3.947 \(\int \cos (c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=224 \[ \frac {\sin (c+d x) \left (5 a^2 (3 A+2 C)+20 a b B+2 b^2 (5 A+4 C)\right )}{15 d}+\frac {\sin (c+d x) \cos ^2(c+d x) \left (2 a^2 C+10 a b B+5 A b^2+4 b^2 C\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )}{8 d}+\frac {1}{8} x \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )+\frac {b (2 a C+5 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

1/8*(8*A*a*b+4*B*a^2+3*B*b^2+6*C*a*b)*x+1/15*(20*a*b*B+5*a^2*(3*A+2*C)+2*b^2*(5*A+4*C))*sin(d*x+c)/d+1/8*(8*A*
a*b+4*B*a^2+3*B*b^2+6*C*a*b)*cos(d*x+c)*sin(d*x+c)/d+1/15*(5*A*b^2+10*B*a*b+2*C*a^2+4*C*b^2)*cos(d*x+c)^2*sin(
d*x+c)/d+1/20*b*(5*B*b+2*C*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*C*cos(d*x+c)^2*(a+b*cos(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.33, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3049, 3033, 3023, 2734} \[ \frac {\sin (c+d x) \left (5 a^2 (3 A+2 C)+20 a b B+2 b^2 (5 A+4 C)\right )}{15 d}+\frac {\sin (c+d x) \cos ^2(c+d x) \left (2 a^2 C+10 a b B+5 A b^2+4 b^2 C\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )}{8 d}+\frac {1}{8} x \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )+\frac {b (2 a C+5 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*A*b + 4*a^2*B + 3*b^2*B + 6*a*b*C)*x)/8 + ((20*a*b*B + 5*a^2*(3*A + 2*C) + 2*b^2*(5*A + 4*C))*Sin[c + d*
x])/(15*d) + ((8*a*A*b + 4*a^2*B + 3*b^2*B + 6*a*b*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + ((5*A*b^2 + 10*a*b*B
+ 2*a^2*C + 4*b^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(15*d) + (b*(5*b*B + 2*a*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20
*d) + (C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*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 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]

Rule 3033

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
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[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 3049

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] + Dist[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])))

Rubi steps

\begin {align*} \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos (c+d x) (a+b \cos (c+d x)) \left (a (5 A+2 C)+(5 A b+5 a B+4 b C) \cos (c+d x)+(5 b B+2 a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b (5 b B+2 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos (c+d x) \left (4 a^2 (5 A+2 C)+5 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)+4 \left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {\left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {b (5 b B+2 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{60} \int \cos (c+d x) \left (4 \left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right )+15 \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {1}{8} \left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) x+\frac {\left (20 a b B+5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac {\left (8 a A b+4 a^2 B+3 b^2 B+6 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {\left (5 A b^2+10 a b B+2 a^2 C+4 b^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {b (5 b B+2 a C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.84, size = 169, normalized size = 0.75 \[ \frac {60 (c+d x) \left (4 a^2 B+8 a A b+6 a b C+3 b^2 B\right )+60 \sin (c+d x) \left (a^2 (8 A+6 C)+12 a b B+b^2 (6 A+5 C)\right )+120 \sin (2 (c+d x)) \left (a^2 B+2 a b (A+C)+b^2 B\right )+10 \sin (3 (c+d x)) \left (4 a^2 C+8 a b B+4 A b^2+5 b^2 C\right )+15 b (2 a C+b B) \sin (4 (c+d x))+6 b^2 C \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 171, normalized size = 0.76 \[ \frac {15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} d x + {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + 40 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 160 \, B a b + 16 \, {\left (5 \, A + 4 \, C\right )} b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, B a^{2} + 2 \, {\left (4 \, A + 3 \, C\right )} a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*B*a^2 + 2*(4*A + 3*C)*a*b + 3*B*b^2)*d*x + (24*C*b^2*cos(d*x + c)^4 + 30*(2*C*a*b + B*b^2)*cos(d*
x + c)^3 + 40*(3*A + 2*C)*a^2 + 160*B*a*b + 16*(5*A + 4*C)*b^2 + 8*(5*C*a^2 + 10*B*a*b + (5*A + 4*C)*b^2)*cos(
d*x + c)^2 + 15*(4*B*a^2 + 2*(4*A + 3*C)*a*b + 3*B*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.22, size = 184, normalized size = 0.82 \[ \frac {C b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, B a^{2} + 8 \, A a b + 6 \, C a b + 3 \, B b^{2}\right )} x + \frac {{\left (2 \, C a b + B b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, C a^{2} + 8 \, B a b + 4 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{2} + 2 \, A a b + 2 \, C a b + B b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{2} + 6 \, C a^{2} + 12 \, B a b + 6 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*C*b^2*sin(5*d*x + 5*c)/d + 1/8*(4*B*a^2 + 8*A*a*b + 6*C*a*b + 3*B*b^2)*x + 1/32*(2*C*a*b + B*b^2)*sin(4*d
*x + 4*c)/d + 1/48*(4*C*a^2 + 8*B*a*b + 4*A*b^2 + 5*C*b^2)*sin(3*d*x + 3*c)/d + 1/4*(B*a^2 + 2*A*a*b + 2*C*a*b
+ B*b^2)*sin(2*d*x + 2*c)/d + 1/8*(8*A*a^2 + 6*C*a^2 + 12*B*a*b + 6*A*b^2 + 5*C*b^2)*sin(d*x + c)/d

