3.445 \(\int (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=164 \[ \frac{a^3 \left (c^2+6 c d+5 d^2\right ) \cos ^3(e+f x)}{3 f}-\frac{a^3 \left (12 c^2+30 c d+13 d^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a^3 x \left (20 c^2+30 c d+13 d^2\right )-\frac{4 a^3 (c+d)^2 \cos (e+f x)}{f}-\frac{a^3 d (2 c+3 d) \sin ^3(e+f x) \cos (e+f x)}{4 f}-\frac{a^3 d^2 \cos ^5(e+f x)}{5 f} \]

[Out]

(a^3*(20*c^2 + 30*c*d + 13*d^2)*x)/8 - (4*a^3*(c + d)^2*Cos[e + f*x])/f + (a^3*(c^2 + 6*c*d + 5*d^2)*Cos[e + f
*x]^3)/(3*f) - (a^3*d^2*Cos[e + f*x]^5)/(5*f) - (a^3*(12*c^2 + 30*c*d + 13*d^2)*Cos[e + f*x]*Sin[e + f*x])/(8*
f) - (a^3*d*(2*c + 3*d)*Cos[e + f*x]*Sin[e + f*x]^3)/(4*f)

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Rubi [A]  time = 0.259766, antiderivative size = 189, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2761, 2751, 2645, 2638, 2635, 8, 2633} \[ \frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos ^3(e+f x)}{60 f}-\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos (e+f x)}{5 f}-\frac{3 a^3 \left (20 c^2+30 c d+13 d^2\right ) \sin (e+f x) \cos (e+f x)}{40 f}+\frac{1}{8} a^3 x \left (20 c^2+30 c d+13 d^2\right )-\frac{d (10 c-d) \cos (e+f x) (a \sin (e+f x)+a)^3}{20 f}-\frac{d^2 \cos (e+f x) (a \sin (e+f x)+a)^4}{5 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(20*c^2 + 30*c*d + 13*d^2)*x)/8 - (a^3*(20*c^2 + 30*c*d + 13*d^2)*Cos[e + f*x])/(5*f) + (a^3*(20*c^2 + 30
*c*d + 13*d^2)*Cos[e + f*x]^3)/(60*f) - (3*a^3*(20*c^2 + 30*c*d + 13*d^2)*Cos[e + f*x]*Sin[e + f*x])/(40*f) -
((10*c - d)*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^3)/(20*f) - (d^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^4)/(5*a*f)

Rule 2761

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[(
d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x]
)^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d
, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 2751

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

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

Rule 8

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

Rule 2633

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2 \, dx &=-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^4}{5 a f}+\frac{\int (a+a \sin (e+f x))^3 \left (a \left (5 c^2+4 d^2\right )+a (10 c-d) d \sin (e+f x)\right ) \, dx}{5 a}\\ &=-\frac{(10 c-d) d \cos (e+f x) (a+a \sin (e+f x))^3}{20 f}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^4}{5 a f}+\frac{1}{20} \left (20 c^2+30 c d+13 d^2\right ) \int (a+a \sin (e+f x))^3 \, dx\\ &=-\frac{(10 c-d) d \cos (e+f x) (a+a \sin (e+f x))^3}{20 f}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^4}{5 a f}+\frac{1}{20} \left (20 c^2+30 c d+13 d^2\right ) \int \left (a^3+3 a^3 \sin (e+f x)+3 a^3 \sin ^2(e+f x)+a^3 \sin ^3(e+f x)\right ) \, dx\\ &=\frac{1}{20} a^3 \left (20 c^2+30 c d+13 d^2\right ) x-\frac{(10 c-d) d \cos (e+f x) (a+a \sin (e+f x))^3}{20 f}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^4}{5 a f}+\frac{1}{20} \left (a^3 \left (20 c^2+30 c d+13 d^2\right )\right ) \int \sin ^3(e+f x) \, dx+\frac{1}{20} \left (3 a^3 \left (20 c^2+30 c d+13 d^2\right )\right ) \int \sin (e+f x) \, dx+\frac{1}{20} \left (3 a^3 \left (20 c^2+30 c d+13 d^2\right )\right ) \int \sin ^2(e+f x) \, dx\\ &=\frac{1}{20} a^3 \left (20 c^2+30 c d+13 d^2\right ) x-\frac{3 a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos (e+f x)}{20 f}-\frac{3 a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos (e+f x) \sin (e+f x)}{40 f}-\frac{(10 c-d) d \cos (e+f x) (a+a \sin (e+f x))^3}{20 f}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^4}{5 a f}+\frac{1}{40} \left (3 a^3 \left (20 c^2+30 c d+13 d^2\right )\right ) \int 1 \, dx-\frac{\left (a^3 \left (20 c^2+30 c d+13 d^2\right )\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{20 f}\\ &=\frac{1}{8} a^3 \left (20 c^2+30 c d+13 d^2\right ) x-\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos (e+f x)}{5 f}+\frac{a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos ^3(e+f x)}{60 f}-\frac{3 a^3 \left (20 c^2+30 c d+13 d^2\right ) \cos (e+f x) \sin (e+f x)}{40 f}-\frac{(10 c-d) d \cos (e+f x) (a+a \sin (e+f x))^3}{20 f}-\frac{d^2 \cos (e+f x) (a+a \sin (e+f x))^4}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.729981, size = 177, normalized size = 1.08 \[ -\frac{a^3 \cos (e+f x) \left (30 \left (20 c^2+30 c d+13 d^2\right ) \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\sqrt{\cos ^2(e+f x)} \left (8 \left (5 c^2+30 c d+19 d^2\right ) \sin ^2(e+f x)+15 \left (12 c^2+30 c d+13 d^2\right ) \sin (e+f x)+8 \left (55 c^2+90 c d+38 d^2\right )+30 d (2 c+3 d) \sin ^3(e+f x)+24 d^2 \sin ^4(e+f x)\right )\right )}{120 f \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Cos[e + f*x]*(30*(20*c^2 + 30*c*d + 13*d^2)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x]^2
]*(8*(55*c^2 + 90*c*d + 38*d^2) + 15*(12*c^2 + 30*c*d + 13*d^2)*Sin[e + f*x] + 8*(5*c^2 + 30*c*d + 19*d^2)*Sin
[e + f*x]^2 + 30*d*(2*c + 3*d)*Sin[e + f*x]^3 + 24*d^2*Sin[e + f*x]^4)))/(120*f*Sqrt[Cos[e + f*x]^2])

