3.425 \(\int (a+a \sin (e+f x)) (c+d \sin (e+f x))^4 \, dx\)

Optimal. Leaf size=227 \[ -\frac{a \left (112 c^2 d^2+95 c^3 d+12 c^4+80 c d^3+16 d^4\right ) \cos (e+f x)}{30 f}-\frac{a \left (12 c^2+35 c d+16 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 f}-\frac{a d \left (130 c^2 d+24 c^3+116 c d^2+45 d^3\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac{1}{8} a x \left (24 c^2 d^2+16 c^3 d+8 c^4+12 c d^3+3 d^4\right )-\frac{a \cos (e+f x) (c+d \sin (e+f x))^4}{5 f}-\frac{a (4 c+5 d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 f} \]

[Out]

(a*(8*c^4 + 16*c^3*d + 24*c^2*d^2 + 12*c*d^3 + 3*d^4)*x)/8 - (a*(12*c^4 + 95*c^3*d + 112*c^2*d^2 + 80*c*d^3 +
16*d^4)*Cos[e + f*x])/(30*f) - (a*d*(24*c^3 + 130*c^2*d + 116*c*d^2 + 45*d^3)*Cos[e + f*x]*Sin[e + f*x])/(120*
f) - (a*(12*c^2 + 35*c*d + 16*d^2)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(60*f) - (a*(4*c + 5*d)*Cos[e + f*x]*(
c + d*Sin[e + f*x])^3)/(20*f) - (a*Cos[e + f*x]*(c + d*Sin[e + f*x])^4)/(5*f)

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Rubi [A]  time = 0.281751, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ -\frac{a \left (112 c^2 d^2+95 c^3 d+12 c^4+80 c d^3+16 d^4\right ) \cos (e+f x)}{30 f}-\frac{a \left (12 c^2+35 c d+16 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 f}-\frac{a d \left (130 c^2 d+24 c^3+116 c d^2+45 d^3\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac{1}{8} a x \left (24 c^2 d^2+16 c^3 d+8 c^4+12 c d^3+3 d^4\right )-\frac{a \cos (e+f x) (c+d \sin (e+f x))^4}{5 f}-\frac{a (4 c+5 d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^4,x]

[Out]

(a*(8*c^4 + 16*c^3*d + 24*c^2*d^2 + 12*c*d^3 + 3*d^4)*x)/8 - (a*(12*c^4 + 95*c^3*d + 112*c^2*d^2 + 80*c*d^3 +
16*d^4)*Cos[e + f*x])/(30*f) - (a*d*(24*c^3 + 130*c^2*d + 116*c*d^2 + 45*d^3)*Cos[e + f*x]*Sin[e + f*x])/(120*
f) - (a*(12*c^2 + 35*c*d + 16*d^2)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(60*f) - (a*(4*c + 5*d)*Cos[e + f*x]*(
c + d*Sin[e + f*x])^3)/(20*f) - (a*Cos[e + f*x]*(c + d*Sin[e + f*x])^4)/(5*f)

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 (a+a \sin (e+f x)) (c+d \sin (e+f x))^4 \, dx &=-\frac{a \cos (e+f x) (c+d \sin (e+f x))^4}{5 f}+\frac{1}{5} \int (c+d \sin (e+f x))^3 (a (5 c+4 d)+a (4 c+5 d) \sin (e+f x)) \, dx\\ &=-\frac{a (4 c+5 d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 f}-\frac{a \cos (e+f x) (c+d \sin (e+f x))^4}{5 f}+\frac{1}{20} \int (c+d \sin (e+f x))^2 \left (a \left (20 c^2+28 c d+15 d^2\right )+a \left (12 c^2+35 c d+16 d^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac{a \left (12 c^2+35 c d+16 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 f}-\frac{a (4 c+5 d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 f}-\frac{a \cos (e+f x) (c+d \sin (e+f x))^4}{5 f}+\frac{1}{60} \int (c+d \sin (e+f x)) \left (a \left (60 c^3+108 c^2 d+115 c d^2+32 d^3\right )+a \left (24 c^3+130 c^2 d+116 c d^2+45 d^3\right ) \sin (e+f x)\right ) \, dx\\ &=\frac{1}{8} a \left (8 c^4+16 c^3 d+24 c^2 d^2+12 c d^3+3 d^4\right ) x-\frac{a \left (12 c^4+95 c^3 d+112 c^2 d^2+80 c d^3+16 d^4\right ) \cos (e+f x)}{30 f}-\frac{a d \left (24 c^3+130 c^2 d+116 c d^2+45 d^3\right ) \cos (e+f x) \sin (e+f x)}{120 f}-\frac{a \left (12 c^2+35 c d+16 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^2}{60 f}-\frac{a (4 c+5 d) \cos (e+f x) (c+d \sin (e+f x))^3}{20 f}-\frac{a \cos (e+f x) (c+d \sin (e+f x))^4}{5 f}\\ \end{align*}