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maple [A]  time = 0.29, size = 244, normalized size = 1.09 \[ \frac {a^{2} A \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 B a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 C a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{2} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {b^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*A*sin(d*x+c)+B*a^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*a^2*C*(2+cos(d*x+c)^2)*sin(d*x+c)+2*
A*a*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2/3*B*a*b*(2+cos(d*x+c)^2)*sin(d*x+c)+2*C*a*b*(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*A*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+b^2*B*(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^2*C*(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 = 0.34, size = 233, normalized size = 1.04 \[ \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{2} + 480 \, A a^{2} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 + 240*(2*d*x +
2*c + sin(2*d*x + 2*c))*A*a*b - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a*b + 30*(12*d*x + 12*c + sin(4*d*x +
4*c) + 8*sin(2*d*x + 2*c))*C*a*b - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^2 + 15*(12*d*x + 12*c + sin(4*d*
x + 4*c) + 8*sin(2*d*x + 2*c))*B*b^2 + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*b^2 + 480
*A*a^2*sin(d*x + c))/d

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mupad [B]  time = 2.44, size = 256, normalized size = 1.14 \[ \frac {30\,B\,a^2\,\sin \left (2\,c+2\,d\,x\right )+10\,A\,b^2\,\sin \left (3\,c+3\,d\,x\right )+30\,B\,b^2\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,a^2\,\sin \left (3\,c+3\,d\,x\right )+\frac {15\,B\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{4}+\frac {25\,C\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,C\,b^2\,\sin \left (5\,c+5\,d\,x\right )}{2}+120\,A\,a^2\,\sin \left (c+d\,x\right )+90\,A\,b^2\,\sin \left (c+d\,x\right )+90\,C\,a^2\,\sin \left (c+d\,x\right )+75\,C\,b^2\,\sin \left (c+d\,x\right )+60\,A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+20\,B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )+60\,C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+\frac {15\,C\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{2}+60\,B\,a^2\,d\,x+45\,B\,b^2\,d\,x+180\,B\,a\,b\,\sin \left (c+d\,x\right )+120\,A\,a\,b\,d\,x+90\,C\,a\,b\,d\,x}{120\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(30*B*a^2*sin(2*c + 2*d*x) + 10*A*b^2*sin(3*c + 3*d*x) + 30*B*b^2*sin(2*c + 2*d*x) + 10*C*a^2*sin(3*c + 3*d*x)
+ (15*B*b^2*sin(4*c + 4*d*x))/4 + (25*C*b^2*sin(3*c + 3*d*x))/2 + (3*C*b^2*sin(5*c + 5*d*x))/2 + 120*A*a^2*si
n(c + d*x) + 90*A*b^2*sin(c + d*x) + 90*C*a^2*sin(c + d*x) + 75*C*b^2*sin(c + d*x) + 60*A*a*b*sin(2*c + 2*d*x)
+ 20*B*a*b*sin(3*c + 3*d*x) + 60*C*a*b*sin(2*c + 2*d*x) + (15*C*a*b*sin(4*c + 4*d*x))/2 + 60*B*a^2*d*x + 45*B
*b^2*d*x + 180*B*a*b*sin(c + d*x) + 120*A*a*b*d*x + 90*C*a*b*d*x)/(120*d)

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sympy [A]  time = 3.15, size = 570, normalized size = 2.54 \[ \begin {cases} \frac {A a^{2} \sin {\left (c + d x \right )}}{d} + A a b x \sin ^{2}{\left (c + d x \right )} + A a b x \cos ^{2}{\left (c + d x \right )} + \frac {A a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 A b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 B a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 B a b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {5 C a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {8 C b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b^{2} \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 )^{2} \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \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))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

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

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