3.245 \(\int (a+a \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=154 \[ -\frac {a^3 (15 B+13 C) \sin ^3(c+d x)}{60 d}+\frac {a^3 (15 B+13 C) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 B+13 C) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (15 B+13 C)+\frac {(5 B-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 a d} \]

[Out]

1/8*a^3*(15*B+13*C)*x+1/5*a^3*(15*B+13*C)*sin(d*x+c)/d+3/40*a^3*(15*B+13*C)*cos(d*x+c)*sin(d*x+c)/d+1/20*(5*B-
C)*(a+a*cos(d*x+c))^3*sin(d*x+c)/d+1/5*C*(a+a*cos(d*x+c))^4*sin(d*x+c)/a/d-1/60*a^3*(15*B+13*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.19, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3023, 2751, 2645, 2637, 2635, 8, 2633} \[ -\frac {a^3 (15 B+13 C) \sin ^3(c+d x)}{60 d}+\frac {a^3 (15 B+13 C) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 B+13 C) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (15 B+13 C)+\frac {(5 B-C) \sin (c+d x) (a \cos (c+d x)+a)^3}{20 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 a d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a^3*(15*B + 13*C)*x)/8 + (a^3*(15*B + 13*C)*Sin[c + d*x])/(5*d) + (3*a^3*(15*B + 13*C)*Cos[c + d*x]*Sin[c + d
*x])/(40*d) + ((5*B - C)*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) + (C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])
/(5*a*d) - (a^3*(15*B + 13*C)*Sin[c + d*x]^3)/(60*d)

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]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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[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]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {\int (a+a \cos (c+d x))^3 (4 a C+a (5 B-C) \cos (c+d x)) \, dx}{5 a}\\ &=\frac {(5 B-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} (15 B+13 C) \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac {(5 B-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} (15 B+13 C) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac {1}{20} a^3 (15 B+13 C) x+\frac {(5 B-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{20} \left (a^3 (15 B+13 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 B+13 C)\right ) \int \cos (c+d x) \, dx+\frac {1}{20} \left (3 a^3 (15 B+13 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {1}{20} a^3 (15 B+13 C) x+\frac {3 a^3 (15 B+13 C) \sin (c+d x)}{20 d}+\frac {3 a^3 (15 B+13 C) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 B-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}+\frac {1}{40} \left (3 a^3 (15 B+13 C)\right ) \int 1 \, dx-\frac {\left (a^3 (15 B+13 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac {1}{8} a^3 (15 B+13 C) x+\frac {a^3 (15 B+13 C) \sin (c+d x)}{5 d}+\frac {3 a^3 (15 B+13 C) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(5 B-C) (a+a \cos (c+d x))^3 \sin (c+d x)}{20 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 a d}-\frac {a^3 (15 B+13 C) \sin ^3(c+d x)}{60 d}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 108, normalized size = 0.70 \[ \frac {a^3 (60 (26 B+23 C) \sin (c+d x)+480 (B+C) \sin (2 (c+d x))+120 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+900 B d x+170 C \sin (3 (c+d x))+45 C \sin (4 (c+d x))+6 C \sin (5 (c+d x))+780 c C+780 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(a^3*(780*c*C + 900*B*d*x + 780*C*d*x + 60*(26*B + 23*C)*Sin[c + d*x] + 480*(B + C)*Sin[2*(c + d*x)] + 120*B*S
in[3*(c + d*x)] + 170*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 45*C*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)
]))/(480*d)

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fricas [A]  time = 0.44, size = 110, normalized size = 0.71 \[ \frac {15 \, {\left (15 \, B + 13 \, C\right )} a^{3} d x + {\left (24 \, C a^{3} \cos \left (d x + c\right )^{4} + 30 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, B + 19 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (45 \, B + 38 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(15*B + 13*C)*a^3*d*x + (24*C*a^3*cos(d*x + c)^4 + 30*(B + 3*C)*a^3*cos(d*x + c)^3 + 8*(15*B + 19*C)
*a^3*cos(d*x + c)^2 + 15*(15*B + 13*C)*a^3*cos(d*x + c) + 8*(45*B + 38*C)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.68, size = 136, normalized size = 0.88 \[ \frac {C a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (15 \, B a^{3} + 13 \, C a^{3}\right )} x + \frac {{\left (B a^{3} + 3 \, C a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (12 \, B a^{3} + 17 \, C a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{3} + C a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{d} + \frac {{\left (26 \, B a^{3} + 23 \, C a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/80*C*a^3*sin(5*d*x + 5*c)/d + 1/8*(15*B*a^3 + 13*C*a^3)*x + 1/32*(B*a^3 + 3*C*a^3)*sin(4*d*x + 4*c)/d + 1/48
*(12*B*a^3 + 17*C*a^3)*sin(3*d*x + 3*c)/d + (B*a^3 + C*a^3)*sin(2*d*x + 2*c)/d + 1/8*(26*B*a^3 + 23*C*a^3)*sin
(d*x + c)/d