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Maple [B]  time = 0.045, size = 319, normalized size = 2. \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{3}{c}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+2\,{a}^{3}cd \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}^{3}{d}^{2}\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 ) }+3\,{a}^{3}{c}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -2\,{a}^{3}cd \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +3\,{a}^{3}{d}^{2} \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 ) -3\,{a}^{3}{c}^{2}\cos \left ( fx+e \right ) +6\,{a}^{3}cd \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{a}^{3}{d}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +{a}^{3}{c}^{2} \left ( fx+e \right ) -2\,{a}^{3}cd\cos \left ( fx+e \right ) +{a}^{3}{d}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16491, size = 416, normalized size = 2.54 \begin{align*} \frac{160 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} c^{2} + 360 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c^{2} + 480 \,{\left (f x + e\right )} a^{3} c^{2} + 960 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} c d + 30 \,{\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^{3} c d + 720 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c d - 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^{3} d^{2} + 480 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} d^{2} + 45 \,{\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^{3} d^{2} + 120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} d^{2} - 1440 \, a^{3} c^{2} \cos \left (f x + e\right ) - 960 \, a^{3} c 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))^3*(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/480*(160*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^3*c^2 + 360*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3*c^2 + 480*(f*x
+ e)*a^3*c^2 + 960*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^3*c*d + 30*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2
*f*x + 2*e))*a^3*c*d + 720*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3*c*d - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3
+ 15*cos(f*x + e))*a^3*d^2 + 480*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^3*d^2 + 45*(12*f*x + 12*e + sin(4*f*x +
4*e) - 8*sin(2*f*x + 2*e))*a^3*d^2 + 120*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3*d^2 - 1440*a^3*c^2*cos(f*x + e)
- 960*a^3*c*d*cos(f*x + e))/f

