\(\int \cos (c+d x) (a+b \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \, dx\) [549]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [B] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 31, antiderivative size = 345 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{4} a b \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) x+\frac {\left (35 a^4 (3 A+2 C)+84 a^2 b^2 (5 A+4 C)+8 b^4 (7 A+6 C)\right ) \sin (c+d x)}{105 d}+\frac {a b \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {\left (4 a^4 C+4 b^4 (7 A+6 C)+3 a^2 b^2 (63 A+50 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{105 d}+\frac {a b \left (126 A b^2+6 a^2 C+103 b^2 C\right ) \cos ^3(c+d x) \sin (c+d x)}{210 d}+\frac {\left (2 a^2 C+b^2 (7 A+6 C)\right ) \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{35 d}+\frac {2 a C \cos ^2(c+d x) (a+b \cos (c+d x))^3 \sin (c+d x)}{21 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^4 \sin (c+d x)}{7 d} \] Output:

1/4*a*b*(b^2*(6*A+5*C)+a^2*(8*A+6*C))*x+1/105*(35*a^4*(3*A+2*C)+84*a^2*b^2 
*(5*A+4*C)+8*b^4*(7*A+6*C))*sin(d*x+c)/d+1/4*a*b*(b^2*(6*A+5*C)+a^2*(8*A+6 
*C))*cos(d*x+c)*sin(d*x+c)/d+1/105*(4*a^4*C+4*b^4*(7*A+6*C)+3*a^2*b^2*(63* 
A+50*C))*cos(d*x+c)^2*sin(d*x+c)/d+1/210*a*b*(126*A*b^2+6*C*a^2+103*C*b^2) 
*cos(d*x+c)^3*sin(d*x+c)/d+1/35*(2*a^2*C+b^2*(7*A+6*C))*cos(d*x+c)^2*(a+b* 
cos(d*x+c))^2*sin(d*x+c)/d+2/21*a*C*cos(d*x+c)^2*(a+b*cos(d*x+c))^3*sin(d* 
x+c)/d+1/7*C*cos(d*x+c)^2*(a+b*cos(d*x+c))^4*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.02 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {13440 a^3 A b c+10080 a A b^3 c+10080 a^3 b c C+8400 a b^3 c C+13440 a^3 A b d x+10080 a A b^3 d x+10080 a^3 b C d x+8400 a b^3 C d x+105 \left (16 a^4 (4 A+3 C)+48 a^2 b^2 (6 A+5 C)+5 b^4 (8 A+7 C)\right ) \sin (c+d x)+420 a b \left (16 a^2 (A+C)+b^2 (16 A+15 C)\right ) \sin (2 (c+d x))+3360 a^2 A b^2 \sin (3 (c+d x))+700 A b^4 \sin (3 (c+d x))+560 a^4 C \sin (3 (c+d x))+4200 a^2 b^2 C \sin (3 (c+d x))+735 b^4 C \sin (3 (c+d x))+840 a A b^3 \sin (4 (c+d x))+840 a^3 b C \sin (4 (c+d x))+1260 a b^3 C \sin (4 (c+d x))+84 A b^4 \sin (5 (c+d x))+504 a^2 b^2 C \sin (5 (c+d x))+147 b^4 C \sin (5 (c+d x))+140 a b^3 C \sin (6 (c+d x))+15 b^4 C \sin (7 (c+d x))}{6720 d} \] Input:

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

Output:

(13440*a^3*A*b*c + 10080*a*A*b^3*c + 10080*a^3*b*c*C + 8400*a*b^3*c*C + 13 
440*a^3*A*b*d*x + 10080*a*A*b^3*d*x + 10080*a^3*b*C*d*x + 8400*a*b^3*C*d*x 
 + 105*(16*a^4*(4*A + 3*C) + 48*a^2*b^2*(6*A + 5*C) + 5*b^4*(8*A + 7*C))*S 
in[c + d*x] + 420*a*b*(16*a^2*(A + C) + b^2*(16*A + 15*C))*Sin[2*(c + d*x) 
] + 3360*a^2*A*b^2*Sin[3*(c + d*x)] + 700*A*b^4*Sin[3*(c + d*x)] + 560*a^4 
*C*Sin[3*(c + d*x)] + 4200*a^2*b^2*C*Sin[3*(c + d*x)] + 735*b^4*C*Sin[3*(c 
 + d*x)] + 840*a*A*b^3*Sin[4*(c + d*x)] + 840*a^3*b*C*Sin[4*(c + d*x)] + 1 
260*a*b^3*C*Sin[4*(c + d*x)] + 84*A*b^4*Sin[5*(c + d*x)] + 504*a^2*b^2*C*S 
in[5*(c + d*x)] + 147*b^4*C*Sin[5*(c + d*x)] + 140*a*b^3*C*Sin[6*(c + d*x) 
] + 15*b^4*C*Sin[7*(c + d*x)])/(6720*d)
 

