\(\int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\) [78]

Optimal result

Integrand size = 26, antiderivative size = 165 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {4 a b^3 \cos ^3(c+d x)}{3 d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}+\frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {b^4 \sin ^5(c+d x)}{5 d} \] Output:

-4/3*a*b^3*cos(d*x+c)^3/d-4/5*a^3*b*cos(d*x+c)^5/d+4/5*a*b^3*cos(d*x+c)^5/ 
d+a^4*sin(d*x+c)/d-2/3*a^4*sin(d*x+c)^3/d+2*a^2*b^2*sin(d*x+c)^3/d+1/5*a^4 
*sin(d*x+c)^5/d-6/5*a^2*b^2*sin(d*x+c)^5/d+1/5*b^4*sin(d*x+c)^5/d
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.78 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {-12 a^3 b \cos ^5(c+d x)+15 a^4 \sin (c+d x)-10 a^2 \left (a^2-3 b^2\right ) \sin ^3(c+d x)+3 \left (a^4-6 a^2 b^2+b^4\right ) \sin ^5(c+d x)+4 a b^3 \cos (c+d x) \left (-2+\frac {2}{\sqrt {\cos ^2(c+d x)}}-\sin ^2(c+d x)+3 \sin ^4(c+d x)\right )}{15 d} \] Input:

Integrate[Cos[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]
 

Output:

(-12*a^3*b*Cos[c + d*x]^5 + 15*a^4*Sin[c + d*x] - 10*a^2*(a^2 - 3*b^2)*Sin 
[c + d*x]^3 + 3*(a^4 - 6*a^2*b^2 + b^4)*Sin[c + d*x]^5 + 4*a*b^3*Cos[c + d 
*x]*(-2 + 2/Sqrt[Cos[c + d*x]^2] - Sin[c + d*x]^2 + 3*Sin[c + d*x]^4))/(15 
*d)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3042, 3569, 2009}

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.

\(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4dx\)

\(\Big \downarrow \) 3569

\(\displaystyle \int \left (a^4 \cos ^5(c+d x)+4 a^3 b \sin (c+d x) \cos ^4(c+d x)+6 a^2 b^2 \sin ^2(c+d x) \cos ^3(c+d x)+4 a b^3 \sin ^3(c+d x) \cos ^2(c+d x)+b^4 \sin ^4(c+d x) \cos (c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^4 \sin ^5(c+d x)}{5 d}-\frac {2 a^4 \sin ^3(c+d x)}{3 d}+\frac {a^4 \sin (c+d x)}{d}-\frac {4 a^3 b \cos ^5(c+d x)}{5 d}-\frac {6 a^2 b^2 \sin ^5(c+d x)}{5 d}+\frac {2 a^2 b^2 \sin ^3(c+d x)}{d}+\frac {4 a b^3 \cos ^5(c+d x)}{5 d}-\frac {4 a b^3 \cos ^3(c+d x)}{3 d}+\frac {b^4 \sin ^5(c+d x)}{5 d}\)

Input:

Int[Cos[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]
 

Output:

(-4*a*b^3*Cos[c + d*x]^3)/(3*d) - (4*a^3*b*Cos[c + d*x]^5)/(5*d) + (4*a*b^ 
3*Cos[c + d*x]^5)/(5*d) + (a^4*Sin[c + d*x])/d - (2*a^4*Sin[c + d*x]^3)/(3 
*d) + (2*a^2*b^2*Sin[c + d*x]^3)/d + (a^4*Sin[c + d*x]^5)/(5*d) - (6*a^2*b 
^2*Sin[c + d*x]^5)/(5*d) + (b^4*Sin[c + d*x]^5)/(5*d)
 

Defintions of rubi rules used

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 3569

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si 
n[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*(a 
*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] && Inte 
gerQ[m] && IGtQ[n, 0]
 

