3.986 \(\int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=134 \[ -\frac {B (a \sin (c+d x)+a)^{11}}{11 a^8 d}-\frac {(A-7 B) (a \sin (c+d x)+a)^{10}}{10 a^7 d}+\frac {2 (A-3 B) (a \sin (c+d x)+a)^9}{3 a^6 d}-\frac {(3 A-5 B) (a \sin (c+d x)+a)^8}{2 a^5 d}+\frac {8 (A-B) (a \sin (c+d x)+a)^7}{7 a^4 d} \]

[Out]

8/7*(A-B)*(a+a*sin(d*x+c))^7/a^4/d-1/2*(3*A-5*B)*(a+a*sin(d*x+c))^8/a^5/d+2/3*(A-3*B)*(a+a*sin(d*x+c))^9/a^6/d
-1/10*(A-7*B)*(a+a*sin(d*x+c))^10/a^7/d-1/11*B*(a+a*sin(d*x+c))^11/a^8/d

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Rubi [A]  time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2836, 77} \[ -\frac {(A-7 B) (a \sin (c+d x)+a)^{10}}{10 a^7 d}+\frac {2 (A-3 B) (a \sin (c+d x)+a)^9}{3 a^6 d}-\frac {(3 A-5 B) (a \sin (c+d x)+a)^8}{2 a^5 d}+\frac {8 (A-B) (a \sin (c+d x)+a)^7}{7 a^4 d}-\frac {B (a \sin (c+d x)+a)^{11}}{11 a^8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(8*(A - B)*(a + a*Sin[c + d*x])^7)/(7*a^4*d) - ((3*A - 5*B)*(a + a*Sin[c + d*x])^8)/(2*a^5*d) + (2*(A - 3*B)*(
a + a*Sin[c + d*x])^9)/(3*a^6*d) - ((A - 7*B)*(a + a*Sin[c + d*x])^10)/(10*a^7*d) - (B*(a + a*Sin[c + d*x])^11
)/(11*a^8*d)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a-x)^3 (a+x)^6 \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (8 a^3 (A-B) (a+x)^6-4 a^2 (3 A-5 B) (a+x)^7+6 a (A-3 B) (a+x)^8+(-A+7 B) (a+x)^9-\frac {B (a+x)^{10}}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac {8 (A-B) (a+a \sin (c+d x))^7}{7 a^4 d}-\frac {(3 A-5 B) (a+a \sin (c+d x))^8}{2 a^5 d}+\frac {2 (A-3 B) (a+a \sin (c+d x))^9}{3 a^6 d}-\frac {(A-7 B) (a+a \sin (c+d x))^{10}}{10 a^7 d}-\frac {B (a+a \sin (c+d x))^{11}}{11 a^8 d}\\ \end {align*}

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Mathematica [A]  time = 1.53, size = 86, normalized size = 0.64 \[ -\frac {a^3 (\sin (c+d x)+1)^7 \left (21 (11 A-37 B) \sin ^3(c+d x)+(1029 B-847 A) \sin ^2(c+d x)+14 (77 A-39 B) \sin (c+d x)-484 A+210 B \sin ^4(c+d x)+78 B\right )}{2310 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-1/2310*(a^3*(1 + Sin[c + d*x])^7*(-484*A + 78*B + 14*(77*A - 39*B)*Sin[c + d*x] + (-847*A + 1029*B)*Sin[c + d
*x]^2 + 21*(11*A - 37*B)*Sin[c + d*x]^3 + 210*B*Sin[c + d*x]^4))/d

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fricas [A]  time = 0.75, size = 155, normalized size = 1.16 \[ \frac {231 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{10} - 1155 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{8} + 2 \, {\left (105 \, B a^{3} \cos \left (d x + c\right )^{10} - 35 \, {\left (11 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} + 20 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 24 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 32 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 64 \, {\left (11 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2310 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2310*(231*(A + 3*B)*a^3*cos(d*x + c)^10 - 1155*(A + B)*a^3*cos(d*x + c)^8 + 2*(105*B*a^3*cos(d*x + c)^10 - 3
5*(11*A + 15*B)*a^3*cos(d*x + c)^8 + 20*(11*A + 3*B)*a^3*cos(d*x + c)^6 + 24*(11*A + 3*B)*a^3*cos(d*x + c)^4 +
32*(11*A + 3*B)*a^3*cos(d*x + c)^2 + 64*(11*A + 3*B)*a^3)*sin(d*x + c))/d