Rubi [A] (verified)

Time = 1.72 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3529, 3042, 3528, 27, 3042, 3528, 3042, 3512, 27, 3042, 3502, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3529

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

\(\displaystyle \frac {1}{7} \left (\frac {1}{6} \int 2 \cos (c+d x) (a+b \cos (c+d x))^2 \left ((21 A+10 C) a^2+2 b (21 A+17 C) \cos (c+d x) a+3 \left (2 C a^2+b^2 (7 A+6 C)\right ) \cos ^2(c+d x)\right )dx+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \cos (c+d x) (a+b \cos (c+d x))^2 \left ((21 A+10 C) a^2+2 b (21 A+17 C) \cos (c+d x) a+3 \left (2 C a^2+b^2 (7 A+6 C)\right ) \cos ^2(c+d x)\right )dx+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left ((21 A+10 C) a^2+2 b (21 A+17 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+3 \left (2 C a^2+b^2 (7 A+6 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3528

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \int \cos (c+d x) (a+b \cos (c+d x)) \left (2 a \left (6 C a^2+126 A b^2+103 b^2 C\right ) \cos ^2(c+d x)+b \left ((315 A+244 C) a^2+12 b^2 (7 A+6 C)\right ) \cos (c+d x)+a \left ((105 A+62 C) a^2+6 b^2 (7 A+6 C)\right )\right )dx+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (2 a \left (6 C a^2+126 A b^2+103 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b \left ((315 A+244 C) a^2+12 b^2 (7 A+6 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left ((105 A+62 C) a^2+6 b^2 (7 A+6 C)\right )\right )dx+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3512

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{4} \int 2 \cos (c+d x) \left (2 \left ((105 A+62 C) a^2+6 b^2 (7 A+6 C)\right ) a^2+105 b \left ((8 A+6 C) a^2+b^2 (6 A+5 C)\right ) \cos (c+d x) a+6 \left (4 C a^4+3 b^2 (63 A+50 C) a^2+4 b^4 (7 A+6 C)\right ) \cos ^2(c+d x)\right )dx+\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{2} \int \cos (c+d x) \left (2 \left ((105 A+62 C) a^2+6 b^2 (7 A+6 C)\right ) a^2+105 b \left ((8 A+6 C) a^2+b^2 (6 A+5 C)\right ) \cos (c+d x) a+6 \left (4 C a^4+3 b^2 (63 A+50 C) a^2+4 b^4 (7 A+6 C)\right ) \cos ^2(c+d x)\right )dx+\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{2} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 \left ((105 A+62 C) a^2+6 b^2 (7 A+6 C)\right ) a^2+105 b \left ((8 A+6 C) a^2+b^2 (6 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+6 \left (4 C a^4+3 b^2 (63 A+50 C) a^2+4 b^4 (7 A+6 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \int 3 \cos (c+d x) \left (2 \left (35 (3 A+2 C) a^4+84 b^2 (5 A+4 C) a^2+8 b^4 (7 A+6 C)\right )+105 a b \left ((8 A+6 C) a^2+b^2 (6 A+5 C)\right ) \cos (c+d x)\right )dx+\frac {2 \left (4 a^4 C+3 a^2 b^2 (63 A+50 C)+4 b^4 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{2} \left (\int \cos (c+d x) \left (2 \left (35 (3 A+2 C) a^4+84 b^2 (5 A+4 C) a^2+8 b^4 (7 A+6 C)\right )+105 a b \left ((8 A+6 C) a^2+b^2 (6 A+5 C)\right ) \cos (c+d x)\right )dx+\frac {2 \left (4 a^4 C+3 a^2 b^2 (63 A+50 C)+4 b^4 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {1}{5} \left (\frac {1}{2} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 \left (35 (3 A+2 C) a^4+84 b^2 (5 A+4 C) a^2+8 b^4 (7 A+6 C)\right )+105 a b \left ((8 A+6 C) a^2+b^2 (6 A+5 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 \left (4 a^4 C+3 a^2 b^2 (63 A+50 C)+4 b^4 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{d}\right )+\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