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Fricas [A]  time = 2.03207, size = 413, normalized size = 2.52 \begin{align*} -\frac{24 \, a^{3} d^{2} \cos \left (f x + e\right )^{5} - 40 \,{\left (a^{3} c^{2} + 6 \, a^{3} c d + 5 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (20 \, a^{3} c^{2} + 30 \, a^{3} c d + 13 \, a^{3} d^{2}\right )} f x + 480 \,{\left (a^{3} c^{2} + 2 \, a^{3} c d + a^{3} d^{2}\right )} \cos \left (f x + e\right ) - 15 \,{\left (2 \,{\left (2 \, a^{3} c d + 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{3} -{\left (12 \, a^{3} c^{2} + 34 \, a^{3} c d + 19 \, a^{3} d^{2}\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))^3*(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/120*(24*a^3*d^2*cos(f*x + e)^5 - 40*(a^3*c^2 + 6*a^3*c*d + 5*a^3*d^2)*cos(f*x + e)^3 - 15*(20*a^3*c^2 + 30*
a^3*c*d + 13*a^3*d^2)*f*x + 480*(a^3*c^2 + 2*a^3*c*d + a^3*d^2)*cos(f*x + e) - 15*(2*(2*a^3*c*d + 3*a^3*d^2)*c
os(f*x + e)^3 - (12*a^3*c^2 + 34*a^3*c*d + 19*a^3*d^2)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 4.03135, size = 702, normalized size = 4.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**3*c**2*x*sin(e + f*x)**2/2 + 3*a**3*c**2*x*cos(e + f*x)**2/2 + a**3*c**2*x - a**3*c**2*sin(e +
f*x)**2*cos(e + f*x)/f - 3*a**3*c**2*sin(e + f*x)*cos(e + f*x)/(2*f) - 2*a**3*c**2*cos(e + f*x)**3/(3*f) - 3*
a**3*c**2*cos(e + f*x)/f + 3*a**3*c*d*x*sin(e + f*x)**4/4 + 3*a**3*c*d*x*sin(e + f*x)**2*cos(e + f*x)**2/2 + 3
*a**3*c*d*x*sin(e + f*x)**2 + 3*a**3*c*d*x*cos(e + f*x)**4/4 + 3*a**3*c*d*x*cos(e + f*x)**2 - 5*a**3*c*d*sin(e
+ f*x)**3*cos(e + f*x)/(4*f) - 6*a**3*c*d*sin(e + f*x)**2*cos(e + f*x)/f - 3*a**3*c*d*sin(e + f*x)*cos(e + f*
x)**3/(4*f) - 3*a**3*c*d*sin(e + f*x)*cos(e + f*x)/f - 4*a**3*c*d*cos(e + f*x)**3/f - 2*a**3*c*d*cos(e + f*x)/
f + 9*a**3*d**2*x*sin(e + f*x)**4/8 + 9*a**3*d**2*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + a**3*d**2*x*sin(e + f*
x)**2/2 + 9*a**3*d**2*x*cos(e + f*x)**4/8 + a**3*d**2*x*cos(e + f*x)**2/2 - a**3*d**2*sin(e + f*x)**4*cos(e +
f*x)/f - 15*a**3*d**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 4*a**3*d**2*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) -
3*a**3*d**2*sin(e + f*x)**2*cos(e + f*x)/f - 9*a**3*d**2*sin(e + f*x)*cos(e + f*x)**3/(8*f) - a**3*d**2*sin(e
+ f*x)*cos(e + f*x)/(2*f) - 8*a**3*d**2*cos(e + f*x)**5/(15*f) - 2*a**3*d**2*cos(e + f*x)**3/f, Ne(f, 0)), (x
*(c + d*sin(e))**2*(a*sin(e) + a)**3, True))

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Giac [A]  time = 1.31324, size = 339, normalized size = 2.07 \begin{align*} -\frac{a^{3} d^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac{2 \, a^{3} c d \cos \left (f x + e\right )}{f} - \frac{a^{3} d^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac{3}{8} \,{\left (4 \, a^{3} c^{2} + 10 \, a^{3} c d + 3 \, a^{3} d^{2}\right )} x + \frac{1}{2} \,{\left (2 \, a^{3} c^{2} + a^{3} d^{2}\right )} x + \frac{{\left (4 \, a^{3} c^{2} + 24 \, a^{3} c d + 17 \, a^{3} d^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (30 \, a^{3} c^{2} + 36 \, a^{3} c d + 23 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )}{8 \, f} + \frac{{\left (2 \, a^{3} c d + 3 \, a^{3} d^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac{{\left (3 \, a^{3} c^{2} + 8 \, a^{3} c d + 3 \, a^{3} d^{2}\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))^3*(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/80*a^3*d^2*cos(5*f*x + 5*e)/f - 2*a^3*c*d*cos(f*x + e)/f - 1/4*a^3*d^2*sin(2*f*x + 2*e)/f + 3/8*(4*a^3*c^2
+ 10*a^3*c*d + 3*a^3*d^2)*x + 1/2*(2*a^3*c^2 + a^3*d^2)*x + 1/48*(4*a^3*c^2 + 24*a^3*c*d + 17*a^3*d^2)*cos(3*f
*x + 3*e)/f - 1/8*(30*a^3*c^2 + 36*a^3*c*d + 23*a^3*d^2)*cos(f*x + e)/f + 1/32*(2*a^3*c*d + 3*a^3*d^2)*sin(4*f
*x + 4*e)/f - 1/4*(3*a^3*c^2 + 8*a^3*c*d + 3*a^3*d^2)*sin(2*f*x + 2*e)/f