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.84

method	result	size
parts	\(\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {b^{4} \sin \left (d x +c \right )^{5}}{5 d}+\frac {6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right )^{5}}{5}+\frac {\sin \left (d x +c \right )^{3}}{3}\right )}{d}+\frac {4 b^{3} a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )}{d}-\frac {4 a^{3} b \cos \left (d x +c \right )^{5}}{5 d}\)	\(139\)
derivativedivides	\(\frac {\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {4 a^{3} b \cos \left (d x +c \right )^{5}}{5}+6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+4 b^{3} a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+\frac {b^{4} \sin \left (d x +c \right )^{5}}{5}}{d}\)	\(142\)
default	\(\frac {\frac {a^{4} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {4 a^{3} b \cos \left (d x +c \right )^{5}}{5}+6 a^{2} b^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )+4 b^{3} a \left (-\frac {\sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{3}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+\frac {b^{4} \sin \left (d x +c \right )^{5}}{5}}{d}\)	\(142\)
parallelrisch	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a^{4}-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} a^{3} b +\frac {8 \left (a^{4}+6 a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3}-16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a \,b^{3}+\frac {4 \left (29 a^{4}-24 a^{2} b^{2}+24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15}+\frac {16 \left (-3 a^{3} b +b^{3} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3}+\frac {8 \left (a^{4}+6 a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{3}}{3}+2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 a^{3} b}{5}-\frac {16 b^{3} a}{15}}{d \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(222\)
norman	\(\frac {-\frac {24 a^{3} b +16 b^{3} a}{15 d}+\frac {2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {16 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}-\frac {16 b^{3} a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {8 a^{3} b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {4 \left (29 a^{4}-24 a^{2} b^{2}+24 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {2 \left (24 a^{3} b -8 b^{3} a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}+\frac {8 a^{2} \left (a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {8 a^{2} \left (a^{2}+6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\)	\(252\)
risch	\(-\frac {a^{3} b \cos \left (d x +c \right )}{2 d}-\frac {a \,b^{3} \cos \left (d x +c \right )}{2 d}+\frac {5 a^{4} \sin \left (d x +c \right )}{8 d}+\frac {3 a^{2} b^{2} \sin \left (d x +c \right )}{4 d}+\frac {b^{4} \sin \left (d x +c \right )}{8 d}-\frac {a^{3} b \cos \left (5 d x +5 c \right )}{20 d}+\frac {a \,b^{3} \cos \left (5 d x +5 c \right )}{20 d}+\frac {\sin \left (5 d x +5 c \right ) a^{4}}{80 d}-\frac {3 \sin \left (5 d x +5 c \right ) a^{2} b^{2}}{40 d}+\frac {\sin \left (5 d x +5 c \right ) b^{4}}{80 d}-\frac {a^{3} b \cos \left (3 d x +3 c \right )}{4 d}-\frac {a \,b^{3} \cos \left (3 d x +3 c \right )}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) a^{4}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) a^{2} b^{2}}{8 d}-\frac {\sin \left (3 d x +3 c \right ) b^{4}}{16 d}\)	\(257\)
orering	\(\text {Expression too large to display}\)	\(1490\)

Input:

int(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
 

Output:

1/5*a^4/d*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+1/5*b^4*sin(d*x+c 
)^5/d+6*a^2*b^2/d*(-1/5*sin(d*x+c)^5+1/3*sin(d*x+c)^3)+4*b^3*a/d*(-1/5*sin 
(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)-4/5*a^3*b*cos(d*x+c)^5/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {20 \, a b^{3} \cos \left (d x + c\right )^{3} + 12 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{4} + 12 \, a^{2} b^{2} + 3 \, b^{4} + 2 \, {\left (2 \, a^{4} + 3 \, a^{2} b^{2} - 3 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")
 

Output:

-1/15*(20*a*b^3*cos(d*x + c)^3 + 12*(a^3*b - a*b^3)*cos(d*x + c)^5 - (3*(a 
^4 - 6*a^2*b^2 + b^4)*cos(d*x + c)^4 + 8*a^4 + 12*a^2*b^2 + 3*b^4 + 2*(2*a 
^4 + 3*a^2*b^2 - 3*b^4)*cos(d*x + c)^2)*sin(d*x + c))/d
 

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.25 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\begin {cases} \frac {8 a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {4 a^{3} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {4 a^{2} b^{2} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {2 a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {4 a b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {8 a b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {b^{4} \sin ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + b \sin {\left (c \right )}\right )^{4} \cos {\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {12 \, a^{3} b \cos \left (d x + c\right )^{5} - 3 \, b^{4} \sin \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} + 6 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} b^{2} - 4 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a b^{3}}{15 \, d} \] Input:

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")
 

Output:

-1/15*(12*a^3*b*cos(d*x + c)^5 - 3*b^4*sin(d*x + c)^5 - (3*sin(d*x + c)^5 
- 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4 + 6*(3*sin(d*x + c)^5 - 5*sin(d 
*x + c)^3)*a^2*b^2 - 4*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a*b^3)/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=-\frac {{\left (a^{3} b - a b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{20 \, d} - \frac {{\left (3 \, a^{3} b + a b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {{\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )}{2 \, d} + \frac {{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (5 \, a^{4} - 6 \, a^{2} b^{2} - 3 \, b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (5 \, a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:

integrate(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")
 

Output:

-1/20*(a^3*b - a*b^3)*cos(5*d*x + 5*c)/d - 1/12*(3*a^3*b + a*b^3)*cos(3*d* 
x + 3*c)/d - 1/2*(a^3*b + a*b^3)*cos(d*x + c)/d + 1/80*(a^4 - 6*a^2*b^2 + 
b^4)*sin(5*d*x + 5*c)/d + 1/48*(5*a^4 - 6*a^2*b^2 - 3*b^4)*sin(3*d*x + 3*c 
)/d + 1/8*(5*a^4 + 6*a^2*b^2 + b^4)*sin(d*x + c)/d
 

Mupad [B] (verification not implemented)

Time = 16.84 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.24 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^4\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^4\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^4-6\,a^3\,b\,{\cos \left (c+d\,x\right )}^5-9\,\sin \left (c+d\,x\right )\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^4+3\,\sin \left (c+d\,x\right )\,a^2\,b^2\,{\cos \left (c+d\,x\right )}^2+6\,\sin \left (c+d\,x\right )\,a^2\,b^2+6\,a\,b^3\,{\cos \left (c+d\,x\right )}^5-10\,a\,b^3\,{\cos \left (c+d\,x\right )}^3+\frac {3\,\sin \left (c+d\,x\right )\,b^4\,{\cos \left (c+d\,x\right )}^4}{2}-3\,\sin \left (c+d\,x\right )\,b^4\,{\cos \left (c+d\,x\right )}^2+\frac {3\,\sin \left (c+d\,x\right )\,b^4}{2}\right )}{15\,d} \] Input:

int(cos(c + d*x)*(a*cos(c + d*x) + b*sin(c + d*x))^4,x)
 

Output:

(2*(4*a^4*sin(c + d*x) + (3*b^4*sin(c + d*x))/2 - 10*a*b^3*cos(c + d*x)^3 
+ 6*a*b^3*cos(c + d*x)^5 - 6*a^3*b*cos(c + d*x)^5 + 2*a^4*cos(c + d*x)^2*s 
in(c + d*x) + (3*a^4*cos(c + d*x)^4*sin(c + d*x))/2 + 6*a^2*b^2*sin(c + d* 
x) - 3*b^4*cos(c + d*x)^2*sin(c + d*x) + (3*b^4*cos(c + d*x)^4*sin(c + d*x 
))/2 + 3*a^2*b^2*cos(c + d*x)^2*sin(c + d*x) - 9*a^2*b^2*cos(c + d*x)^4*si 
n(c + d*x)))/(15*d)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.98 \[ \int \cos (c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx=\frac {-12 \cos \left (d x +c \right )^{5} a^{3} b -8 \cos \left (d x +c \right )^{5} a \,b^{3}+15 \cos \left (d x +c \right )^{4} \sin \left (d x +c \right ) a^{4}-20 \cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2} a \,b^{3}+20 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} a^{4}+30 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3} a^{2} b^{2}+8 \sin \left (d x +c \right )^{5} a^{4}+12 \sin \left (d x +c \right )^{5} a^{2} b^{2}+3 \sin \left (d x +c \right )^{5} b^{4}}{15 d} \] Input:

int(cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)
 

Output:

( - 12*cos(c + d*x)**5*a**3*b - 8*cos(c + d*x)**5*a*b**3 + 15*cos(c + d*x) 
**4*sin(c + d*x)*a**4 - 20*cos(c + d*x)**3*sin(c + d*x)**2*a*b**3 + 20*cos 
(c + d*x)**2*sin(c + d*x)**3*a**4 + 30*cos(c + d*x)**2*sin(c + d*x)**3*a** 
2*b**2 + 8*sin(c + d*x)**5*a**4 + 12*sin(c + d*x)**5*a**2*b**2 + 3*sin(c + 
 d*x)**5*b**4)/(15*d)