3.3.33 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\) [233]

Optimal. Leaf size=171 \[ \frac {1}{8} \left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) x+\frac {\left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac {b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d} \]

[Out]

1/8*(8*A*a^3+12*A*a*b^2+12*B*a^2*b+3*B*b^3)*x+1/6*(16*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*sin(d*x+c)/d+1/24*b*
(20*A*a*b+6*B*a^2+9*B*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/12*(4*A*b+3*B*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*B*(a
+b*cos(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]
time = 0.13, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2832, 2813} \begin {gather*} \frac {b \left (6 a^2 B+20 a A b+9 b^2 B\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {\left (3 a^3 B+16 a^2 A b+12 a b^2 B+4 A b^3\right ) \sin (c+d x)}{6 d}+\frac {1}{8} x \left (8 a^3 A+12 a^2 b B+12 a A b^2+3 b^3 B\right )+\frac {(3 a B+4 A b) \sin (c+d x) (a+b \cos (c+d x))^2}{12 d}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 3*b^3*B)*x)/8 + ((16*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*Sin[c + d
*x])/(6*d) + (b*(20*a*A*b + 6*a^2*B + 9*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(24*d) + ((4*A*b + 3*a*B)*(a + b*Cos
[c + d*x])^2*Sin[c + d*x])/(12*d) + (B*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d)

Rule 2813

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 2832

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)*Sim
p[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]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\frac {B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x))^2 (4 a A+3 b B+(4 A b+3 a B) \cos (c+d x)) \, dx\\ &=\frac {(4 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac {1}{12} \int (a+b \cos (c+d x)) \left (12 a^2 A+8 A b^2+15 a b B+\left (20 a A b+6 a^2 B+9 b^2 B\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {1}{8} \left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) x+\frac {\left (16 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sin (c+d x)}{6 d}+\frac {b \left (20 a A b+6 a^2 B+9 b^2 B\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 A b+3 a B) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {B (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.45, size = 140, normalized size = 0.82 \begin {gather*} \frac {12 \left (8 a^3 A+12 a A b^2+12 a^2 b B+3 b^3 B\right ) (c+d x)+24 \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \sin (c+d x)+24 b \left (3 a A b+3 a^2 B+b^2 B\right ) \sin (2 (c+d x))+8 b^2 (A b+3 a B) \sin (3 (c+d x))+3 b^3 B \sin (4 (c+d x))}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(8*a^3*A + 12*a*A*b^2 + 12*a^2*b*B + 3*b^3*B)*(c + d*x) + 24*(12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*
Sin[c + d*x] + 24*b*(3*a*A*b + 3*a^2*B + b^2*B)*Sin[2*(c + d*x)] + 8*b^2*(A*b + 3*a*B)*Sin[3*(c + d*x)] + 3*b^
3*B*Sin[4*(c + d*x)])/(96*d)

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Maple [A]
time = 0.13, size = 180, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {b^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+B a \,b^{2} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )+3 A a \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{2} b \sin \left (d x +c \right )+a^{3} B \sin \left (d x +c \right )+A \,a^{3} \left (d x +c \right )}{d}\)	\(180\)
default	\(\frac {b^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{3} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}+B a \,b^{2} \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )+3 A a \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b B \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{2} b \sin \left (d x +c \right )+a^{3} B \sin \left (d x +c \right )+A \,a^{3} \left (d x +c \right )}{d}\)	\(180\)
risch	\(a^{3} x A +\frac {3 x A a \,b^{2}}{2}+\frac {3 x \,a^{2} b B}{2}+\frac {3 b^{3} B x}{8}+\frac {3 \sin \left (d x +c \right ) A \,a^{2} b}{d}+\frac {3 \sin \left (d x +c \right ) A \,b^{3}}{4 d}+\frac {a^{3} B \sin \left (d x +c \right )}{d}+\frac {9 \sin \left (d x +c \right ) B a \,b^{2}}{4 d}+\frac {\sin \left (4 d x +4 c \right ) b^{3} B}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,b^{3}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B a \,b^{2}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) A a \,b^{2}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b B}{4 d}+\frac {\sin \left (2 d x +2 c \right ) b^{3} B}{4 d}\)	\(203\)
norman	\(\frac {\left (A \,a^{3}+\frac {3}{2} A a \,b^{2}+\frac {3}{2} a^{2} b B +\frac {3}{8} b^{3} B \right ) x +\left (A \,a^{3}+\frac {3}{2} A a \,b^{2}+\frac {3}{2} a^{2} b B +\frac {3}{8} b^{3} B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A \,a^{3}+6 A a \,b^{2}+6 a^{2} b B +\frac {3}{2} b^{3} B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A \,a^{3}+6 A a \,b^{2}+6 a^{2} b B +\frac {3}{2} b^{3} B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A \,a^{3}+9 A a \,b^{2}+9 a^{2} b B +\frac {9}{4} b^{3} B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (24 A \,a^{2} b -12 A a \,b^{2}+8 A \,b^{3}+8 a^{3} B -12 a^{2} b B +24 B a \,b^{2}-5 b^{3} B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (24 A \,a^{2} b +12 A a \,b^{2}+8 A \,b^{3}+8 a^{3} B +12 a^{2} b B +24 B a \,b^{2}+5 b^{3} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (216 A \,a^{2} b -36 A a \,b^{2}+40 A \,b^{3}+72 a^{3} B -36 a^{2} b B +120 B a \,b^{2}+9 b^{3} B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (216 A \,a^{2} b +36 A a \,b^{2}+40 A \,b^{3}+72 a^{3} B +36 a^{2} b B +120 B a \,b^{2}-9 b^{3} B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\)	\(455\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(b^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)+1/3*A*b^3*(cos(d*x+c)^2+2)*sin(d*x+c)+
B*a*b^2*(cos(d*x+c)^2+2)*sin(d*x+c)+3*A*a*b^2*(1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+3*a^2*b*B*(1/2*sin(d*x
+c)*cos(d*x+c)+1/2*d*x+1/2*c)+3*A*a^2*b*sin(d*x+c)+a^3*B*sin(d*x+c)+A*a^3*(d*x+c))