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giac [B]  time = 0.89, size = 283, normalized size = 2.11 \[ \frac {B a^{3} \sin \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {{\left (A a^{3} - B a^{3}\right )} \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac {{\left (23 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {{\left (11 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {7 \, {\left (13 \, A a^{3} + 7 \, B a^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {{\left (4 \, A a^{3} + 3 \, B a^{3}\right )} \sin \left (9 \, d x + 9 \, c\right )}{3072 \, d} - \frac {{\left (44 \, A a^{3} + 61 \, B a^{3}\right )} \sin \left (7 \, d x + 7 \, c\right )}{7168 \, d} + \frac {{\left (16 \, A a^{3} - 107 \, B a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{5120 \, d} + \frac {{\left (56 \, A a^{3} - B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{512 \, d} + \frac {91 \, {\left (4 \, A a^{3} + B a^{3}\right )} \sin \left (d x + c\right )}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/11264*B*a^3*sin(11*d*x + 11*c)/d + 1/5120*(A*a^3 + 3*B*a^3)*cos(10*d*x + 10*c)/d - 1/512*(A*a^3 - B*a^3)*cos
(8*d*x + 8*c)/d - 1/1024*(23*A*a^3 + 5*B*a^3)*cos(6*d*x + 6*c)/d - 1/128*(11*A*a^3 + 5*B*a^3)*cos(4*d*x + 4*c)
/d - 7/512*(13*A*a^3 + 7*B*a^3)*cos(2*d*x + 2*c)/d - 1/3072*(4*A*a^3 + 3*B*a^3)*sin(9*d*x + 9*c)/d - 1/7168*(4
4*A*a^3 + 61*B*a^3)*sin(7*d*x + 7*c)/d + 1/5120*(16*A*a^3 - 107*B*a^3)*sin(5*d*x + 5*c)/d + 1/512*(56*A*a^3 -
B*a^3)*sin(3*d*x + 3*c)/d + 91/512*(4*A*a^3 + B*a^3)*sin(d*x + c)/d

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maple [B]  time = 0.49, size = 345, normalized size = 2.57 \[ \frac {a^{3} A \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{8}\left (d x +c \right )\right )}{40}\right )+B \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{11}-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{33}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{231}\right )+3 a^{3} A \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+3 B \,a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{8}\left (d x +c \right )\right )}{40}\right )-\frac {3 a^{3} A \left (\cos ^{8}\left (d x +c \right )\right )}{8}+3 B \,a^{3} \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+\frac {a^{3} A \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}-\frac {B \,a^{3} \left (\cos ^{8}\left (d x +c \right )\right )}{8}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(-1/10*sin(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8)+B*a^3*(-1/11*sin(d*x+c)^3*cos(d*x+c)^8-1/33*cos
(d*x+c)^8*sin(d*x+c)+1/231*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+3*a^3*A*(-1/9*cos
(d*x+c)^8*sin(d*x+c)+1/63*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+3*B*a^3*(-1/10*sin
(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8)-3/8*a^3*A*cos(d*x+c)^8+3*B*a^3*(-1/9*cos(d*x+c)^8*sin(d*x+c)+1/63*(1
6/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+1/7*a^3*A*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+
8/5*cos(d*x+c)^2)*sin(d*x+c)-1/8*B*a^3*cos(d*x+c)^8)

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maxima [A]  time = 0.51, size = 182, normalized size = 1.36 \[ -\frac {210 \, B a^{3} \sin \left (d x + c\right )^{11} + 231 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{10} + 770 \, A a^{3} \sin \left (d x + c\right )^{9} - 2310 \, B a^{3} \sin \left (d x + c\right )^{8} - 660 \, {\left (4 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{7} - 2310 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{6} + 924 \, {\left (3 \, A + 4 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 4620 \, A a^{3} \sin \left (d x + c\right )^{4} - 2310 \, B a^{3} \sin \left (d x + c\right )^{3} - 1155 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 2310 \, A a^{3} \sin \left (d x + c\right )}{2310 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2310*(210*B*a^3*sin(d*x + c)^11 + 231*(A + 3*B)*a^3*sin(d*x + c)^10 + 770*A*a^3*sin(d*x + c)^9 - 2310*B*a^3
*sin(d*x + c)^8 - 660*(4*A + 3*B)*a^3*sin(d*x + c)^7 - 2310*(A - B)*a^3*sin(d*x + c)^6 + 924*(3*A + 4*B)*a^3*s
in(d*x + c)^5 + 4620*A*a^3*sin(d*x + c)^4 - 2310*B*a^3*sin(d*x + c)^3 - 1155*(3*A + B)*a^3*sin(d*x + c)^2 - 23
10*A*a^3*sin(d*x + c))/d

