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

Optimal. Leaf size=243 \[ \frac {\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}+\frac {\left (-6 a^3 B+30 a^2 A b+71 a b^2 B+45 A b^3\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac {1}{8} x \left (4 a^3 B+12 a^2 A b+9 a b^2 B+3 A b^3\right )+\frac {\left (-3 a^4 B+15 a^3 A b+52 a^2 b^2 B+60 a A b^3+16 b^4 B\right ) \sin (c+d x)}{30 b d}+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2968, 3023, 2753, 2734} \[ \frac {\left (15 a^3 A b+52 a^2 b^2 B-3 a^4 B+60 a A b^3+16 b^4 B\right ) \sin (c+d x)}{30 b d}+\frac {\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}+\frac {\left (30 a^2 A b-6 a^3 B+71 a b^2 B+45 A b^3\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac {1}{8} x \left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right )+\frac {(5 A b-a B) \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^4}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*x)/8 + ((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])/(30*b*d) + ((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])/(120
*d) + ((15*a*A*b - 3*a^2*B + 16*b^2*B)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(60*b*d) + ((5*A*b - a*B)*(a + b*C
os[c + d*x])^3*Sin[c + d*x])/(20*b*d) + (B*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*b*d)

Rule 2734

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)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2753

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] + Dist[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 2968

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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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 + 1))/(b*f*(m + 2)), x] + Dist[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]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\int (a+b \cos (c+d x))^3 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {B (a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\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 {(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}+\frac {\int (a+b \cos (c+d x))^2 \left (b (15 A b+13 a B)+\left (15 a A b-3 a^2 B+16 b^2 B\right ) \cos (c+d x)\right ) \, dx}{20 b}\\ &=\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}+\frac {\int (a+b \cos (c+d x)) \left (b \left (75 a A b+33 a^2 B+32 b^2 B\right )+\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \cos (c+d x)\right ) \, dx}{60 b}\\ &=\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}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 176, normalized size = 0.72 \[ \frac {10 b \left (12 a^2 B+12 a A b+5 b^2 B\right ) \sin (3 (c+d x))+60 (c+d x) \left (4 a^3 B+12 a^2 A b+9 a b^2 B+3 A b^3\right )+60 \left (8 a^3 A+18 a^2 b B+18 a A b^2+5 b^3 B\right ) \sin (c+d x)+120 \left (a^3 B+3 a^2 A b+3 a b^2 B+A b^3\right ) \sin (2 (c+d x))+15 b^2 (3 a B+A b) \sin (4 (c+d x))+6 b^3 B \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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fricas [A]  time = 0.70, size = 174, normalized size = 0.72 \[ \frac {15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} d x + {\left (24 \, B b^{3} \cos \left (d x + c\right )^{4} + 120 \, A a^{3} + 240 \, B a^{2} b + 240 \, A a b^{2} + 64 \, B b^{3} + 30 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, B a^{2} b + 15 \, A a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.51, size = 188, normalized size = 0.77 \[ \frac {B b^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} x + \frac {{\left (3 \, B a b^{2} + A b^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, B a^{2} b + 12 \, A a b^{2} + 5 \, B b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{3} + 3 \, A a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{3} + 18 \, B a^{2} b + 18 \, A a b^{2} + 5 \, B b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 227, normalized size = 0.93 \[ \frac {A \,a^{3} \sin \left (d x +c \right )+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} b B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,b^{2} a \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{3} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 217, normalized size = 0.89 \[ \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B b^{3} + 480 \, A a^{3} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 - 480*(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

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mupad [B]  time = 0.78, size = 277, normalized size = 1.14 \[ \frac {3\,A\,b^3\,x}{8}+\frac {B\,a^3\,x}{2}+\frac {3\,A\,a^2\,b\,x}{2}+\frac {9\,B\,a\,b^2\,x}{8}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,b^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {5\,B\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,b^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,B\,a\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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*sin(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)

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sympy [A]  time = 2.76, size = 551, normalized size = 2.27 \[ \begin {cases} \frac {A a^{3} \sin {\left (c + d x \right )}}{d} + \frac {3 A a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a^{2} b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 A a b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A b^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 A b^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 B a^{2} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 B a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 B a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 B a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 B a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 B b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {B b^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a + b \cos {\relax (c )}\right )^{3} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*co
s(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*cos(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)
)

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