\(\int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [965]

Optimal result

Integrand size = 33, antiderivative size = 375 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{16} \left (32 a^3 b B+24 a b^3 B+8 a^4 (2 A+C)+12 a^2 b^2 (4 A+3 C)+b^4 (6 A+5 C)\right ) x+\frac {\left (24 a^4 b B+224 a^2 b^3 B+32 b^5 B-4 a^5 C+32 a b^4 (5 A+4 C)+a^3 b^2 (190 A+121 C)\right ) \sin (c+d x)}{60 b d}+\frac {\left (48 a^3 b B+232 a b^3 B-8 a^4 C+15 b^4 (6 A+5 C)+2 a^2 b^2 (130 A+89 C)\right ) \cos (c+d x) \sin (c+d x)}{240 d}+\frac {\left (24 a^2 b B+32 b^3 B-4 a^3 C+a b^2 (70 A+53 C)\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}+\frac {\left (5 b^2 (6 A+5 C)+4 a (6 b B-a C)\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}+\frac {(6 b B-a C) (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac {C (a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d} \] Output:

1/16*(32*B*a^3*b+24*B*a*b^3+8*a^4*(2*A+C)+12*a^2*b^2*(4*A+3*C)+b^4*(6*A+5* 
C))*x+1/60*(24*B*a^4*b+224*B*a^2*b^3+32*B*b^5-4*C*a^5+32*a*b^4*(5*A+4*C)+a 
^3*b^2*(190*A+121*C))*sin(d*x+c)/b/d+1/240*(48*B*a^3*b+232*B*a*b^3-8*a^4*C 
+15*b^4*(6*A+5*C)+2*a^2*b^2*(130*A+89*C))*cos(d*x+c)*sin(d*x+c)/d+1/120*(2 
4*B*a^2*b+32*B*b^3-4*a^3*C+a*b^2*(70*A+53*C))*(a+b*cos(d*x+c))^2*sin(d*x+c 
)/b/d+1/120*(5*b^2*(6*A+5*C)+4*a*(6*B*b-C*a))*(a+b*cos(d*x+c))^3*sin(d*x+c 
)/b/d+1/30*(6*B*b-C*a)*(a+b*cos(d*x+c))^4*sin(d*x+c)/b/d+1/6*C*(a+b*cos(d* 
x+c))^5*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 4.99 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {960 a^4 A c+2880 a^2 A b^2 c+360 A b^4 c+1920 a^3 b B c+1440 a b^3 B c+480 a^4 c C+2160 a^2 b^2 c C+300 b^4 c C+960 a^4 A d x+2880 a^2 A b^2 d x+360 A b^4 d x+1920 a^3 b B d x+1440 a b^3 B d x+480 a^4 C d x+2160 a^2 b^2 C d x+300 b^4 C d x+120 \left (8 a^4 B+36 a^2 b^2 B+5 b^4 B+8 a^3 b (4 A+3 C)+4 a b^3 (6 A+5 C)\right ) \sin (c+d x)+15 \left (64 a^3 b B+64 a b^3 B+16 a^4 C+96 a^2 b^2 (A+C)+b^4 (16 A+15 C)\right ) \sin (2 (c+d x))+320 a A b^3 \sin (3 (c+d x))+480 a^2 b^2 B \sin (3 (c+d x))+100 b^4 B \sin (3 (c+d x))+320 a^3 b C \sin (3 (c+d x))+400 a b^3 C \sin (3 (c+d x))+30 A b^4 \sin (4 (c+d x))+120 a b^3 B \sin (4 (c+d x))+180 a^2 b^2 C \sin (4 (c+d x))+45 b^4 C \sin (4 (c+d x))+12 b^4 B \sin (5 (c+d x))+48 a b^3 C \sin (5 (c+d x))+5 b^4 C \sin (6 (c+d x))}{960 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x 
]
 

Output:

(960*a^4*A*c + 2880*a^2*A*b^2*c + 360*A*b^4*c + 1920*a^3*b*B*c + 1440*a*b^ 
3*B*c + 480*a^4*c*C + 2160*a^2*b^2*c*C + 300*b^4*c*C + 960*a^4*A*d*x + 288 
0*a^2*A*b^2*d*x + 360*A*b^4*d*x + 1920*a^3*b*B*d*x + 1440*a*b^3*B*d*x + 48 
0*a^4*C*d*x + 2160*a^2*b^2*C*d*x + 300*b^4*C*d*x + 120*(8*a^4*B + 36*a^2*b 
^2*B + 5*b^4*B + 8*a^3*b*(4*A + 3*C) + 4*a*b^3*(6*A + 5*C))*Sin[c + d*x] + 
 15*(64*a^3*b*B + 64*a*b^3*B + 16*a^4*C + 96*a^2*b^2*(A + C) + b^4*(16*A + 
 15*C))*Sin[2*(c + d*x)] + 320*a*A*b^3*Sin[3*(c + d*x)] + 480*a^2*b^2*B*Si 
n[3*(c + d*x)] + 100*b^4*B*Sin[3*(c + d*x)] + 320*a^3*b*C*Sin[3*(c + d*x)] 
 + 400*a*b^3*C*Sin[3*(c + d*x)] + 30*A*b^4*Sin[4*(c + d*x)] + 120*a*b^3*B* 
Sin[4*(c + d*x)] + 180*a^2*b^2*C*Sin[4*(c + d*x)] + 45*b^4*C*Sin[4*(c + d* 
x)] + 12*b^4*B*Sin[5*(c + d*x)] + 48*a*b^3*C*Sin[5*(c + d*x)] + 5*b^4*C*Si 
n[6*(c + d*x)])/(960*d)
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3502, 3042, 3232, 3042, 3232, 27, 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.

\(\displaystyle \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\int (a+b \cos (c+d x))^4 (b (6 A+5 C)+(6 b B-a C) \cos (c+d x))dx}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (b (6 A+5 C)+(6 b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {1}{5} \int (a+b \cos (c+d x))^3 \left (3 b (10 a A+8 b B+7 a C)+\left (5 (6 A+5 C) b^2+4 a (6 b B-a C)\right ) \cos (c+d x)\right )dx+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (3 b (10 a A+8 b B+7 a C)+\left (5 (6 A+5 C) b^2+4 a (6 b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{4} \int 3 (a+b \cos (c+d x))^2 \left (b \left (8 (5 A+3 C) a^2+56 b B a+5 b^2 (6 A+5 C)\right )+\left (-4 C a^3+24 b B a^2+b^2 (70 A+53 C) a+32 b^3 B\right ) \cos (c+d x)\right )dx+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{4 d}\right )+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \int (a+b \cos (c+d x))^2 \left (b \left (8 (5 A+3 C) a^2+56 b B a+5 b^2 (6 A+5 C)\right )+\left (-4 C a^3+24 b B a^2+b^2 (70 A+53 C) a+32 b^3 B\right ) \cos (c+d x)\right )dx+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{4 d}\right )+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b \left (8 (5 A+3 C) a^2+56 b B a+5 b^2 (6 A+5 C)\right )+\left (-4 C a^3+24 b B a^2+b^2 (70 A+53 C) a+32 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{4 d}\right )+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \int (a+b \cos (c+d x)) \left (b \left (8 (15 A+8 C) a^3+216 b B a^2+b^2 (230 A+181 C) a+64 b^3 B\right )+\left (-8 C a^4+48 b B a^3+2 b^2 (130 A+89 C) a^2+232 b^3 B a+15 b^4 (6 A+5 C)\right ) \cos (c+d x)\right )dx+\frac {\sin (c+d x) \left (-4 a^3 C+24 a^2 b B+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{4 d}\right )+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {1}{3} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b \left (8 (15 A+8 C) a^3+216 b B a^2+b^2 (230 A+181 C) a+64 b^3 B\right )+\left (-8 C a^4+48 b B a^3+2 b^2 (130 A+89 C) a^2+232 b^3 B a+15 b^4 (6 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {\sin (c+d x) \left (-4 a^3 C+24 a^2 b B+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{4 d}\right )+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {\frac {1}{5} \left (\frac {3}{4} \left (\frac {\sin (c+d x) \left (-4 a^3 C+24 a^2 b B+a b^2 (70 A+53 C)+32 b^3 B\right ) (a+b \cos (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {b \sin (c+d x) \cos (c+d x) \left (-8 a^4 C+48 a^3 b B+2 a^2 b^2 (130 A+89 C)+232 a b^3 B+15 b^4 (6 A+5 C)\right )}{2 d}+\frac {15}{2} b x \left (8 a^4 (2 A+C)+32 a^3 b B+12 a^2 b^2 (4 A+3 C)+24 a b^3 B+b^4 (6 A+5 C)\right )+\frac {2 \sin (c+d x) \left (-4 a^5 C+24 a^4 b B+a^3 b^2 (190 A+121 C)+224 a^2 b^3 B+32 a b^4 (5 A+4 C)+32 b^5 B\right )}{d}\right )\right )+\frac {\sin (c+d x) \left (4 a (6 b B-a C)+5 b^2 (6 A+5 C)\right ) (a+b \cos (c+d x))^3}{4 d}\right )+\frac {(6 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}}{6 b}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}\)