Mathematica [A]  time = 1.39053, size = 207, normalized size = 0.91 \[ \frac{a (\sin (e+f x)+1) \left (15 \left (-8 d \left (6 c^2 d+4 c^3+4 c d^2+d^3\right ) \sin (2 (e+f x))+4 f x \left (24 c^2 d^2+16 c^3 d+8 c^4+12 c d^3+3 d^4\right )+d^3 (4 c+d) \sin (4 (e+f x))\right )+10 d^2 \left (24 c^2+16 c d+5 d^2\right ) \cos (3 (e+f x))-60 \left (36 c^2 d^2+32 c^3 d+8 c^4+24 c d^3+5 d^4\right ) \cos (e+f x)-6 d^4 \cos (5 (e+f x))\right )}{480 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^4,x]

[Out]

(a*(1 + Sin[e + f*x])*(-60*(8*c^4 + 32*c^3*d + 36*c^2*d^2 + 24*c*d^3 + 5*d^4)*Cos[e + f*x] + 10*d^2*(24*c^2 +
16*c*d + 5*d^2)*Cos[3*(e + f*x)] - 6*d^4*Cos[5*(e + f*x)] + 15*(4*(8*c^4 + 16*c^3*d + 24*c^2*d^2 + 12*c*d^3 +
3*d^4)*f*x - 8*d*(4*c^3 + 6*c^2*d + 4*c*d^2 + d^3)*Sin[2*(e + f*x)] + d^3*(4*c + d)*Sin[4*(e + f*x)])))/(480*f
*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)

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Maple [A]  time = 0.046, size = 259, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( -a{c}^{4}\cos \left ( fx+e \right ) +4\,a{c}^{3}d \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -2\,a{c}^{2}{d}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +4\,ac{d}^{3} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{a{d}^{4}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+a{c}^{4} \left ( fx+e \right ) -4\,a{c}^{3}d\cos \left ( fx+e \right ) +6\,a{c}^{2}{d}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{\frac{4\,ac{d}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+a{d}^{4} \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(c+d*sin(f*x+e))^4,x)

[Out]

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

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Maxima [A]  time = 1.05633, size = 338, normalized size = 1.49 \begin{align*} \frac{480 \,{\left (f x + e\right )} a c^{4} + 480 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c^{3} d + 960 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c^{2} d^{2} + 720 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c^{2} d^{2} + 640 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c d^{3} + 60 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a c d^{3} - 32 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a d^{4} + 15 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a d^{4} - 480 \, a c^{4} \cos \left (f x + e\right ) - 1920 \, a c^{3} d \cos \left (f x + e\right )}{480 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(480*(f*x + e)*a*c^4 + 480*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*c^3*d + 960*(cos(f*x + e)^3 - 3*cos(f*x +
e))*a*c^2*d^2 + 720*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*c^2*d^2 + 640*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*c*d^3
+ 60*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a*c*d^3 - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)
^3 + 15*cos(f*x + e))*a*d^4 + 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a*d^4 - 480*a*c^4*cos
(f*x + e) - 1920*a*c^3*d*cos(f*x + e))/f

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Fricas [A]  time = 1.1881, size = 481, normalized size = 2.12 \begin{align*} -\frac{24 \, a d^{4} \cos \left (f x + e\right )^{5} - 80 \,{\left (3 \, a c^{2} d^{2} + 2 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (8 \, a c^{4} + 16 \, a c^{3} d + 24 \, a c^{2} d^{2} + 12 \, a c d^{3} + 3 \, a d^{4}\right )} f x + 120 \,{\left (a c^{4} + 4 \, a c^{3} d + 6 \, a c^{2} d^{2} + 4 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right ) - 15 \,{\left (2 \,{\left (4 \, a c d^{3} + a d^{4}\right )} \cos \left (f x + e\right )^{3} -{\left (16 \, a c^{3} d + 24 \, a c^{2} d^{2} + 20 \, a c d^{3} + 5 \, a d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(24*a*d^4*cos(f*x + e)^5 - 80*(3*a*c^2*d^2 + 2*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - 15*(8*a*c^4 + 16*a*c^3
*d + 24*a*c^2*d^2 + 12*a*c*d^3 + 3*a*d^4)*f*x + 120*(a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4)*cos(
f*x + e) - 15*(2*(4*a*c*d^3 + a*d^4)*cos(f*x + e)^3 - (16*a*c^3*d + 24*a*c^2*d^2 + 20*a*c*d^3 + 5*a*d^4)*cos(f
*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 4.01411, size = 580, normalized size = 2.56 \begin{align*} \begin{cases} a c^{4} x - \frac{a c^{4} \cos{\left (e + f x \right )}}{f} + 2 a c^{3} d x \sin ^{2}{\left (e + f x \right )} + 2 a c^{3} d x \cos ^{2}{\left (e + f x \right )} - \frac{2 a c^{3} d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 a c^{3} d \cos{\left (e + f x \right )}}{f} + 3 a c^{2} d^{2} x \sin ^{2}{\left (e + f x \right )} + 3 a c^{2} d^{2} x \cos ^{2}{\left (e + f x \right )} - \frac{6 a c^{2} d^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a c^{2} d^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 a c^{2} d^{2} \cos ^{3}{\left (e + f x \right )}}{f} + \frac{3 a c d^{3} x \sin ^{4}{\left (e + f x \right )}}{2} + 3 a c d^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )} + \frac{3 a c d^{3} x \cos ^{4}{\left (e + f x \right )}}{2} - \frac{5 a c d^{3} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{4 a c d^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a c d^{3} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{2 f} - \frac{8 a c d^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{3 a d^{4} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 a d^{4} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{3 a d^{4} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{a d^{4} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{5 a d^{4} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{4 a d^{4} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{3 a d^{4} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{8 a d^{4} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text{for}\: f \neq 0 \\x \left (c + d \sin{\left (e \right )}\right )^{4} \left (a \sin{\left (e \right )} + a\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(c+d*sin(f*x+e))**4,x)

