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

[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)

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Rubi [A]  time = 0.333154, antiderivative size = 243, normalized size of antiderivative = 1., 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 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 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 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]

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*}

Mathematica [A]  time = 0.671014, size = 176, normalized size = 0.72 \[ \frac{60 (c+d x) \left (12 a^2 A b+4 a^3 B+9 a b^2 B+3 A b^3\right )+10 b \left (12 a^2 B+12 a A b+5 b^2 B\right ) \sin (3 (c+d x))+60 \left (8 a^3 A+18 a^2 b B+18 a A b^2+5 b^3 B\right ) \sin (c+d x)+120 \left (3 a^2 A b+a^3 B+3 a b^2 B+A b^3\right ) \sin (2 (c+d x))+15 b^2 (3 a B+A b) \sin (4 (c+d x))+6 b^3 B \sin (5 (c+d x))}{480 d} \]

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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Maple [A]  time = 0.125, size = 227, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( A{a}^{3}\sin \left ( dx+c \right ) +{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{a}^{2}b \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{2}bB \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +Aa{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,Ba{b}^{2} \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{b}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{B{b}^{3}\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+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*a*b^2*(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*b^3*(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 = 1.11803, size = 293, normalized size = 1.21 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.5275, size = 423, normalized size = 1.74 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 3.85204, size = 551, normalized size = 2.27 \begin{align*} \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{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right )^{3} \cos{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.34094, size = 254, normalized size = 1.05 \begin{align*} \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} \end{align*}

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