3.244 \(\int \cos (c+d x) (a+a \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=201 \[ -\frac{a^3 (19 B+17 C) \sin ^3(c+d x)}{15 d}+\frac{a^3 (19 B+17 C) \sin (c+d x)}{5 d}+\frac{a^3 (22 B+21 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac{(3 B+4 C) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac{a^3 (26 B+23 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a^3 x (26 B+23 C)+\frac{a C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^2}{6 d} \]

[Out]

(a^3*(26*B + 23*C)*x)/16 + (a^3*(19*B + 17*C)*Sin[c + d*x])/(5*d) + (a^3*(26*B + 23*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^3*(22*B + 21*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (a*C*Cos[c + d*x]^3*(a + a*Cos[c + d*x])
^2*Sin[c + d*x])/(6*d) + ((3*B + 4*C)*Cos[c + d*x]^3*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d) - (a^3*(19*
B + 17*C)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.48687, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3029, 2976, 2968, 3023, 2748, 2635, 8, 2633} \[ -\frac{a^3 (19 B+17 C) \sin ^3(c+d x)}{15 d}+\frac{a^3 (19 B+17 C) \sin (c+d x)}{5 d}+\frac{a^3 (22 B+21 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac{(3 B+4 C) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{15 d}+\frac{a^3 (26 B+23 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a^3 x (26 B+23 C)+\frac{a C \sin (c+d x) \cos ^3(c+d x) (a \cos (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a^3*(26*B + 23*C)*x)/16 + (a^3*(19*B + 17*C)*Sin[c + d*x])/(5*d) + (a^3*(26*B + 23*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^3*(22*B + 21*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (a*C*Cos[c + d*x]^3*(a + a*Cos[c + d*x])
^2*Sin[c + d*x])/(6*d) + ((3*B + 4*C)*Cos[c + d*x]^3*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(15*d) - (a^3*(19*
B + 17*C)*Sin[c + d*x]^3)/(15*d)

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]

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 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]

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 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 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]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+a \cos (c+d x))^3 \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))^3 (B+C \cos (c+d x)) \, dx\\ &=\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^2(c+d x) (a+a \cos (c+d x))^2 (3 a (2 B+C)+2 a (3 B+4 C) \cos (c+d x)) \, dx\\ &=\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{30} \int \cos ^2(c+d x) (a+a \cos (c+d x)) \left (3 a^2 (16 B+13 C)+3 a^2 (22 B+21 C) \cos (c+d x)\right ) \, dx\\ &=\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{30} \int \cos ^2(c+d x) \left (3 a^3 (16 B+13 C)+\left (3 a^3 (16 B+13 C)+3 a^3 (22 B+21 C)\right ) \cos (c+d x)+3 a^3 (22 B+21 C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{120} \int \cos ^2(c+d x) \left (15 a^3 (26 B+23 C)+24 a^3 (19 B+17 C) \cos (c+d x)\right ) \, dx\\ &=\frac{a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{5} \left (a^3 (19 B+17 C)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{8} \left (a^3 (26 B+23 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{a^3 (26 B+23 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}+\frac{1}{16} \left (a^3 (26 B+23 C)\right ) \int 1 \, dx-\frac{\left (a^3 (19 B+17 C)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{1}{16} a^3 (26 B+23 C) x+\frac{a^3 (19 B+17 C) \sin (c+d x)}{5 d}+\frac{a^3 (26 B+23 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^3 (22 B+21 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac{a C \cos ^3(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{6 d}+\frac{(3 B+4 C) \cos ^3(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{15 d}-\frac{a^3 (19 B+17 C) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.414656, size = 130, normalized size = 0.65 \[ \frac{a^3 (120 (23 B+21 C) \sin (c+d x)+15 (64 B+63 C) \sin (2 (c+d x))+340 B \sin (3 (c+d x))+90 B \sin (4 (c+d x))+12 B \sin (5 (c+d x))+1560 B d x+380 C \sin (3 (c+d x))+135 C \sin (4 (c+d x))+36 C \sin (5 (c+d x))+5 C \sin (6 (c+d x))+1380 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a^3*(1560*B*d*x + 1380*C*d*x + 120*(23*B + 21*C)*Sin[c + d*x] + 15*(64*B + 63*C)*Sin[2*(c + d*x)] + 340*B*Sin
[3*(c + d*x)] + 380*C*Sin[3*(c + d*x)] + 90*B*Sin[4*(c + d*x)] + 135*C*Sin[4*(c + d*x)] + 12*B*Sin[5*(c + d*x)
] + 36*C*Sin[5*(c + d*x)] + 5*C*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.027, size = 266, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{3}B \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +3\,{a}^{3}C \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{a}^{3}B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{3\,{a}^{3}C\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+3\,{a}^{3}B \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{a}^{3}C \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{{a}^{3}B\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.48159, size = 354, normalized size = 1.76 \begin{align*} \frac{64 \,{\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 a^{3} - 960 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 90 \,{\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^{3} + 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 192 \,{\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^{3} - 5 \,{\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^{3} - 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 90 \,{\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}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/960*(64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^3 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c
))*B*a^3 + 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^3 + 240*(2*d*x + 2*c + sin(2*d*x + 2
*c))*B*a^3 + 192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^3 - 5*(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^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 +
90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3)/d