[Out]

Piecewise((a*c**4*x - a*c**4*cos(e + f*x)/f + 2*a*c**3*d*x*sin(e + f*x)**2 + 2*a*c**3*d*x*cos(e + f*x)**2 - 2*
a*c**3*d*sin(e + f*x)*cos(e + f*x)/f - 4*a*c**3*d*cos(e + f*x)/f + 3*a*c**2*d**2*x*sin(e + f*x)**2 + 3*a*c**2*
d**2*x*cos(e + f*x)**2 - 6*a*c**2*d**2*sin(e + f*x)**2*cos(e + f*x)/f - 3*a*c**2*d**2*sin(e + f*x)*cos(e + f*x
)/f - 4*a*c**2*d**2*cos(e + f*x)**3/f + 3*a*c*d**3*x*sin(e + f*x)**4/2 + 3*a*c*d**3*x*sin(e + f*x)**2*cos(e +
f*x)**2 + 3*a*c*d**3*x*cos(e + f*x)**4/2 - 5*a*c*d**3*sin(e + f*x)**3*cos(e + f*x)/(2*f) - 4*a*c*d**3*sin(e +
f*x)**2*cos(e + f*x)/f - 3*a*c*d**3*sin(e + f*x)*cos(e + f*x)**3/(2*f) - 8*a*c*d**3*cos(e + f*x)**3/(3*f) + 3*
a*d**4*x*sin(e + f*x)**4/8 + 3*a*d**4*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + 3*a*d**4*x*cos(e + f*x)**4/8 - a*d
**4*sin(e + f*x)**4*cos(e + f*x)/f - 5*a*d**4*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 4*a*d**4*sin(e + f*x)**2*co
s(e + f*x)**3/(3*f) - 3*a*d**4*sin(e + f*x)*cos(e + f*x)**3/(8*f) - 8*a*d**4*cos(e + f*x)**5/(15*f), Ne(f, 0))
, (x*(c + d*sin(e))**4*(a*sin(e) + a), True))

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Giac [A]  time = 1.29283, size = 367, normalized size = 1.62 \begin{align*} -\frac{a d^{4} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{a c d^{3} \cos \left (3 \, f x + 3 \, e\right )}{3 \, f} + \frac{a c d^{3} \sin \left (4 \, f x + 4 \, e\right )}{8 \, f} + \frac{a d^{4} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{1}{8} \,{\left (8 \, a c^{4} + 24 \, a c^{2} d^{2} + 3 \, a d^{4}\right )} x + \frac{1}{2} \,{\left (4 \, a c^{3} d + 3 \, a c d^{3}\right )} x + \frac{{\left (24 \, a c^{2} d^{2} + 5 \, a d^{4}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (8 \, a c^{4} + 36 \, a c^{2} d^{2} + 5 \, a d^{4}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (4 \, a c^{3} d + 3 \, a c d^{3}\right )} \cos \left (f x + e\right )}{f} - \frac{{\left (a c^{3} d + a c d^{3}\right )} \sin \left (2 \, f x + 2 \, e\right )}{f} - \frac{{\left (6 \, a c^{2} d^{2} + a d^{4}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*a*d^4*cos(5*f*x + 5*e)/f + 1/3*a*c*d^3*cos(3*f*x + 3*e)/f + 1/8*a*c*d^3*sin(4*f*x + 4*e)/f + 1/32*a*d^4*
sin(4*f*x + 4*e)/f + 1/8*(8*a*c^4 + 24*a*c^2*d^2 + 3*a*d^4)*x + 1/2*(4*a*c^3*d + 3*a*c*d^3)*x + 1/48*(24*a*c^2
*d^2 + 5*a*d^4)*cos(3*f*x + 3*e)/f - 1/8*(8*a*c^4 + 36*a*c^2*d^2 + 5*a*d^4)*cos(f*x + e)/f - (4*a*c^3*d + 3*a*
c*d^3)*cos(f*x + e)/f - (a*c^3*d + a*c*d^3)*sin(2*f*x + 2*e)/f - 1/4*(6*a*c^2*d^2 + a*d^4)*sin(2*f*x + 2*e)/f