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mupad [B]  time = 0.20, size = 177, normalized size = 1.32 \[ \frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}-\frac {A\,a^3\,{\sin \left (c+d\,x\right )}^9}{3}-2\,A\,a^3\,{\sin \left (c+d\,x\right )}^4+a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A-B\right )-\frac {a^3\,{\sin \left (c+d\,x\right )}^{10}\,\left (A+3\,B\right )}{10}+B\,a^3\,{\sin \left (c+d\,x\right )}^3+B\,a^3\,{\sin \left (c+d\,x\right )}^8-\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^{11}}{11}-\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^5\,\left (3\,A+4\,B\right )}{5}+\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^7\,\left (4\,A+3\,B\right )}{7}+A\,a^3\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)

[Out]

((a^3*sin(c + d*x)^2*(3*A + B))/2 - (A*a^3*sin(c + d*x)^9)/3 - 2*A*a^3*sin(c + d*x)^4 + a^3*sin(c + d*x)^6*(A
- B) - (a^3*sin(c + d*x)^10*(A + 3*B))/10 + B*a^3*sin(c + d*x)^3 + B*a^3*sin(c + d*x)^8 - (B*a^3*sin(c + d*x)^
11)/11 - (2*a^3*sin(c + d*x)^5*(3*A + 4*B))/5 + (2*a^3*sin(c + d*x)^7*(4*A + 3*B))/7 + A*a^3*sin(c + d*x))/d

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sympy [A]  time = 38.22, size = 636, normalized size = 4.75 \[ \begin {cases} \frac {A a^{3} \sin ^{10}{\left (c + d x \right )}}{40 d} + \frac {16 A a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac {A a^{3} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} + \frac {24 A a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {16 A a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {A a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {6 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {3 A a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {16 B a^{3} \sin ^{11}{\left (c + d x \right )}}{1155 d} + \frac {3 B a^{3} \sin ^{10}{\left (c + d x \right )}}{40 d} + \frac {8 B a^{3} \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{105 d} + \frac {16 B a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac {3 B a^{3} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} + \frac {6 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{35 d} + \frac {24 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {3 B a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{5 d} + \frac {6 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {B a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{7}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((A*a**3*sin(c + d*x)**10/(40*d) + 16*A*a**3*sin(c + d*x)**9/(105*d) + A*a**3*sin(c + d*x)**8*cos(c +
d*x)**2/(8*d) + 24*A*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 16*A*a**3*sin(c + d*x)**7/(35*d) + A*a**3*
sin(c + d*x)**6*cos(c + d*x)**4/(4*d) + 6*A*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 8*A*a**3*sin(c + d*x)
**5*cos(c + d*x)**2/(5*d) + A*a**3*sin(c + d*x)**4*cos(c + d*x)**6/(4*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)
**6/d + 2*A*a**3*sin(c + d*x)**3*cos(c + d*x)**4/d + A*a**3*sin(c + d*x)*cos(c + d*x)**6/d - 3*A*a**3*cos(c +
d*x)**8/(8*d) + 16*B*a**3*sin(c + d*x)**11/(1155*d) + 3*B*a**3*sin(c + d*x)**10/(40*d) + 8*B*a**3*sin(c + d*x)
**9*cos(c + d*x)**2/(105*d) + 16*B*a**3*sin(c + d*x)**9/(105*d) + 3*B*a**3*sin(c + d*x)**8*cos(c + d*x)**2/(8*
d) + 6*B*a**3*sin(c + d*x)**7*cos(c + d*x)**4/(35*d) + 24*B*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 3*B*
a**3*sin(c + d*x)**6*cos(c + d*x)**4/(4*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)**6/(5*d) + 6*B*a**3*sin(c + d
*x)**5*cos(c + d*x)**4/(5*d) + 3*B*a**3*sin(c + d*x)**4*cos(c + d*x)**6/(4*d) + B*a**3*sin(c + d*x)**3*cos(c +
d*x)**6/d - B*a**3*cos(c + d*x)**8/(8*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c)**7, True))

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