )+\left (-6 B a^3+30 A b a^2+71 b^2 B a+45 A b^3\right ) \cos (c+d x)\right )dx+\frac {\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b \left (33 B a^2+75 A b a+32 b^2 B\right )+\left (-6 B a^3+30 A b a^2+71 b^2 B a+45 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {\frac {1}{4} \left (\frac {\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {b \left (-6 a^3 B+30 a^2 A b+71 a b^2 B+45 A b^3\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {15}{2} b x \left (4 a^3 B+12 a^2 A b+9 a b^2 B+3 A b^3\right )+\frac {2 \left (-3 a^4 B+15 a^3 A b+52 a^2 b^2 B+60 a A b^3+16 b^4 B\right ) \sin (c+d x)}{d}\right )\right )+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{5 b}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}\)

Input:

Int[Cos[c + d*x]*(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x]),x]
 

Output:

(B*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*b*d) + (((5*A*b - a*B)*(a + b*C 
os[c + d*x])^3*Sin[c + d*x])/(4*d) + (((15*a*A*b - 3*a^2*B + 16*b^2*B)*(a 
+ b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + ((15*b*(12*a^2*A*b + 3*A*b^3 + 4 
*a^3*B + 9*a*b^2*B)*x)/2 + (2*(15*a^3*A*b + 60*a*A*b^3 - 3*a^4*B + 52*a^2* 
b^2*B + 16*b^4*B)*Sin[c + d*x])/d + (b*(30*a^2*A*b + 45*A*b^3 - 6*a^3*B + 
71*a*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3)/4)/(5*b)
 

Defintions of rubi rules used

rule 3042

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

rule 3213

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co 
s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free 
Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
 

rule 3232

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

rule 3447

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

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 42.58 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.72

method	result	size
parts	\(\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (3 A \,a^{2} b +a^{3} B \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{3} A \sin \left (d x +c \right )}{d}+\frac {b^{3} B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}\)	\(176\)
parallelrisch	\(\frac {120 \left (3 A \,a^{2} b +A \,b^{3}+a^{3} B +3 B a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+10 \left (12 A a \,b^{2}+12 B \,a^{2} b +5 b^{3} B \right ) \sin \left (3 d x +3 c \right )+15 \left (A \,b^{3}+3 B a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+6 b^{3} B \sin \left (5 d x +5 c \right )+60 \left (8 a^{3} A +18 A a \,b^{2}+18 B \,a^{2} b +5 b^{3} B \right ) \sin \left (d x +c \right )+720 x d \left (A \,a^{2} b +\frac {1}{4} A \,b^{3}+\frac {1}{3} a^{3} B +\frac {3}{4} B a \,b^{2}\right )}{480 d}\)	\(179\)
derivativedivides	\(\frac {a^{3} A \sin \left (d x +c \right )+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )+A a \,b^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{3} B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(227\)
default	\(\frac {a^{3} A \sin \left (d x +c \right )+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )+A a \,b^{2} \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{3} B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(227\)
risch	\(\frac {3 x A \,a^{2} b}{2}+\frac {3 x A \,b^{3}}{8}+\frac {a^{3} B x}{2}+\frac {9 x B a \,b^{2}}{8}+\frac {a^{3} A \sin \left (d x +c \right )}{d}+\frac {9 \sin \left (d x +c \right ) A a \,b^{2}}{4 d}+\frac {9 \sin \left (d x +c \right ) B \,a^{2} b}{4 d}+\frac {5 \sin \left (d x +c \right ) b^{3} B}{8 d}+\frac {b^{3} B \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) A \,b^{3}}{32 d}+\frac {3 \sin \left (4 d x +4 c \right ) B a \,b^{2}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A a \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2} b}{4 d}+\frac {5 \sin \left (3 d x +3 c \right ) b^{3} B}{48 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} B}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B a \,b^{2}}{4 d}\)	\(278\)
norman	\(\frac {\left (\frac {3}{2} A \,a^{2} b +\frac {3}{8} A \,b^{3}+\frac {1}{2} a^{3} B +\frac {9}{8} B a \,b^{2}\right ) x +\left (15 A \,a^{2} b +\frac {15}{4} A \,b^{3}+5 a^{3} B +\frac {45}{4} B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (15 A \,a^{2} b +\frac {15}{4} A \,b^{3}+5 a^{3} B +\frac {45}{4} B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3}{2} A \,a^{2} b +\frac {3}{8} A \,b^{3}+\frac {1}{2} a^{3} B +\frac {9}{8} B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {15}{2} A \,a^{2} b +\frac {15}{8} A \,b^{3}+\frac {5}{2} a^{3} B +\frac {45}{8} B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {15}{2} A \,a^{2} b +\frac {15}{8} A \,b^{3}+\frac {5}{2} a^{3} B +\frac {45}{8} B a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {4 \left (45 a^{3} A +75 A a \,b^{2}+75 B \,a^{2} b +29 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {\left (8 a^{3} A -12 A \,a^{2} b +24 A a \,b^{2}-5 A \,b^{3}-4 a^{3} B +24 B \,a^{2} b -15 B a \,b^{2}+8 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {\left (8 a^{3} A +12 A \,a^{2} b +24 A a \,b^{2}+5 A \,b^{3}+4 a^{3} B +24 B \,a^{2} b +15 B a \,b^{2}+8 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (48 a^{3} A -36 A \,a^{2} b +96 A a \,b^{2}-3 A \,b^{3}-12 a^{3} B +96 B \,a^{2} b -9 B a \,b^{2}+16 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (48 a^{3} A +36 A \,a^{2} b +96 A a \,b^{2}+3 A \,b^{3}+12 a^{3} B +96 B \,a^{2} b +9 B a \,b^{2}+16 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(564\)
orering	\(\text {Expression too large to display}\)	\(4041\)