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Maxima [A]
time = 0.28, size = 171, normalized size = 1.00 \begin {gather*} \frac {96 \, {\left (d x + c\right )} A a^{3} + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b + 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} - 96 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b^{2} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{3} + 3 \, {\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^{3} + 96 \, B a^{3} \sin \left (d x + c\right ) + 288 \, A a^{2} b \sin \left (d x + c\right )}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(96*(d*x + c)*A*a^3 + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2*b + 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*A
*a*b^2 - 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a*b^2 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^3 + 3*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b^3 + 96*B*a^3*sin(d*x + c) + 288*A*a^2*b*sin(d*x + c))/d

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Fricas [A]
time = 0.35, size = 136, normalized size = 0.80 \begin {gather*} \frac {3 \, {\left (8 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 3 \, B b^{3}\right )} d x + {\left (6 \, B b^{3} \cos \left (d x + c\right )^{3} + 24 \, B a^{3} + 72 \, A a^{2} b + 48 \, B a b^{2} + 16 \, A b^{3} + 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (4 \, B a^{2} b + 4 \, A a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*A*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 3*B*b^3)*d*x + (6*B*b^3*cos(d*x + c)^3 + 24*B*a^3 + 72*A*a^2*b +
48*B*a*b^2 + 16*A*b^3 + 8*(3*B*a*b^2 + A*b^3)*cos(d*x + c)^2 + 9*(4*B*a^2*b + 4*A*a*b^2 + B*b^3)*cos(d*x + c))
*sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (170) = 340\).
time = 0.23, size = 386, normalized size = 2.26 \begin {gather*} \begin {cases} A a^{3} x + \frac {3 A a^{2} b \sin {\left (c + d x \right )}}{d} + \frac {3 A a b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 A a b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{3} \sin {\left (c + d x \right )}}{d} + \frac {3 B a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 B a b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)),x)

[Out]

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

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Giac [A]
time = 0.46, size = 148, normalized size = 0.87 \begin {gather*} \frac {B b^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (8 \, A a^{3} + 12 \, B a^{2} b + 12 \, A a b^{2} + 3 \, B b^{3}\right )} x + \frac {{\left (3 \, B a b^{2} + A b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (3 \, B a^{2} b + 3 \, A a b^{2} + B b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*B*b^3*sin(4*d*x + 4*c)/d + 1/8*(8*A*a^3 + 12*B*a^2*b + 12*A*a*b^2 + 3*B*b^3)*x + 1/12*(3*B*a*b^2 + A*b^3)
*sin(3*d*x + 3*c)/d + 1/4*(3*B*a^2*b + 3*A*a*b^2 + B*b^3)*sin(2*d*x + 2*c)/d + 1/4*(4*B*a^3 + 12*A*a^2*b + 9*B
*a*b^2 + 3*A*b^3)*sin(d*x + c)/d

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Mupad [B]
time = 0.57, size = 202, normalized size = 1.18 \begin {gather*} A\,a^3\,x+\frac {3\,B\,b^3\,x}{8}+\frac {3\,A\,a\,b^2\,x}{2}+\frac {3\,B\,a^2\,b\,x}{2}+\frac {3\,A\,b^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {A\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,A\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,A\,a^2\,b\,\sin \left (c+d\,x\right )}{d}+\frac {9\,B\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^3,x)

[Out]

A*a^3*x + (3*B*b^3*x)/8 + (3*A*a*b^2*x)/2 + (3*B*a^2*b*x)/2 + (3*A*b^3*sin(c + d*x))/(4*d) + (B*a^3*sin(c + d*
x))/d + (A*b^3*sin(3*c + 3*d*x))/(12*d) + (B*b^3*sin(2*c + 2*d*x))/(4*d) + (B*b^3*sin(4*c + 4*d*x))/(32*d) + (
3*A*a*b^2*sin(2*c + 2*d*x))/(4*d) + (3*B*a^2*b*sin(2*c + 2*d*x))/(4*d) + (B*a*b^2*sin(3*c + 3*d*x))/(4*d) + (3
*A*a^2*b*sin(c + d*x))/d + (9*B*a*b^2*sin(c + d*x))/(4*d)

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