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

Optimal. Leaf size=189 \[ \frac {\left (4 a^2 B+6 a b C+3 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (4 a^2 B+6 a b C+3 b^2 B\right )-\frac {\left (5 a (a C+2 b B)+4 b^2 C\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (5 a (a C+2 b B)+4 b^2 C\right ) \sin (c+d x)}{5 d}+\frac {b (6 a C+5 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {b C \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3029, 2990, 3023, 2748, 2635, 8, 2633} \[ \frac {\left (4 a^2 B+6 a b C+3 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (4 a^2 B+6 a b C+3 b^2 B\right )-\frac {\left (5 a (a C+2 b B)+4 b^2 C\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (5 a (a C+2 b B)+4 b^2 C\right ) \sin (c+d x)}{5 d}+\frac {b (6 a C+5 b B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {b C \sin (c+d x) \cos ^3(c+d x) (a+b \cos (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2990

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 3029

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] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 146, normalized size = 0.77 \[ \frac {60 (c+d x) \left (4 a^2 B+6 a b C+3 b^2 B\right )+60 \left (6 a^2 C+12 a b B+5 b^2 C\right ) \sin (c+d x)+120 \left (a^2 B+2 a b C+b^2 B\right ) \sin (2 (c+d x))+10 \left (4 a^2 C+8 a b B+5 b^2 C\right ) \sin (3 (c+d x))+15 b (2 a C+b B) \sin (4 (c+d x))+6 b^2 C \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*(4*a^2*B + 3*b^2*B + 6*a*b*C)*(c + d*x) + 60*(12*a*b*B + 6*a^2*C + 5*b^2*C)*Sin[c + d*x] + 120*(a^2*B + b^
2*B + 2*a*b*C)*Sin[2*(c + d*x)] + 10*(8*a*b*B + 4*a^2*C + 5*b^2*C)*Sin[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 = 0.43, size = 142, normalized size = 0.75 \[ \frac {15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} d x + {\left (24 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} + 80 \, C a^{2} + 160 \, B a b + 64 \, C b^{2} + 8 \, {\left (5 \, C a^{2} + 10 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*B*a^2 + 6*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 +
80*C*a^2 + 160*B*a*b + 64*C*b^2 + 8*(5*C*a^2 + 10*B*a*b + 4*C*b^2)*cos(d*x + c)^2 + 15*(4*B*a^2 + 6*C*a*b + 3
*B*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.29, size = 156, normalized size = 0.83 \[ \frac {C b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (4 \, B a^{2} + 6 \, C a b + 3 \, B b^{2}\right )} x + \frac {{\left (2 \, C a b + B b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, C a^{2} + 8 \, B a b + 5 \, C b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{2} + 2 \, C a b + B b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, C a^{2} + 12 \, B a b + 5 \, C b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^2*(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 + 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 + 5*C*b^2)*sin(3*d*x + 3*c)/d + 1/4*(B*a^2 + 2*C*a*b + B*b^2)*sin(2*d*x + 2*c)/d +
1/8*(6*C*a^2 + 12*B*a*b + 5*C*b^2)*sin(d*x + c)/d

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maple [A]  time = 0.29, size = 184, normalized size = 0.97 \[ \frac {\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B a b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {b^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+b^{2} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 176, normalized size = 0.93 \[ \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{2}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^2*(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 - 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 + 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)/d

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mupad [B]  time = 5.41, size = 307, normalized size = 1.62 \[ \frac {x\,\left (B\,a^2+\frac {3\,C\,a\,b}{2}+\frac {3\,B\,b^2}{4}\right )}{2}+\frac {\left (2\,C\,a^2-\frac {5\,B\,b^2}{4}-B\,a^2+2\,C\,b^2+4\,B\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {16\,C\,a^2}{3}-\frac {B\,b^2}{2}-2\,B\,a^2+\frac {8\,C\,b^2}{3}+\frac {32\,B\,a\,b}{3}-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,C\,a^2}{3}+\frac {40\,B\,a\,b}{3}+\frac {116\,C\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,B\,a^2+\frac {B\,b^2}{2}+\frac {16\,C\,a^2}{3}+\frac {8\,C\,b^2}{3}+\frac {32\,B\,a\,b}{3}+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B\,a^2+\frac {5\,B\,b^2}{4}+2\,C\,a^2+2\,C\,b^2+4\,B\,a\,b+\frac {5\,C\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(B*a^2 + (3*B*b^2)/4 + (3*C*a*b)/2))/2 + (tan(c/2 + (d*x)/2)^5*((20*C*a^2)/3 + (116*C*b^2)/15 + (40*B*a*b)/
3) - tan(c/2 + (d*x)/2)^9*(B*a^2 + (5*B*b^2)/4 - 2*C*a^2 - 2*C*b^2 - 4*B*a*b + (5*C*a*b)/2) + tan(c/2 + (d*x)/
2)^3*(2*B*a^2 + (B*b^2)/2 + (16*C*a^2)/3 + (8*C*b^2)/3 + (32*B*a*b)/3 + C*a*b) - tan(c/2 + (d*x)/2)^7*(2*B*a^2
+ (B*b^2)/2 - (16*C*a^2)/3 - (8*C*b^2)/3 - (32*B*a*b)/3 + C*a*b) + tan(c/2 + (d*x)/2)*(B*a^2 + (5*B*b^2)/4 +
2*C*a^2 + 2*C*b^2 + 4*B*a*b + (5*C*a*b)/2))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2
+ (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1))

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sympy [A]  time = 2.91, size = 462, normalized size = 2.44 \[ \begin {cases} \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 B a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 B a b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {5 C a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {8 C b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{2} \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((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*(B*cos(c) + C*cos(c)**2)*cos(c), True))

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