Input:

int(cos(d*x+c)*(a+cos(d*x+c)*b)^3*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

(A*b^3+3*B*a*b^2)/d*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+ 
3/8*c)+1/3*(3*A*a*b^2+3*B*a^2*b)/d*(cos(d*x+c)^2+2)*sin(d*x+c)+(3*A*a^2*b+ 
B*a^3)/d*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^3*A*sin(d*x+c)/d+1/5* 
b^3*B/d*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.72 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} d x + {\left (24 \, B b^{3} \cos \left (d x + c\right )^{4} + 120 \, A a^{3} + 240 \, B a^{2} b + 240 \, A a b^{2} + 64 \, B b^{3} + 30 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, B a^{2} b + 15 \, A a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="fri 
cas")
 

Output:

1/120*(15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*d*x + (24*B*b^3*cos 
(d*x + c)^4 + 120*A*a^3 + 240*B*a^2*b + 240*A*a*b^2 + 64*B*b^3 + 30*(3*B*a 
*b^2 + A*b^3)*cos(d*x + c)^3 + 8*(15*B*a^2*b + 15*A*a*b^2 + 4*B*b^3)*cos(d 
*x + c)^2 + 15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c))* 
sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 551 vs. \(2 (241) = 482\).

Time = 0.31 (sec) , antiderivative size = 551, normalized size of antiderivative = 2.27 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a^{3} \sin {\left (c + d x \right )}}{d} + \frac {3 A a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 A a b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A b^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 A b^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 B a^{2} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 B a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 B a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 B a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 B b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {B b^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right )^{3} \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)*(a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)),x)
 

Output:

Piecewise((A*a**3*sin(c + d*x)/d + 3*A*a**2*b*x*sin(c + d*x)**2/2 + 3*A*a* 
*2*b*x*cos(c + d*x)**2/2 + 3*A*a**2*b*sin(c + d*x)*cos(c + d*x)/(2*d) + 2* 
A*a*b**2*sin(c + d*x)**3/d + 3*A*a*b**2*sin(c + d*x)*cos(c + d*x)**2/d + 3 
*A*b**3*x*sin(c + d*x)**4/8 + 3*A*b**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 
 + 3*A*b**3*x*cos(c + d*x)**4/8 + 3*A*b**3*sin(c + d*x)**3*cos(c + d*x)/(8 
*d) + 5*A*b**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + B*a**3*x*sin(c + d*x)* 
*2/2 + B*a**3*x*cos(c + d*x)**2/2 + B*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) 
 + 2*B*a**2*b*sin(c + d*x)**3/d + 3*B*a**2*b*sin(c + d*x)*cos(c + d*x)**2/ 
d + 9*B*a*b**2*x*sin(c + d*x)**4/8 + 9*B*a*b**2*x*sin(c + d*x)**2*cos(c + 
d*x)**2/4 + 9*B*a*b**2*x*cos(c + d*x)**4/8 + 9*B*a*b**2*sin(c + d*x)**3*co 
s(c + d*x)/(8*d) + 15*B*a*b**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 8*B*b* 
*3*sin(c + d*x)**5/(15*d) + 4*B*b**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) 
 + B*b**3*sin(c + d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(A + B*cos(c))*(a 
+ b*cos(c))**3*cos(c), True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.89 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{3} + 480 \, A a^{3} \sin \left (d x + c\right )}{480 \, d} \] Input:

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="max 
ima")
 

Output:

1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 360*(2*d*x + 2*c + sin 
(2*d*x + 2*c))*A*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2*b - 4 
80*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b^2 + 45*(12*d*x + 12*c + sin(4*d 
*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a*b^2 + 15*(12*d*x + 12*c + sin(4*d*x + 
4*c) + 8*sin(2*d*x + 2*c))*A*b^3 + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^ 
3 + 15*sin(d*x + c))*B*b^3 + 480*A*a^3*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {B b^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} x + \frac {{\left (3 \, B a b^{2} + A b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, B a^{2} b + 12 \, A a b^{2} + 5 \, B b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{3} + 3 \, A a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 5 \, B b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="gia 
c")
 

Output:

1/80*B*b^3*sin(5*d*x + 5*c)/d + 1/8*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3* 
A*b^3)*x + 1/32*(3*B*a*b^2 + A*b^3)*sin(4*d*x + 4*c)/d + 1/48*(12*B*a^2*b 
+ 12*A*a*b^2 + 5*B*b^3)*sin(3*d*x + 3*c)/d + 1/4*(B*a^3 + 3*A*a^2*b + 3*B* 
a*b^2 + A*b^3)*sin(2*d*x + 2*c)/d + 1/8*(8*A*a^3 + 18*B*a^2*b + 18*A*a*b^2 
 + 5*B*b^3)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 42.56 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.14 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {3\,A\,b^3\,x}{8}+\frac {B\,a^3\,x}{2}+\frac {3\,A\,a^2\,b\,x}{2}+\frac {9\,B\,a\,b^2\,x}{8}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,b^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {5\,B\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \] Input:

int(cos(c + d*x)*(A + B*cos(c + d*x))*(a + b*cos(c + d*x))^3,x)
 

Output:

(3*A*b^3*x)/8 + (B*a^3*x)/2 + (3*A*a^2*b*x)/2 + (9*B*a*b^2*x)/8 + (A*a^3*s 
in(c + d*x))/d + (5*B*b^3*sin(c + d*x))/(8*d) + (A*b^3*sin(2*c + 2*d*x))/( 
4*d) + (B*a^3*sin(2*c + 2*d*x))/(4*d) + (A*b^3*sin(4*c + 4*d*x))/(32*d) + 
(5*B*b^3*sin(3*c + 3*d*x))/(48*d) + (B*b^3*sin(5*c + 5*d*x))/(80*d) + (3*A 
*a^2*b*sin(2*c + 2*d*x))/(4*d) + (A*a*b^2*sin(3*c + 3*d*x))/(4*d) + (3*B*a 
*b^2*sin(2*c + 2*d*x))/(4*d) + (B*a^2*b*sin(3*c + 3*d*x))/(4*d) + (3*B*a*b 
^2*sin(4*c + 4*d*x))/(32*d) + (9*A*a*b^2*sin(c + d*x))/(4*d) + (9*B*a^2*b* 
sin(c + d*x))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.64 \[ \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{3}+60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}+6 \sin \left (d x +c \right )^{5} b^{4}-60 \sin \left (d x +c \right )^{3} a^{2} b^{2}-20 \sin \left (d x +c \right )^{3} b^{4}+30 \sin \left (d x +c \right ) a^{4}+180 \sin \left (d x +c \right ) a^{2} b^{2}+30 \sin \left (d x +c \right ) b^{4}+60 a^{3} b d x +45 a \,b^{3} d x}{30 d} \] Input:

int(cos(d*x+c)*(a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x)
 

Output:

( - 30*cos(c + d*x)*sin(c + d*x)**3*a*b**3 + 60*cos(c + d*x)*sin(c + d*x)* 
a**3*b + 75*cos(c + d*x)*sin(c + d*x)*a*b**3 + 6*sin(c + d*x)**5*b**4 - 60 
*sin(c + d*x)**3*a**2*b**2 - 20*sin(c + d*x)**3*b**4 + 30*sin(c + d*x)*a** 
4 + 180*sin(c + d*x)*a**2*b**2 + 30*sin(c + d*x)*b**4 + 60*a**3*b*d*x + 45 
*a*b**3*d*x)/(30*d)