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Fricas [A]  time = 1.7361, size = 332, normalized size = 1.65 \begin{align*} \frac{15 \,{\left (26 \, B + 23 \, C\right )} a^{3} d x +{\left (40 \, C a^{3} \cos \left (d x + c\right )^{5} + 48 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 10 \,{\left (18 \, B + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \,{\left (19 \, B + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \,{\left (26 \, B + 23 \, C\right )} a^{3} \cos \left (d x + c\right ) + 32 \,{\left (19 \, B + 17 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(26*B + 23*C)*a^3*d*x + (40*C*a^3*cos(d*x + c)^5 + 48*(B + 3*C)*a^3*cos(d*x + c)^4 + 10*(18*B + 23*C
)*a^3*cos(d*x + c)^3 + 16*(19*B + 17*C)*a^3*cos(d*x + c)^2 + 15*(26*B + 23*C)*a^3*cos(d*x + c) + 32*(19*B + 17
*C)*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 6.1858, size = 699, normalized size = 3.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((9*B*a**3*x*sin(c + d*x)**4/8 + 9*B*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + B*a**3*x*sin(c + d*x)
**2/2 + 9*B*a**3*x*cos(c + d*x)**4/8 + B*a**3*x*cos(c + d*x)**2/2 + 8*B*a**3*sin(c + d*x)**5/(15*d) + 4*B*a**3
*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 9*B*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*B*a**3*sin(c + d*x)**
3/d + B*a**3*sin(c + d*x)*cos(c + d*x)**4/d + 15*B*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*B*a**3*sin(c +
d*x)*cos(c + d*x)**2/d + B*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) + 5*C*a**3*x*sin(c + d*x)**6/16 + 15*C*a**3*x*
sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*C*a**3*x*sin(c + d*x)**4/8 + 15*C*a**3*x*sin(c + d*x)**2*cos(c + d*x)**
4/16 + 9*C*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 5*C*a**3*x*cos(c + d*x)**6/16 + 9*C*a**3*x*cos(c + d*x)*
*4/8 + 5*C*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 8*C*a**3*sin(c + d*x)**5/(5*d) + 5*C*a**3*sin(c + d*x)**
3*cos(c + d*x)**3/(6*d) + 4*C*a**3*sin(c + d*x)**3*cos(c + d*x)**2/d + 9*C*a**3*sin(c + d*x)**3*cos(c + d*x)/(
8*d) + 2*C*a**3*sin(c + d*x)**3/(3*d) + 11*C*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 3*C*a**3*sin(c + d*x)*
cos(c + d*x)**4/d + 15*C*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + C*a**3*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)**3*cos(c), True))

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Giac [A]  time = 1.48267, size = 224, normalized size = 1.11 \begin{align*} \frac{C a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{1}{16} \,{\left (26 \, B a^{3} + 23 \, C a^{3}\right )} x + \frac{{\left (B a^{3} + 3 \, C a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{3 \,{\left (2 \, B a^{3} + 3 \, C a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (17 \, B a^{3} + 19 \, C a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (64 \, B a^{3} + 63 \, C a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{{\left (23 \, B a^{3} + 21 \, C a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/192*C*a^3*sin(6*d*x + 6*c)/d + 1/16*(26*B*a^3 + 23*C*a^3)*x + 1/80*(B*a^3 + 3*C*a^3)*sin(5*d*x + 5*c)/d + 3/
64*(2*B*a^3 + 3*C*a^3)*sin(4*d*x + 4*c)/d + 1/48*(17*B*a^3 + 19*C*a^3)*sin(3*d*x + 3*c)/d + 1/64*(64*B*a^3 + 6
3*C*a^3)*sin(2*d*x + 2*c)/d + 1/8*(23*B*a^3 + 21*C*a^3)*sin(d*x + c)/d