\(\int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx\) [301]

Optimal result

Integrand size = 24, antiderivative size = 157 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=4 a \left (5 a^2+3 c^2\right ) x-\frac {4 c \left (15 a^2+4 c^2\right ) \cos (d+e x)}{3 e}-\frac {4 a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{3 e}-\frac {20 \left (a c \cos (d+e x)+a^2 \sin (d+e x)\right ) (a-a \cos (d+e x)+c \sin (d+e x))}{3 e}-\frac {8 (c \cos (d+e x)+a \sin (d+e x)) (a-a \cos (d+e x)+c \sin (d+e x))^2}{3 e} \] Output:

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

Mathematica [A] (verified)

Time = 1.64 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=\frac {2 \left (6 a \left (5 a^2+3 c^2\right ) (d+e x)-9 c \left (5 a^2+c^2\right ) \cos (d+e x)+18 a^2 c \cos (2 (d+e x))+c \left (-3 a^2+c^2\right ) \cos (3 (d+e x))-9 a \left (5 a^2+c^2\right ) \sin (d+e x)+9 a \left (a^2-c^2\right ) \sin (2 (d+e x))-a \left (a^2-3 c^2\right ) \sin (3 (d+e x))\right )}{3 e} \] Input:

Integrate[(2*a - 2*a*Cos[d + e*x] + 2*c*Sin[d + e*x])^3,x]
 

Output:

(2*(6*a*(5*a^2 + 3*c^2)*(d + e*x) - 9*c*(5*a^2 + c^2)*Cos[d + e*x] + 18*a^ 
2*c*Cos[2*(d + e*x)] + c*(-3*a^2 + c^2)*Cos[3*(d + e*x)] - 9*a*(5*a^2 + c^ 
2)*Sin[d + e*x] + 9*a*(a^2 - c^2)*Sin[2*(d + e*x)] - a*(a^2 - 3*c^2)*Sin[3 
*(d + e*x)]))/(3*e)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3599, 27, 3042, 3625, 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.

\(\displaystyle \int (-2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (-2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^3dx\)

\(\Big \downarrow \) 3599

\(\displaystyle \frac {1}{3} \int 8 (-\cos (d+e x) a+a+c \sin (d+e x)) \left (-5 \cos (d+e x) a^2+5 a^2+5 c \sin (d+e x) a+2 c^2\right )dx-\frac {8 (a \sin (d+e x)+c \cos (d+e x)) (a (-\cos (d+e x))+a+c \sin (d+e x))^2}{3 e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {8}{3} \int (-\cos (d+e x) a+a+c \sin (d+e x)) \left (-5 \cos (d+e x) a^2+5 a^2+5 c \sin (d+e x) a+2 c^2\right )dx-\frac {8 (a \sin (d+e x)+c \cos (d+e x)) (a (-\cos (d+e x))+a+c \sin (d+e x))^2}{3 e}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {8}{3} \int (-\cos (d+e x) a+a+c \sin (d+e x)) \left (-5 \cos (d+e x) a^2+5 a^2+5 c \sin (d+e x) a+2 c^2\right )dx-\frac {8 (a \sin (d+e x)+c \cos (d+e x)) (a (-\cos (d+e x))+a+c \sin (d+e x))^2}{3 e}\)

\(\Big \downarrow \) 3625

\(\displaystyle \frac {8}{3} \left (\frac {\int \left (3 \left (5 a^2+3 c^2\right ) a^2-\left (15 a^2+4 c^2\right ) \cos (d+e x) a^2+c \left (15 a^2+4 c^2\right ) \sin (d+e x) a\right )dx}{2 a}-\frac {5 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right ) (a (-\cos (d+e x))+a+c \sin (d+e x))}{2 e}\right )-\frac {8 (a \sin (d+e x)+c \cos (d+e x)) (a (-\cos (d+e x))+a+c \sin (d+e x))^2}{3 e}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {8}{3} \left (\frac {-\frac {a^2 \left (15 a^2+4 c^2\right ) \sin (d+e x)}{e}-\frac {a c \left (15 a^2+4 c^2\right ) \cos (d+e x)}{e}+3 a^2 x \left (5 a^2+3 c^2\right )}{2 a}-\frac {5 \left (a^2 \sin (d+e x)+a c \cos (d+e x)\right ) (a (-\cos (d+e x))+a+c \sin (d+e x))}{2 e}\right )-\frac {8 (a \sin (d+e x)+c \cos (d+e x)) (a (-\cos (d+e x))+a+c \sin (d+e x))^2}{3 e}\)