\(\Big \downarrow \) 3213

\(\displaystyle \frac {1}{7} \left (\frac {1}{3} \left (\frac {3 \left (2 a^2 C+b^2 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}+\frac {1}{5} \left (\frac {a b \left (6 a^2 C+126 A b^2+103 b^2 C\right ) \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac {1}{2} \left (\frac {105 a b \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {105}{2} a b x \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right )+\frac {2 \left (35 a^4 (3 A+2 C)+84 a^2 b^2 (5 A+4 C)+8 b^4 (7 A+6 C)\right ) \sin (c+d x)}{d}+\frac {2 \left (4 a^4 C+3 a^2 b^2 (63 A+50 C)+4 b^4 (7 A+6 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{d}\right )\right )\right )+\frac {2 a C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^3}{3 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^4}{7 d}\)

Input:

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

Output:

(C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(7*d) + ((2*a*C*Cos 
[c + d*x]^2*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(3*d) + ((3*(2*a^2*C + b^ 
2*(7*A + 6*C))*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d) + 
 ((a*b*(126*A*b^2 + 6*a^2*C + 103*b^2*C)*Cos[c + d*x]^3*Sin[c + d*x])/(2*d 
) + ((105*a*b*(b^2*(6*A + 5*C) + a^2*(8*A + 6*C))*x)/2 + (2*(35*a^4*(3*A + 
 2*C) + 84*a^2*b^2*(5*A + 4*C) + 8*b^4*(7*A + 6*C))*Sin[c + d*x])/d + (105 
*a*b*(b^2*(6*A + 5*C) + a^2*(8*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(2*d) 
+ (2*(4*a^4*C + 4*b^4*(7*A + 6*C) + 3*a^2*b^2*(63*A + 50*C))*Cos[c + d*x]^ 
2*Sin[c + d*x])/d)/2)/5)/3)/7
 

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 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]
 

rule 3512
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f 
_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si 
n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[(a + b*Si 
n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + 
A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 
0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 3528
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x 
])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + 
n + 2))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* 
d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a 
*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + 
 n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ 
m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
 

rule 3529
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] : 
> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 
1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2))   Int[(a + b*Sin[e + f*x 
])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( 
n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* 
(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 
0])))
 
Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.96

\[\frac {A \,a^{4} \sin \left (d x +c \right )+\frac {a^{4} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+4 A \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 A \,a^{2} b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {6 C \,a^{2} b^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+4 a A \,b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 C a \,b^{3} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {A \,b^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {C \,b^{4} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\]

Input:

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

Output:

1/d*(A*a^4*sin(d*x+c)+1/3*a^4*C*(2+cos(d*x+c)^2)*sin(d*x+c)+4*A*a^3*b*(1/2 
*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*a^3*b*C*(1/4*(cos(d*x+c)^3+3/2*cos 
(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+2*A*a^2*b^2*(2+cos(d*x+c)^2)*sin(d*x+c) 
+6/5*C*a^2*b^2*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+4*a*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)+4*C*a*b^3*(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*A*b^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+1/7*C*b^4*(16/5 
+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.73 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {105 \, {\left (2 \, {\left (4 \, A + 3 \, C\right )} a^{3} b + {\left (6 \, A + 5 \, C\right )} a b^{3}\right )} d x + {\left (60 \, C b^{4} \cos \left (d x + c\right )^{6} + 280 \, C a b^{3} \cos \left (d x + c\right )^{5} + 140 \, {\left (3 \, A + 2 \, C\right )} a^{4} + 336 \, {\left (5 \, A + 4 \, C\right )} a^{2} b^{2} + 32 \, {\left (7 \, A + 6 \, C\right )} b^{4} + 12 \, {\left (42 \, C a^{2} b^{2} + {\left (7 \, A + 6 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 70 \, {\left (6 \, C a^{3} b + {\left (6 \, A + 5 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (35 \, C a^{4} + 42 \, {\left (5 \, A + 4 \, C\right )} a^{2} b^{2} + 4 \, {\left (7 \, A + 6 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} + 105 \, {\left (2 \, {\left (4 \, A + 3 \, C\right )} a^{3} b + {\left (6 \, A + 5 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{420 \, d} \] Input:

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

Output:

1/420*(105*(2*(4*A + 3*C)*a^3*b + (6*A + 5*C)*a*b^3)*d*x + (60*C*b^4*cos(d 
*x + c)^6 + 280*C*a*b^3*cos(d*x + c)^5 + 140*(3*A + 2*C)*a^4 + 336*(5*A + 
4*C)*a^2*b^2 + 32*(7*A + 6*C)*b^4 + 12*(42*C*a^2*b^2 + (7*A + 6*C)*b^4)*co 
s(d*x + c)^4 + 70*(6*C*a^3*b + (6*A + 5*C)*a*b^3)*cos(d*x + c)^3 + 4*(35*C 
*a^4 + 42*(5*A + 4*C)*a^2*b^2 + 4*(7*A + 6*C)*b^4)*cos(d*x + c)^2 + 105*(2 
*(4*A + 3*C)*a^3*b + (6*A + 5*C)*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. 850 vs. \(2 (332) = 664\).

Time = 0.55 (sec) , antiderivative size = 850, normalized size of antiderivative = 2.46 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((A*a**4*sin(c + d*x)/d + 2*A*a**3*b*x*sin(c + d*x)**2 + 2*A*a**3 
*b*x*cos(c + d*x)**2 + 2*A*a**3*b*sin(c + d*x)*cos(c + d*x)/d + 4*A*a**2*b 
**2*sin(c + d*x)**3/d + 6*A*a**2*b**2*sin(c + d*x)*cos(c + d*x)**2/d + 3*A 
*a*b**3*x*sin(c + d*x)**4/2 + 3*A*a*b**3*x*sin(c + d*x)**2*cos(c + d*x)**2 
 + 3*A*a*b**3*x*cos(c + d*x)**4/2 + 3*A*a*b**3*sin(c + d*x)**3*cos(c + d*x 
)/(2*d) + 5*A*a*b**3*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 8*A*b**4*sin(c + 
 d*x)**5/(15*d) + 4*A*b**4*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*b**4* 
sin(c + d*x)*cos(c + d*x)**4/d + 2*C*a**4*sin(c + d*x)**3/(3*d) + C*a**4*s 
in(c + d*x)*cos(c + d*x)**2/d + 3*C*a**3*b*x*sin(c + d*x)**4/2 + 3*C*a**3* 
b*x*sin(c + d*x)**2*cos(c + d*x)**2 + 3*C*a**3*b*x*cos(c + d*x)**4/2 + 3*C 
*a**3*b*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 5*C*a**3*b*sin(c + d*x)*cos(c 
 + d*x)**3/(2*d) + 16*C*a**2*b**2*sin(c + d*x)**5/(5*d) + 8*C*a**2*b**2*si 
n(c + d*x)**3*cos(c + d*x)**2/d + 6*C*a**2*b**2*sin(c + d*x)*cos(c + d*x)* 
*4/d + 5*C*a*b**3*x*sin(c + d*x)**6/4 + 15*C*a*b**3*x*sin(c + d*x)**4*cos( 
c + d*x)**2/4 + 15*C*a*b**3*x*sin(c + d*x)**2*cos(c + d*x)**4/4 + 5*C*a*b* 
*3*x*cos(c + d*x)**6/4 + 5*C*a*b**3*sin(c + d*x)**5*cos(c + d*x)/(4*d) + 1 
0*C*a*b**3*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) + 11*C*a*b**3*sin(c + d*x 
)*cos(c + d*x)**5/(4*d) + 16*C*b**4*sin(c + d*x)**7/(35*d) + 8*C*b**4*sin( 
c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*C*b**4*sin(c + d*x)**3*cos(c + d*x)* 
*4/d + C*b**4*sin(c + d*x)*cos(c + d*x)**6/d, Ne(d, 0)), (x*(A + C*cos(...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 329, normalized size of antiderivative = 0.95 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {560 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 1680 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} b + 3360 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b^{2} - 672 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{2} b^{2} - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{3} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} - 112 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A b^{4} + 48 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} C b^{4} - 1680 \, A a^{4} \sin \left (d x + c\right )}{1680 \, d} \] Input:

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

Output:

-1/1680*(560*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 - 1680*(2*d*x + 2*c + 
 sin(2*d*x + 2*c))*A*a^3*b - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin 
(2*d*x + 2*c))*C*a^3*b + 3360*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b^2 
- 672*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^2*b^2 - 
 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a*b^3 + 35* 
(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*a*b^3 - 112*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + 
 c))*A*b^4 + 48*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 
- 35*sin(d*x + c))*C*b^4 - 1680*A*a^4*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.84 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C b^{4} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {C a b^{3} \sin \left (6 \, d x + 6 \, c\right )}{48 \, d} + \frac {1}{4} \, {\left (8 \, A a^{3} b + 6 \, C a^{3} b + 6 \, A a b^{3} + 5 \, C a b^{3}\right )} x + \frac {{\left (24 \, C a^{2} b^{2} + 4 \, A b^{4} + 7 \, C b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (2 \, C a^{3} b + 2 \, A a b^{3} + 3 \, C a b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {{\left (16 \, C a^{4} + 96 \, A a^{2} b^{2} + 120 \, C a^{2} b^{2} + 20 \, A b^{4} + 21 \, C b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {{\left (16 \, A a^{3} b + 16 \, C a^{3} b + 16 \, A a b^{3} + 15 \, C a b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{16 \, d} + \frac {{\left (64 \, A a^{4} + 48 \, C a^{4} + 288 \, A a^{2} b^{2} + 240 \, C a^{2} b^{2} + 40 \, A b^{4} + 35 \, C b^{4}\right )} \sin \left (d x + c\right )}{64 \, d} \] Input:

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

Output:

1/448*C*b^4*sin(7*d*x + 7*c)/d + 1/48*C*a*b^3*sin(6*d*x + 6*c)/d + 1/4*(8* 
A*a^3*b + 6*C*a^3*b + 6*A*a*b^3 + 5*C*a*b^3)*x + 1/320*(24*C*a^2*b^2 + 4*A 
*b^4 + 7*C*b^4)*sin(5*d*x + 5*c)/d + 1/16*(2*C*a^3*b + 2*A*a*b^3 + 3*C*a*b 
^3)*sin(4*d*x + 4*c)/d + 1/192*(16*C*a^4 + 96*A*a^2*b^2 + 120*C*a^2*b^2 + 
20*A*b^4 + 21*C*b^4)*sin(3*d*x + 3*c)/d + 1/16*(16*A*a^3*b + 16*C*a^3*b + 
16*A*a*b^3 + 15*C*a*b^3)*sin(2*d*x + 2*c)/d + 1/64*(64*A*a^4 + 48*C*a^4 + 
288*A*a^2*b^2 + 240*C*a^2*b^2 + 40*A*b^4 + 35*C*b^4)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 798, normalized size of antiderivative = 2.31 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*(2*A*a^4 + 2*A*b^4 + 2*C*a^4 + 2*C*b^4 + 12*A*a^2*b^2 
+ 12*C*a^2*b^2 + 5*A*a*b^3 + 4*A*a^3*b + (11*C*a*b^3)/2 + 5*C*a^3*b) + tan 
(c/2 + (d*x)/2)^7*(40*A*a^4 + (104*A*b^4)/5 + 24*C*a^4 + (424*C*b^4)/35 + 
144*A*a^2*b^2 + (624*C*a^2*b^2)/5) + tan(c/2 + (d*x)/2)^13*(2*A*a^4 + 2*A* 
b^4 + 2*C*a^4 + 2*C*b^4 + 12*A*a^2*b^2 + 12*C*a^2*b^2 - 5*A*a*b^3 - 4*A*a^ 
3*b - (11*C*a*b^3)/2 - 5*C*a^3*b) + tan(c/2 + (d*x)/2)^3*(12*A*a^4 + (20*A 
*b^4)/3 + (28*C*a^4)/3 + 4*C*b^4 + 56*A*a^2*b^2 + 40*C*a^2*b^2 + 12*A*a*b^ 
3 + 16*A*a^3*b + (14*C*a*b^3)/3 + 12*C*a^3*b) + tan(c/2 + (d*x)/2)^11*(12* 
A*a^4 + (20*A*b^4)/3 + (28*C*a^4)/3 + 4*C*b^4 + 56*A*a^2*b^2 + 40*C*a^2*b^ 
2 - 12*A*a*b^3 - 16*A*a^3*b - (14*C*a*b^3)/3 - 12*C*a^3*b) + tan(c/2 + (d* 
x)/2)^5*(30*A*a^4 + (226*A*b^4)/15 + (58*C*a^4)/3 + (86*C*b^4)/5 + 116*A*a 
^2*b^2 + (452*C*a^2*b^2)/5 + 9*A*a*b^3 + 20*A*a^3*b + (85*C*a*b^3)/6 + 9*C 
*a^3*b) + tan(c/2 + (d*x)/2)^9*(30*A*a^4 + (226*A*b^4)/15 + (58*C*a^4)/3 + 
 (86*C*b^4)/5 + 116*A*a^2*b^2 + (452*C*a^2*b^2)/5 - 9*A*a*b^3 - 20*A*a^3*b 
 - (85*C*a*b^3)/6 - 9*C*a^3*b))/(d*(7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + 
(d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/ 
2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/2)^12 + tan(c/2 + (d*x)/2)^14 + 1)) - 
(a*b*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(8*A*a^2 + 6*A*b^2 + 6*C*a^2 + 5 
*C*b^2))/(2*d) + (a*b*atan((a*b*tan(c/2 + (d*x)/2)*(8*A*a^2 + 6*A*b^2 + 6* 
C*a^2 + 5*C*b^2))/(2*(3*A*a*b^3 + 4*A*a^3*b + (5*C*a*b^3)/2 + 3*C*a^3*b...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.20 \[ \int \cos (c+d x) (a+b \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {-420 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{2} b^{3}+840 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4} b +1050 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b^{3}+504 \sin \left (d x +c \right )^{5} a^{2} b^{2} c -1680 \sin \left (d x +c \right )^{3} a^{2} b^{2} c +2520 \sin \left (d x +c \right ) a^{2} b^{2} c +840 a^{4} b d x +630 a^{2} b^{3} d x +420 \sin \left (d x +c \right ) a^{5}+280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a \,b^{3} c -420 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b c -910 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,b^{3} c +1050 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b c +1155 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3} c +630 a^{3} b c d x +525 a \,b^{3} c d x -60 \sin \left (d x +c \right )^{7} b^{4} c +84 \sin \left (d x +c \right )^{5} a \,b^{4}+252 \sin \left (d x +c \right )^{5} b^{4} c -140 \sin \left (d x +c \right )^{3} a^{4} c -840 \sin \left (d x +c \right )^{3} a^{3} b^{2}-280 \sin \left (d x +c \right )^{3} a \,b^{4}-420 \sin \left (d x +c \right )^{3} b^{4} c +420 \sin \left (d x +c \right ) a^{4} c +2520 \sin \left (d x +c \right ) a^{3} b^{2}+420 \sin \left (d x +c \right ) a \,b^{4}+420 \sin \left (d x +c \right ) b^{4} c}{420 d} \] Input:

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

Output:

(280*cos(c + d*x)*sin(c + d*x)**5*a*b**3*c - 420*cos(c + d*x)*sin(c + d*x) 
**3*a**3*b*c - 420*cos(c + d*x)*sin(c + d*x)**3*a**2*b**3 - 910*cos(c + d* 
x)*sin(c + d*x)**3*a*b**3*c + 840*cos(c + d*x)*sin(c + d*x)*a**4*b + 1050* 
cos(c + d*x)*sin(c + d*x)*a**3*b*c + 1050*cos(c + d*x)*sin(c + d*x)*a**2*b 
**3 + 1155*cos(c + d*x)*sin(c + d*x)*a*b**3*c - 60*sin(c + d*x)**7*b**4*c 
+ 504*sin(c + d*x)**5*a**2*b**2*c + 84*sin(c + d*x)**5*a*b**4 + 252*sin(c 
+ d*x)**5*b**4*c - 140*sin(c + d*x)**3*a**4*c - 840*sin(c + d*x)**3*a**3*b 
**2 - 1680*sin(c + d*x)**3*a**2*b**2*c - 280*sin(c + d*x)**3*a*b**4 - 420* 
sin(c + d*x)**3*b**4*c + 420*sin(c + d*x)*a**5 + 420*sin(c + d*x)*a**4*c + 
 2520*sin(c + d*x)*a**3*b**2 + 2520*sin(c + d*x)*a**2*b**2*c + 420*sin(c + 
 d*x)*a*b**4 + 420*sin(c + d*x)*b**4*c + 840*a**4*b*d*x + 630*a**3*b*c*d*x 
 + 630*a**2*b**3*d*x + 525*a*b**3*c*d*x)/(420*d)