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maple [A]  time = 0.25, size = 223, normalized size = 1.45 \[ \frac {\frac {C \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

1/d*(1/5*C*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^3*B*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*c)+3*C*a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+a^3*B*(2+cos(d*x+c)^2)
*sin(d*x+c)+C*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*a^3*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+C*a^3*(1/2*cos
(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^3*B*sin(d*x+c))

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maxima [A]  time = 0.48, size = 213, normalized size = 1.38 \[ -\frac {480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{3} + 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} - 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 480 \, B a^{3} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/480*(480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c
))*B*a^3 - 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x
+ c))*C*a^3 + 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 - 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*C*a^3 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 480*B*a^3*sin(d*x + c))/d

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mupad [B]  time = 2.36, size = 277, normalized size = 1.80 \[ \frac {\left (\frac {15\,B\,a^3}{4}+\frac {13\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {35\,B\,a^3}{2}+\frac {91\,C\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (32\,B\,a^3+\frac {416\,C\,a^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {61\,B\,a^3}{2}+\frac {133\,C\,a^3}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,B\,a^3}{4}+\frac {51\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (15\,B+13\,C\right )}{4\,\left (\frac {15\,B\,a^3}{4}+\frac {13\,C\,a^3}{4}\right )}\right )\,\left (15\,B+13\,C\right )}{4\,d}-\frac {a^3\,\left (15\,B+13\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*((49*B*a^3)/4 + (51*C*a^3)/4) + tan(c/2 + (d*x)/2)^9*((15*B*a^3)/4 + (13*C*a^3)/4) + tan(c
/2 + (d*x)/2)^7*((35*B*a^3)/2 + (91*C*a^3)/6) + tan(c/2 + (d*x)/2)^3*((61*B*a^3)/2 + (133*C*a^3)/6) + tan(c/2
+ (d*x)/2)^5*(32*B*a^3 + (416*C*a^3)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 +
(d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (a^3*atan((a^3*tan(c/2 + (d*x)/2)*(15*B +
13*C))/(4*((15*B*a^3)/4 + (13*C*a^3)/4)))*(15*B + 13*C))/(4*d) - (a^3*(15*B + 13*C)*(atan(tan(c/2 + (d*x)/2))
- (d*x)/2))/(4*d)

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sympy [A]  time = 2.63, size = 532, normalized size = 3.45 \[ \begin {cases} \frac {3 B a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {5 B a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a^{3} \sin {\left (c + d x \right )}}{d} + \frac {9 C a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 C a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {C a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {9 C a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {C a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {8 C a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {9 C a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 C a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {C a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 C a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 C a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {C a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Piecewise((3*B*a**3*x*sin(c + d*x)**4/8 + 3*B*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*B*a**3*x*sin(c + d*
x)**2/2 + 3*B*a**3*x*cos(c + d*x)**4/8 + 3*B*a**3*x*cos(c + d*x)**2/2 + 3*B*a**3*sin(c + d*x)**3*cos(c + d*x)/
(8*d) + 2*B*a**3*sin(c + d*x)**3/d + 5*B*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*B*a**3*sin(c + d*x)*cos(c
+ d*x)**2/d + 3*B*a**3*sin(c + d*x)*cos(c + d*x)/(2*d) + B*a**3*sin(c + d*x)/d + 9*C*a**3*x*sin(c + d*x)**4/8
+ 9*C*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + C*a**3*x*sin(c + d*x)**2/2 + 9*C*a**3*x*cos(c + d*x)**4/8 +
C*a**3*x*cos(c + d*x)**2/2 + 8*C*a**3*sin(c + d*x)**5/(15*d) + 4*C*a**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d)
+ 9*C*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*C*a**3*sin(c + d*x)**3/d + C*a**3*sin(c + d*x)*cos(c + d*x)*
*4/d + 15*C*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*C*a**3*sin(c + d*x)*cos(c + d*x)**2/d + C*a**3*sin(c +
d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)*(a*cos(c) + a)**3, True))

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