Input:

Int[(2*a - 2*a*Cos[d + e*x] + 2*c*Sin[d + e*x])^3,x]
 

Output:

(-8*(c*Cos[d + e*x] + a*Sin[d + e*x])*(a - a*Cos[d + e*x] + c*Sin[d + e*x] 
)^2)/(3*e) + (8*((-5*(a*c*Cos[d + e*x] + a^2*Sin[d + e*x])*(a - a*Cos[d + 
e*x] + c*Sin[d + e*x]))/(2*e) + (3*a^2*(5*a^2 + 3*c^2)*x - (a*c*(15*a^2 + 
4*c^2)*Cos[d + e*x])/e - (a^2*(15*a^2 + 4*c^2)*Sin[d + e*x])/e)/(2*a)))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3599

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_), x_Symbol] :> Simp[(-(c*Cos[d + e*x] - b*Sin[d + e*x]))*((a + b*Cos[d 
+ e*x] + c*Sin[d + e*x])^(n - 1)/(e*n)), x] + Simp[1/n   Int[Simp[n*a^2 + ( 
n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x 
], x]*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, 
 c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]
 

rule 3625

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_.)*((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_) 
]), x_Symbol] :> Simp[(B*c - b*C - a*C*Cos[d + e*x] + a*B*Sin[d + e*x])*((a 
 + b*Cos[d + e*x] + c*Sin[d + e*x])^n/(a*e*(n + 1))), x] + Simp[1/(a*(n + 1 
))   Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1)*Simp[a*(b*B + c*C)*n 
 + a^2*A*(n + 1) + (n*(a^2*B - B*c^2 + b*c*C) + a*b*A*(n + 1))*Cos[d + e*x] 
 + (n*(b*B*c + a^2*C - b^2*C) + a*c*A*(n + 1))*Sin[d + e*x], x], x], x] /; 
FreeQ[{a, b, c, d, e, A, B, C}, x] && GtQ[n, 0] && NeQ[a^2 - b^2 - c^2, 0]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.13

\[\frac {-\frac {8 a^{3} \left (2+\cos \left (e x +d \right )^{2}\right ) \sin \left (e x +d \right )}{3}-8 a^{2} c \cos \left (e x +d \right )^{3}-8 a \,c^{2} \sin \left (e x +d \right )^{3}-\frac {8 c^{3} \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3}+24 a^{3} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+24 a^{2} c \cos \left (e x +d \right )^{2}+24 a \,c^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-24 a^{3} \sin \left (e x +d \right )-24 a^{2} c \cos \left (e x +d \right )+8 a^{3} \left (e x +d \right )}{e}\]

Input:

int((2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x)
 

Output:

8/e*(-1/3*a^3*(2+cos(e*x+d)^2)*sin(e*x+d)-a^2*c*cos(e*x+d)^3-a*c^2*sin(e*x 
+d)^3-1/3*c^3*(2+sin(e*x+d)^2)*cos(e*x+d)+3*a^3*(1/2*sin(e*x+d)*cos(e*x+d) 
+1/2*e*x+1/2*d)+3*a^2*c*cos(e*x+d)^2+3*a*c^2*(-1/2*sin(e*x+d)*cos(e*x+d)+1 
/2*e*x+1/2*d)-3*a^3*sin(e*x+d)-3*a^2*c*cos(e*x+d)+a^3*(e*x+d))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.85 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=\frac {4 \, {\left (18 \, a^{2} c \cos \left (e x + d\right )^{2} - 2 \, {\left (3 \, a^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} + 3 \, {\left (5 \, a^{3} + 3 \, a c^{2}\right )} e x - 6 \, {\left (3 \, a^{2} c + c^{3}\right )} \cos \left (e x + d\right ) - {\left (22 \, a^{3} + 6 \, a c^{2} + 2 \, {\left (a^{3} - 3 \, a c^{2}\right )} \cos \left (e x + d\right )^{2} - 9 \, {\left (a^{3} - a c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{3 \, e} \] Input:

integrate((2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x, algorithm="fricas")
 

Output:

4/3*(18*a^2*c*cos(e*x + d)^2 - 2*(3*a^2*c - c^3)*cos(e*x + d)^3 + 3*(5*a^3 
 + 3*a*c^2)*e*x - 6*(3*a^2*c + c^3)*cos(e*x + d) - (22*a^3 + 6*a*c^2 + 2*( 
a^3 - 3*a*c^2)*cos(e*x + d)^2 - 9*(a^3 - a*c^2)*cos(e*x + d))*sin(e*x + d) 
)/e
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.85 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=\begin {cases} 12 a^{3} x \sin ^{2}{\left (d + e x \right )} + 12 a^{3} x \cos ^{2}{\left (d + e x \right )} + 8 a^{3} x - \frac {16 a^{3} \sin ^{3}{\left (d + e x \right )}}{3 e} - \frac {8 a^{3} \sin {\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{e} + \frac {12 a^{3} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {24 a^{3} \sin {\left (d + e x \right )}}{e} - \frac {8 a^{2} c \cos ^{3}{\left (d + e x \right )}}{e} + \frac {24 a^{2} c \cos ^{2}{\left (d + e x \right )}}{e} - \frac {24 a^{2} c \cos {\left (d + e x \right )}}{e} + 12 a c^{2} x \sin ^{2}{\left (d + e x \right )} + 12 a c^{2} x \cos ^{2}{\left (d + e x \right )} - \frac {8 a c^{2} \sin ^{3}{\left (d + e x \right )}}{e} - \frac {12 a c^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {8 c^{3} \sin ^{2}{\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {16 c^{3} \cos ^{3}{\left (d + e x \right )}}{3 e} & \text {for}\: e \neq 0 \\x \left (- 2 a \cos {\left (d \right )} + 2 a + 2 c \sin {\left (d \right )}\right )^{3} & \text {otherwise} \end {cases} \] Input:

integrate((2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))**3,x)
 

Output:

Piecewise((12*a**3*x*sin(d + e*x)**2 + 12*a**3*x*cos(d + e*x)**2 + 8*a**3* 
x - 16*a**3*sin(d + e*x)**3/(3*e) - 8*a**3*sin(d + e*x)*cos(d + e*x)**2/e 
+ 12*a**3*sin(d + e*x)*cos(d + e*x)/e - 24*a**3*sin(d + e*x)/e - 8*a**2*c* 
cos(d + e*x)**3/e + 24*a**2*c*cos(d + e*x)**2/e - 24*a**2*c*cos(d + e*x)/e 
 + 12*a*c**2*x*sin(d + e*x)**2 + 12*a*c**2*x*cos(d + e*x)**2 - 8*a*c**2*si 
n(d + e*x)**3/e - 12*a*c**2*sin(d + e*x)*cos(d + e*x)/e - 8*c**3*sin(d + e 
*x)**2*cos(d + e*x)/e - 16*c**3*cos(d + e*x)**3/(3*e), Ne(e, 0)), (x*(-2*a 
*cos(d) + 2*a + 2*c*sin(d))**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.20 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=-\frac {8 \, a^{2} c \cos \left (e x + d\right )^{3}}{e} - \frac {8 \, a c^{2} \sin \left (e x + d\right )^{3}}{e} + 8 \, a^{3} x + \frac {8 \, {\left (\sin \left (e x + d\right )^{3} - 3 \, \sin \left (e x + d\right )\right )} a^{3}}{3 \, e} + \frac {8 \, {\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} c^{3}}{3 \, e} - 24 \, a^{2} {\left (\frac {c \cos \left (e x + d\right )}{e} + \frac {a \sin \left (e x + d\right )}{e}\right )} + 6 \, {\left (\frac {4 \, a c \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{e}\right )} a \] Input:

integrate((2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x, algorithm="maxima")
 

Output:

-8*a^2*c*cos(e*x + d)^3/e - 8*a*c^2*sin(e*x + d)^3/e + 8*a^3*x + 8/3*(sin( 
e*x + d)^3 - 3*sin(e*x + d))*a^3/e + 8/3*(cos(e*x + d)^3 - 3*cos(e*x + d)) 
*c^3/e - 24*a^2*(c*cos(e*x + d)/e + a*sin(e*x + d)/e) + 6*(4*a*c*cos(e*x + 
 d)^2/e + (2*e*x + 2*d + sin(2*e*x + 2*d))*a^2/e + (2*e*x + 2*d - sin(2*e* 
x + 2*d))*c^2/e)*a
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=\frac {12 \, a^{2} c \cos \left (2 \, e x + 2 \, d\right )}{e} + 4 \, {\left (5 \, a^{3} + 3 \, a c^{2}\right )} x - \frac {2 \, {\left (3 \, a^{2} c - c^{3}\right )} \cos \left (3 \, e x + 3 \, d\right )}{3 \, e} - \frac {6 \, {\left (5 \, a^{2} c + c^{3}\right )} \cos \left (e x + d\right )}{e} - \frac {2 \, {\left (a^{3} - 3 \, a c^{2}\right )} \sin \left (3 \, e x + 3 \, d\right )}{3 \, e} + \frac {6 \, {\left (a^{3} - a c^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{e} - \frac {6 \, {\left (5 \, a^{3} + a c^{2}\right )} \sin \left (e x + d\right )}{e} \] Input:

integrate((2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 16.23 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.64 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=\frac {8\,a\,\mathrm {atan}\left (\frac {8\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (5\,a^2+3\,c^2\right )}{40\,a^3+24\,a\,c^2}\right )\,\left (5\,a^2+3\,c^2\right )}{e}-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (40\,a^3+24\,a\,c^2\right )+64\,a^2\,c-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\left (24\,a\,c^2-88\,a^3\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (\frac {320\,a^3}{3}+64\,a\,c^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (192\,a^2\,c+32\,c^3\right )+\frac {32\,c^3}{3}+192\,a^2\,c\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )}-\frac {8\,a\,\left (5\,a^2+3\,c^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )-\frac {e\,x}{2}\right )}{e} \] Input:

int((2*a - 2*a*cos(d + e*x) + 2*c*sin(d + e*x))^3,x)
 

Output:

(8*a*atan((8*a*tan(d/2 + (e*x)/2)*(5*a^2 + 3*c^2))/(24*a*c^2 + 40*a^3))*(5 
*a^2 + 3*c^2))/e - (tan(d/2 + (e*x)/2)*(24*a*c^2 + 40*a^3) + 64*a^2*c - ta 
n(d/2 + (e*x)/2)^5*(24*a*c^2 - 88*a^3) + tan(d/2 + (e*x)/2)^3*(64*a*c^2 + 
(320*a^3)/3) + tan(d/2 + (e*x)/2)^2*(192*a^2*c + 32*c^3) + (32*c^3)/3 + 19 
2*a^2*c*tan(d/2 + (e*x)/2)^4)/(e*(3*tan(d/2 + (e*x)/2)^2 + 3*tan(d/2 + (e* 
x)/2)^4 + tan(d/2 + (e*x)/2)^6 + 1)) - (8*a*(5*a^2 + 3*c^2)*(atan(tan(d/2 
+ (e*x)/2)) - (e*x)/2))/e
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.52 \[ \int (2 a-2 a \cos (d+e x)+2 c \sin (d+e x))^3 \, dx=\frac {-8 \cos \left (e x +d \right )^{3} a^{2} c -\frac {16 \cos \left (e x +d \right )^{3} c^{3}}{3}-8 \cos \left (e x +d \right )^{2} \sin \left (e x +d \right ) a^{3}+12 \cos \left (e x +d \right )^{2} a^{3} e x +24 \cos \left (e x +d \right )^{2} a^{2} c +12 \cos \left (e x +d \right )^{2} a \,c^{2} e x -8 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2} c^{3}+12 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{3}-12 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,c^{2}-24 \cos \left (e x +d \right ) a^{2} c -\frac {16 \sin \left (e x +d \right )^{3} a^{3}}{3}-8 \sin \left (e x +d \right )^{3} a \,c^{2}+12 \sin \left (e x +d \right )^{2} a^{3} e x +12 \sin \left (e x +d \right )^{2} a \,c^{2} e x -24 \sin \left (e x +d \right ) a^{3}+8 a^{3} e x}{e} \] Input:

int((2*a-2*a*cos(e*x+d)+2*c*sin(e*x+d))^3,x)
 

Output:

(4*( - 6*cos(d + e*x)**3*a**2*c - 4*cos(d + e*x)**3*c**3 - 6*cos(d + e*x)* 
*2*sin(d + e*x)*a**3 + 9*cos(d + e*x)**2*a**3*e*x + 18*cos(d + e*x)**2*a** 
2*c + 9*cos(d + e*x)**2*a*c**2*e*x - 6*cos(d + e*x)*sin(d + e*x)**2*c**3 + 
 9*cos(d + e*x)*sin(d + e*x)*a**3 - 9*cos(d + e*x)*sin(d + e*x)*a*c**2 - 1 
8*cos(d + e*x)*a**2*c - 4*sin(d + e*x)**3*a**3 - 6*sin(d + e*x)**3*a*c**2 
+ 9*sin(d + e*x)**2*a**3*e*x + 9*sin(d + e*x)**2*a*c**2*e*x - 18*sin(d + e 
*x)*a**3 + 6*a**3*e*x))/(3*e)