\(\int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [215]

Optimal result

Integrand size = 29, antiderivative size = 105 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{8} (4 a A+3 b B) x+\frac {(A b+a B) \sin (c+d x)}{d}+\frac {(4 a A+3 b B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {b B \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {(A b+a B) \sin ^3(c+d x)}{3 d} \] Output:

1/8*(4*A*a+3*B*b)*x+(A*b+B*a)*sin(d*x+c)/d+1/8*(4*A*a+3*B*b)*cos(d*x+c)*si 
n(d*x+c)/d+1/4*b*B*cos(d*x+c)^3*sin(d*x+c)/d-1/3*(A*b+B*a)*sin(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.87 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {48 a A c+36 b B c+48 a A d x+36 b B d x+96 (A b+a B) \sin (c+d x)-32 (A b+a B) \sin ^3(c+d x)+24 (a A+b B) \sin (2 (c+d x))+3 b B \sin (4 (c+d x))}{96 d} \] Input:

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

Output:

(48*a*A*c + 36*b*B*c + 48*a*A*d*x + 36*b*B*d*x + 96*(A*b + a*B)*Sin[c + d* 
x] - 32*(A*b + a*B)*Sin[c + d*x]^3 + 24*(a*A + b*B)*Sin[2*(c + d*x)] + 3*b 
*B*Sin[4*(c + d*x)])/(96*d)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 3447, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 3115, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3447

\(\displaystyle \int \cos ^2(c+d x) \left ((a B+A b) \cos (c+d x)+a A+b B \cos ^2(c+d x)\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left ((a B+A b) \sin \left (c+d x+\frac {\pi }{2}\right )+a A+b B \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {1}{4} \int \cos ^2(c+d x) (4 a A+3 b B+4 (A b+a B) \cos (c+d x))dx+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (4 a A+3 b B+4 (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{4} \left (4 (a B+A b) \int \cos ^3(c+d x)dx+(4 a A+3 b B) \int \cos ^2(c+d x)dx\right )+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 (a B+A b) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 (a B+A b) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 (a B+A b) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 (a B+A b) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{4} \left ((4 a A+3 b B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 (a B+A b) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {b B \sin (c+d x) \cos ^3(c+d x)}{4 d}\)

Input:

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

Output:

(b*B*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((4*a*A + 3*b*B)*(x/2 + (Cos[c + 
 d*x]*Sin[c + d*x])/(2*d)) - (4*(A*b + a*B)*(-Sin[c + d*x] + Sin[c + d*x]^ 
3/3))/d)/4
 

Defintions of rubi rules used

rule 24

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3113

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

rule 3115

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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3227

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

rule 3447

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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
 

Maple [A] (verified)

Time = 9.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\frac {\left (24 A a +24 B b \right ) \sin \left (2 d x +2 c \right )+\left (8 A b +8 B a \right ) \sin \left (3 d x +3 c \right )+3 B b \sin \left (4 d x +4 c \right )+\left (72 A b +72 B a \right ) \sin \left (d x +c \right )+48 x d \left (A a +\frac {3 B b}{4}\right )}{96 d}\)	\(86\)
parts	\(\frac {\left (A b +B a \right ) \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3 d}+\frac {A a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {B b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(97\)
derivativedivides	\(\frac {B b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+\frac {B a \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+A a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
default	\(\frac {B b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+\frac {B a \left (\cos \left (d x +c \right )^{2}+2\right ) \sin \left (d x +c \right )}{3}+A a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(107\)
risch	\(\frac {a x A}{2}+\frac {3 b B x}{8}+\frac {3 \sin \left (d x +c \right ) A b}{4 d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {B b \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A b}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B a}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A a}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}\)	\(118\)
norman	\(\frac {\left (\frac {A a}{2}+\frac {3 B b}{8}\right ) x +\left (2 A a +\frac {3 B b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (2 A a +\frac {3 B b}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 A a +\frac {9 B b}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {A a}{2}+\frac {3 B b}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {\left (4 A a -8 A b -8 B a +5 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (4 A a +8 A b +8 B a +5 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 A a -40 A b -40 B a -9 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {\left (12 A a +40 A b +40 B a -9 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(247\)
orering	\(\text {Expression too large to display}\)	\(1614\)

Input:

int(cos(d*x+c)^2*(a+cos(d*x+c)*b)*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE 
)
 

Output:

1/96*((24*A*a+24*B*b)*sin(2*d*x+2*c)+(8*A*b+8*B*a)*sin(3*d*x+3*c)+3*B*b*si 
n(4*d*x+4*c)+(72*A*b+72*B*a)*sin(d*x+c)+48*x*d*(A*a+3/4*B*b))/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (4 \, A a + 3 \, B b\right )} d x + {\left (6 \, B b \cos \left (d x + c\right )^{3} + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{2} + 16 \, B a + 16 \, A b + 3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:

integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="fri 
cas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (97) = 194\).

Time = 0.20 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.40 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right ) \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b}{96 \, d} \] Input:

integrate(cos(d*x+c)^2*(a+b*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="max 
ima")
 

Output:

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 32*(sin(d*x + c)^3 - 3*sin 
(d*x + c))*B*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b + 3*(12*d*x + 12 
*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b)/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {1}{8} \, {\left (4 \, A a + 3 \, B b\right )} x + \frac {B b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (B a + A b\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {3 \, {\left (B a + A b\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 41.57 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.11 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {A\,a\,x}{2}+\frac {3\,B\,b\,x}{8}+\frac {3\,A\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \cos ^2(c+d x) (a+b \cos (c+d x)) (A+B \cos (c+d x)) \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b^{2}+12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2}+15 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}-16 \sin \left (d x +c \right )^{3} a b +48 \sin \left (d x +c \right ) a b +12 a^{2} d x +9 b^{2} d x}{24 d} \] Input:

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

Output:

( - 6*cos(c + d*x)*sin(c + d*x)**3*b**2 + 12*cos(c + d*x)*sin(c + d*x)*a** 
2 + 15*cos(c + d*x)*sin(c + d*x)*b**2 - 16*sin(c + d*x)**3*a*b + 48*sin(c 
+ d*x)*a*b + 12*a**2*d*x + 9*b**2*d*x)/(24*d)