Input:

Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
 

Output:

(C*(a + b*Cos[c + d*x])^5*Sin[c + d*x])/(6*b*d) + (((6*b*B - a*C)*(a + b*C 
os[c + d*x])^4*Sin[c + d*x])/(5*d) + (((5*b^2*(6*A + 5*C) + 4*a*(6*b*B - a 
*C))*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + (3*(((24*a^2*b*B + 32*b^ 
3*B - 4*a^3*C + a*b^2*(70*A + 53*C))*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/ 
(3*d) + ((15*b*(32*a^3*b*B + 24*a*b^3*B + 8*a^4*(2*A + C) + 12*a^2*b^2*(4* 
A + 3*C) + b^4*(6*A + 5*C))*x)/2 + (2*(24*a^4*b*B + 224*a^2*b^3*B + 32*b^5 
*B - 4*a^5*C + 32*a*b^4*(5*A + 4*C) + a^3*b^2*(190*A + 121*C))*Sin[c + d*x 
])/d + (b*(48*a^3*b*B + 232*a*b^3*B - 8*a^4*C + 15*b^4*(6*A + 5*C) + 2*a^2 
*b^2*(130*A + 89*C))*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3))/4)/5)/(6*b)
 

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 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 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 = 0.04 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.15

\[\frac {C \,b^{4} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {B \,b^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {4 C a \,b^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+A \,b^{4} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 B a \,b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 C \,a^{2} b^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a A \,b^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {4 a^{3} b C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+6 A \,a^{2} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{3} b \sin \left (d x +c \right )+B \,a^{4} \sin \left (d x +c \right )+A \,a^{4} \left (d x +c \right )}{d}\]

Input:

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

1/d*(C*b^4*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c) 
+5/16*d*x+5/16*c)+1/5*B*b^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c) 
+4/5*C*a*b^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+A*b^4*(1/4*(co 
s(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+4*B*a*b^3*(1/4*(cos(d 
*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+6*C*a^2*b^2*(1/4*(cos(d* 
x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+4/3*a*A*b^3*(2+cos(d*x+c) 
^2)*sin(d*x+c)+2*B*a^2*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+4/3*a^3*b*C*(2+cos( 
d*x+c)^2)*sin(d*x+c)+6*A*a^2*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c) 
+4*B*a^3*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^4*C*(1/2*cos(d*x+c) 
*sin(d*x+c)+1/2*d*x+1/2*c)+4*A*a^3*b*sin(d*x+c)+B*a^4*sin(d*x+c)+A*a^4*(d* 
x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 292, normalized size of antiderivative = 0.78 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (8 \, {\left (2 \, A + C\right )} a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} d x + {\left (40 \, C b^{4} \cos \left (d x + c\right )^{5} + 240 \, B a^{4} + 320 \, {\left (3 \, A + 2 \, C\right )} a^{3} b + 960 \, B a^{2} b^{2} + 128 \, {\left (5 \, A + 4 \, C\right )} a b^{3} + 128 \, B b^{4} + 48 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (36 \, C a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (10 \, C a^{3} b + 15 \, B a^{2} b^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} a b^{3} + 2 \, B b^{4}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (8 \, C a^{4} + 32 \, B a^{3} b + 12 \, {\left (4 \, A + 3 \, C\right )} a^{2} b^{2} + 24 \, B a b^{3} + {\left (6 \, A + 5 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= 
"fricas")
 

Output:

1/240*(15*(8*(2*A + C)*a^4 + 32*B*a^3*b + 12*(4*A + 3*C)*a^2*b^2 + 24*B*a* 
b^3 + (6*A + 5*C)*b^4)*d*x + (40*C*b^4*cos(d*x + c)^5 + 240*B*a^4 + 320*(3 
*A + 2*C)*a^3*b + 960*B*a^2*b^2 + 128*(5*A + 4*C)*a*b^3 + 128*B*b^4 + 48*( 
4*C*a*b^3 + B*b^4)*cos(d*x + c)^4 + 10*(36*C*a^2*b^2 + 24*B*a*b^3 + (6*A + 
 5*C)*b^4)*cos(d*x + c)^3 + 32*(10*C*a^3*b + 15*B*a^2*b^2 + 2*(5*A + 4*C)* 
a*b^3 + 2*B*b^4)*cos(d*x + c)^2 + 15*(8*C*a^4 + 32*B*a^3*b + 12*(4*A + 3*C 
)*a^2*b^2 + 24*B*a*b^3 + (6*A + 5*C)*b^4)*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. 1066 vs. \(2 (367) = 734\).

Time = 0.45 (sec) , antiderivative size = 1066, normalized size of antiderivative = 2.84 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate((a+b*cos(d*x+c))**4*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)
 

Output:

Piecewise((A*a**4*x + 4*A*a**3*b*sin(c + d*x)/d + 3*A*a**2*b**2*x*sin(c + 
d*x)**2 + 3*A*a**2*b**2*x*cos(c + d*x)**2 + 3*A*a**2*b**2*sin(c + d*x)*cos 
(c + d*x)/d + 8*A*a*b**3*sin(c + d*x)**3/(3*d) + 4*A*a*b**3*sin(c + d*x)*c 
os(c + d*x)**2/d + 3*A*b**4*x*sin(c + d*x)**4/8 + 3*A*b**4*x*sin(c + d*x)* 
*2*cos(c + d*x)**2/4 + 3*A*b**4*x*cos(c + d*x)**4/8 + 3*A*b**4*sin(c + d*x 
)**3*cos(c + d*x)/(8*d) + 5*A*b**4*sin(c + d*x)*cos(c + d*x)**3/(8*d) + B* 
a**4*sin(c + d*x)/d + 2*B*a**3*b*x*sin(c + d*x)**2 + 2*B*a**3*b*x*cos(c + 
d*x)**2 + 2*B*a**3*b*sin(c + d*x)*cos(c + d*x)/d + 4*B*a**2*b**2*sin(c + d 
*x)**3/d + 6*B*a**2*b**2*sin(c + d*x)*cos(c + d*x)**2/d + 3*B*a*b**3*x*sin 
(c + d*x)**4/2 + 3*B*a*b**3*x*sin(c + d*x)**2*cos(c + d*x)**2 + 3*B*a*b**3 
*x*cos(c + d*x)**4/2 + 3*B*a*b**3*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 5*B 
*a*b**3*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 8*B*b**4*sin(c + d*x)**5/(15* 
d) + 4*B*b**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + B*b**4*sin(c + d*x)* 
cos(c + d*x)**4/d + C*a**4*x*sin(c + d*x)**2/2 + C*a**4*x*cos(c + d*x)**2/ 
2 + C*a**4*sin(c + d*x)*cos(c + d*x)/(2*d) + 8*C*a**3*b*sin(c + d*x)**3/(3 
*d) + 4*C*a**3*b*sin(c + d*x)*cos(c + d*x)**2/d + 9*C*a**2*b**2*x*sin(c + 
d*x)**4/4 + 9*C*a**2*b**2*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 9*C*a**2*b 
**2*x*cos(c + d*x)**4/4 + 9*C*a**2*b**2*sin(c + d*x)**3*cos(c + d*x)/(4*d) 
 + 15*C*a**2*b**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 32*C*a*b**3*sin(c + 
 d*x)**5/(15*d) + 16*C*a*b**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 4...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {960 \, {\left (d x + c\right )} A a^{4} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} b + 1440 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} - 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b^{2} + 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b^{2} - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{3} + 120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} + 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a b^{3} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{4} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} + 960 \, B a^{4} \sin \left (d x + c\right ) + 3840 \, A a^{3} b \sin \left (d x + c\right )}{960 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= 
"maxima")
 

Output:

1/960*(960*(d*x + c)*A*a^4 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 + 
960*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3*b - 1280*(sin(d*x + c)^3 - 3*si 
n(d*x + c))*C*a^3*b + 1440*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2*b^2 - 19 
20*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2*b^2 + 180*(12*d*x + 12*c + sin( 
4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2*b^2 - 1280*(sin(d*x + c)^3 - 3*si 
n(d*x + c))*A*a*b^3 + 120*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x 
+ 2*c))*B*a*b^3 + 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + 
 c))*C*a*b^3 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))* 
A*b^4 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*b^4 
- 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2* 
d*x + 2*c))*C*b^4 + 960*B*a^4*sin(d*x + c) + 3840*A*a^3*b*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.87 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {C b^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {1}{16} \, {\left (16 \, A a^{4} + 8 \, C a^{4} + 32 \, B a^{3} b + 48 \, A a^{2} b^{2} + 36 \, C a^{2} b^{2} + 24 \, B a b^{3} + 6 \, A b^{4} + 5 \, C b^{4}\right )} x + \frac {{\left (4 \, C a b^{3} + B b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + 2 \, A b^{4} + 3 \, C b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, C a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 20 \, C a b^{3} + 5 \, B b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (16 \, C a^{4} + 64 \, B a^{3} b + 96 \, A a^{2} b^{2} + 96 \, C a^{2} b^{2} + 64 \, B a b^{3} + 16 \, A b^{4} + 15 \, C b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {{\left (8 \, B a^{4} + 32 \, A a^{3} b + 24 \, C a^{3} b + 36 \, B a^{2} b^{2} + 24 \, A a b^{3} + 20 \, C a b^{3} + 5 \, B b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:

integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= 
"giac")
 

Output:

1/192*C*b^4*sin(6*d*x + 6*c)/d + 1/16*(16*A*a^4 + 8*C*a^4 + 32*B*a^3*b + 4 
8*A*a^2*b^2 + 36*C*a^2*b^2 + 24*B*a*b^3 + 6*A*b^4 + 5*C*b^4)*x + 1/80*(4*C 
*a*b^3 + B*b^4)*sin(5*d*x + 5*c)/d + 1/64*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A* 
b^4 + 3*C*b^4)*sin(4*d*x + 4*c)/d + 1/48*(16*C*a^3*b + 24*B*a^2*b^2 + 16*A 
*a*b^3 + 20*C*a*b^3 + 5*B*b^4)*sin(3*d*x + 3*c)/d + 1/64*(16*C*a^4 + 64*B* 
a^3*b + 96*A*a^2*b^2 + 96*C*a^2*b^2 + 64*B*a*b^3 + 16*A*b^4 + 15*C*b^4)*si 
n(2*d*x + 2*c)/d + 1/8*(8*B*a^4 + 32*A*a^3*b + 24*C*a^3*b + 36*B*a^2*b^2 + 
 24*A*a*b^3 + 20*C*a*b^3 + 5*B*b^4)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 2.42 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.42 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=A\,a^4\,x+\frac {3\,A\,b^4\,x}{8}+\frac {C\,a^4\,x}{2}+\frac {5\,C\,b^4\,x}{16}+\frac {3\,B\,a\,b^3\,x}{2}+2\,B\,a^3\,b\,x+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,b^4\,\sin \left (c+d\,x\right )}{8\,d}+3\,A\,a^2\,b^2\,x+\frac {9\,C\,a^2\,b^2\,x}{4}+\frac {A\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,B\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,b^4\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {15\,C\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,C\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {C\,b^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {A\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {B\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{8\,d}+\frac {9\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {5\,C\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {C\,a\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {3\,A\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2\,d}+\frac {3\,C\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,C\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {3\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a^3\,b\,\sin \left (c+d\,x\right )}{d}+\frac {5\,C\,a\,b^3\,\sin \left (c+d\,x\right )}{2\,d}+\frac {3\,C\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \] Input:

int((a + b*cos(c + d*x))^4*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
 

Output:

A*a^4*x + (3*A*b^4*x)/8 + (C*a^4*x)/2 + (5*C*b^4*x)/16 + (3*B*a*b^3*x)/2 + 
 2*B*a^3*b*x + (B*a^4*sin(c + d*x))/d + (5*B*b^4*sin(c + d*x))/(8*d) + 3*A 
*a^2*b^2*x + (9*C*a^2*b^2*x)/4 + (A*b^4*sin(2*c + 2*d*x))/(4*d) + (A*b^4*s 
in(4*c + 4*d*x))/(32*d) + (C*a^4*sin(2*c + 2*d*x))/(4*d) + (5*B*b^4*sin(3* 
c + 3*d*x))/(48*d) + (B*b^4*sin(5*c + 5*d*x))/(80*d) + (15*C*b^4*sin(2*c + 
 2*d*x))/(64*d) + (3*C*b^4*sin(4*c + 4*d*x))/(64*d) + (C*b^4*sin(6*c + 6*d 
*x))/(192*d) + (A*a*b^3*sin(3*c + 3*d*x))/(3*d) + (B*a*b^3*sin(2*c + 2*d*x 
))/d + (B*a^3*b*sin(2*c + 2*d*x))/d + (B*a*b^3*sin(4*c + 4*d*x))/(8*d) + ( 
9*B*a^2*b^2*sin(c + d*x))/(2*d) + (5*C*a*b^3*sin(3*c + 3*d*x))/(12*d) + (C 
*a^3*b*sin(3*c + 3*d*x))/(3*d) + (C*a*b^3*sin(5*c + 5*d*x))/(20*d) + (3*A* 
a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (B*a^2*b^2*sin(3*c + 3*d*x))/(2*d) + (3* 
C*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (3*C*a^2*b^2*sin(4*c + 4*d*x))/(16*d) 
+ (3*A*a*b^3*sin(c + d*x))/d + (4*A*a^3*b*sin(c + d*x))/d + (5*C*a*b^3*sin 
(c + d*x))/(2*d) + (3*C*a^3*b*sin(c + d*x))/d
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.03 \[ \int (a+b \cos (c+d x))^4 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {-360 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b^{2} c +900 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{2} c +540 a^{2} b^{2} c d x +48 \sin \left (d x +c \right )^{5} b^{5}-160 \sin \left (d x +c \right )^{3} b^{5}+240 \sin \left (d x +c \right ) b^{5}+40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b^{4} c -300 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{4}-130 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{4} c +120 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} c +1200 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b^{2}+750 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{4}+165 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4} c +192 \sin \left (d x +c \right )^{5} a \,b^{3} c -320 \sin \left (d x +c \right )^{3} a^{3} b c -640 \sin \left (d x +c \right )^{3} a \,b^{3} c +960 \sin \left (d x +c \right ) a^{3} b c +960 \sin \left (d x +c \right ) a \,b^{3} c +120 a^{4} c d x +1200 a^{3} b^{2} d x +450 a \,b^{4} d x +75 b^{4} c d x -800 \sin \left (d x +c \right )^{3} a^{2} b^{3}+1200 \sin \left (d x +c \right ) a^{4} b +2400 \sin \left (d x +c \right ) a^{2} b^{3}+240 a^{5} d x}{240 d} \] Input:

int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
 

Output:

(40*cos(c + d*x)*sin(c + d*x)**5*b**4*c - 360*cos(c + d*x)*sin(c + d*x)**3 
*a**2*b**2*c - 300*cos(c + d*x)*sin(c + d*x)**3*a*b**4 - 130*cos(c + d*x)* 
sin(c + d*x)**3*b**4*c + 120*cos(c + d*x)*sin(c + d*x)*a**4*c + 1200*cos(c 
 + d*x)*sin(c + d*x)*a**3*b**2 + 900*cos(c + d*x)*sin(c + d*x)*a**2*b**2*c 
 + 750*cos(c + d*x)*sin(c + d*x)*a*b**4 + 165*cos(c + d*x)*sin(c + d*x)*b* 
*4*c + 192*sin(c + d*x)**5*a*b**3*c + 48*sin(c + d*x)**5*b**5 - 320*sin(c 
+ d*x)**3*a**3*b*c - 800*sin(c + d*x)**3*a**2*b**3 - 640*sin(c + d*x)**3*a 
*b**3*c - 160*sin(c + d*x)**3*b**5 + 1200*sin(c + d*x)*a**4*b + 960*sin(c 
+ d*x)*a**3*b*c + 2400*sin(c + d*x)*a**2*b**3 + 960*sin(c + d*x)*a*b**3*c 
+ 240*sin(c + d*x)*b**5 + 240*a**5*d*x + 120*a**4*c*d*x + 1200*a**3*b**2*d 
*x + 540*a**2*b**2*c*d*x + 450*a*b**4*d*x + 75*b**4*c*d*x)/(240*d)