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

Optimal. Leaf size=160 \[ -\frac {a^2 (10 B+9 C) \sin ^3(c+d x)}{15 d}+\frac {a^2 (10 B+9 C) \sin (c+d x)}{5 d}+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a^2 x (7 B+6 C)+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d} \]

[Out]

1/8*a^2*(7*B+6*C)*x+1/5*a^2*(10*B+9*C)*sin(d*x+c)/d+1/8*a^2*(7*B+6*C)*cos(d*x+c)*sin(d*x+c)/d+1/20*a^2*(5*B+6*
C)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*C*cos(d*x+c)^3*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d-1/15*a^2*(10*B+9*C)*sin(d*x+
c)^3/d

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Rubi [A]  time = 0.34, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3029, 2976, 2968, 3023, 2748, 2635, 8, 2633} \[ -\frac {a^2 (10 B+9 C) \sin ^3(c+d x)}{15 d}+\frac {a^2 (10 B+9 C) \sin (c+d x)}{5 d}+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {a^2 (7 B+6 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a^2 x (7 B+6 C)+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*B + 6*C)*x)/8 + (a^2*(10*B + 9*C)*Sin[c + d*x])/(5*d) + (a^2*(7*B + 6*C)*Cos[c + d*x]*Sin[c + d*x])/(8
*d) + (a^2*(5*B + 6*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (C*Cos[c + d*x]^3*(a^2 + a^2*Cos[c + d*x])*Sin[c
+ d*x])/(5*d) - (a^2*(10*B + 9*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 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 2976

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 - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[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+a \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+a \cos (c+d x))^2 (B+C \cos (c+d x)) \, dx\\ &=\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^2(c+d x) (a+a \cos (c+d x)) (a (5 B+3 C)+a (5 B+6 C) \cos (c+d x)) \, dx\\ &=\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^2(c+d x) \left (a^2 (5 B+3 C)+\left (a^2 (5 B+3 C)+a^2 (5 B+6 C)\right ) \cos (c+d x)+a^2 (5 B+6 C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (5 B+6 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^2(c+d x) \left (5 a^2 (7 B+6 C)+4 a^2 (10 B+9 C) \cos (c+d x)\right ) \, dx\\ &=\frac {a^2 (5 B+6 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{4} \left (a^2 (7 B+6 C)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{5} \left (a^2 (10 B+9 C)\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 B+6 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{8} \left (a^2 (7 B+6 C)\right ) \int 1 \, dx-\frac {\left (a^2 (10 B+9 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{8} a^2 (7 B+6 C) x+\frac {a^2 (10 B+9 C) \sin (c+d x)}{5 d}+\frac {a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 B+6 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {a^2 (10 B+9 C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.33, size = 104, normalized size = 0.65 \[ \frac {a^2 (60 (12 B+11 C) \sin (c+d x)+240 (B+C) \sin (2 (c+d x))+80 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+420 B d x+90 C \sin (3 (c+d x))+30 C \sin (4 (c+d x))+6 C \sin (5 (c+d x))+360 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(420*B*d*x + 360*C*d*x + 60*(12*B + 11*C)*Sin[c + d*x] + 240*(B + C)*Sin[2*(c + d*x)] + 80*B*Sin[3*(c + d
*x)] + 90*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d)

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fricas [A]  time = 0.43, size = 110, normalized size = 0.69 \[ \frac {15 \, {\left (7 \, B + 6 \, C\right )} a^{2} d x + {\left (24 \, C a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \, {\left (10 \, B + 9 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(7*B + 6*C)*a^2*d*x + (24*C*a^2*cos(d*x + c)^4 + 30*(B + 2*C)*a^2*cos(d*x + c)^3 + 8*(10*B + 9*C)*a^
2*cos(d*x + c)^2 + 15*(7*B + 6*C)*a^2*cos(d*x + c) + 16*(10*B + 9*C)*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.30, size = 137, normalized size = 0.86 \[ \frac {C a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (7 \, B a^{2} + 6 \, C a^{2}\right )} x + \frac {{\left (B a^{2} + 2 \, C a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (8 \, B a^{2} + 9 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{2} + C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (12 \, B a^{2} + 11 \, C a^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*C*a^2*sin(5*d*x + 5*c)/d + 1/8*(7*B*a^2 + 6*C*a^2)*x + 1/32*(B*a^2 + 2*C*a^2)*sin(4*d*x + 4*c)/d + 1/48*(
8*B*a^2 + 9*C*a^2)*sin(3*d*x + 3*c)/d + 1/2*(B*a^2 + C*a^2)*sin(2*d*x + 2*c)/d + 1/8*(12*B*a^2 + 11*C*a^2)*sin
(d*x + c)/d

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maple [A]  time = 0.25, size = 186, normalized size = 1.16 \[ \frac {\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 178, normalized size = 1.11 \[ -\frac {320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.36, size = 277, normalized size = 1.73 \[ \frac {\left (\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {49\,B\,a^2}{6}+7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {40\,B\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {79\,B\,a^2}{6}+9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,B\,a^2}{4}+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,B+6\,C\right )}{4\,\left (\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )}\right )\,\left (7\,B+6\,C\right )}{4\,d}-\frac {a^2\,\left (7\,B+6\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]

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 + a*cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)*((25*B*a^2)/4 + (13*C*a^2)/2) + tan(c/2 + (d*x)/2)^9*((7*B*a^2)/4 + (3*C*a^2)/2) + tan(c/2
+ (d*x)/2)^7*((49*B*a^2)/6 + 7*C*a^2) + tan(c/2 + (d*x)/2)^3*((79*B*a^2)/6 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*
((40*B*a^2)/3 + (72*C*a^2)/5))/(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)) + (a^2*atan((a^2*tan(c/2 + (d*x)/2)*(7*B + 6*C))/(4*((7
*B*a^2)/4 + (3*C*a^2)/2)))*(7*B + 6*C))/(4*d) - (a^2*(7*B + 6*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)

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sympy [A]  time = 2.47, size = 462, normalized size = 2.89 \[ \begin {cases} \frac {3 B a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 C a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{2} \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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