3.379 \(\int \cos ^4(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=159 \[ -\frac {a^2 \cos ^9(c+d x)}{9 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{4 d}-\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{8 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{32 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{64 d}+\frac {3 a^2 x}{64} \]

[Out]

3/64*a^2*x-2/5*a^2*cos(d*x+c)^5/d+3/7*a^2*cos(d*x+c)^7/d-1/9*a^2*cos(d*x+c)^9/d+3/64*a^2*cos(d*x+c)*sin(d*x+c)
/d+1/32*a^2*cos(d*x+c)^3*sin(d*x+c)/d-1/8*a^2*cos(d*x+c)^5*sin(d*x+c)/d-1/4*a^2*cos(d*x+c)^5*sin(d*x+c)^3/d

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Rubi [A]  time = 0.25, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2873, 2565, 14, 2568, 2635, 8, 270} \[ -\frac {a^2 \cos ^9(c+d x)}{9 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{4 d}-\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{8 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{32 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{64 d}+\frac {3 a^2 x}{64} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(3*a^2*x)/64 - (2*a^2*Cos[c + d*x]^5)/(5*d) + (3*a^2*Cos[c + d*x]^7)/(7*d) - (a^2*Cos[c + d*x]^9)/(9*d) + (3*a
^2*Cos[c + d*x]*Sin[c + d*x])/(64*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(32*d) - (a^2*Cos[c + d*x]^5*Sin[c +
d*x])/(8*d) - (a^2*Cos[c + d*x]^5*Sin[c + d*x]^3)/(4*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2568

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

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 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^4(c+d x) \sin ^3(c+d x)+2 a^2 \cos ^4(c+d x) \sin ^4(c+d x)+a^2 \cos ^4(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx+a^2 \int \cos ^4(c+d x) \sin ^5(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} \left (3 a^2\right ) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^2 \operatorname {Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} a^2 \int \cos ^4(c+d x) \, dx-\frac {a^2 \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^2 \operatorname {Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \cos ^9(c+d x)}{9 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{32 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{32} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \cos ^9(c+d x)}{9 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{64 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{32 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{64} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac {3 a^2 x}{64}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \cos ^9(c+d x)}{9 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{64 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{32 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 86, normalized size = 0.54 \[ \frac {a^2 (-2520 \sin (4 (c+d x))+315 \sin (8 (c+d x))-11340 \cos (c+d x)-3360 \cos (3 (c+d x))+1008 \cos (5 (c+d x))+450 \cos (7 (c+d x))-70 \cos (9 (c+d x))+7560 c+7560 d x)}{161280 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(7560*c + 7560*d*x - 11340*Cos[c + d*x] - 3360*Cos[3*(c + d*x)] + 1008*Cos[5*(c + d*x)] + 450*Cos[7*(c +
d*x)] - 70*Cos[9*(c + d*x)] - 2520*Sin[4*(c + d*x)] + 315*Sin[8*(c + d*x)]))/(161280*d)

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fricas [A]  time = 0.60, size = 111, normalized size = 0.70 \[ -\frac {2240 \, a^{2} \cos \left (d x + c\right )^{9} - 8640 \, a^{2} \cos \left (d x + c\right )^{7} + 8064 \, a^{2} \cos \left (d x + c\right )^{5} - 945 \, a^{2} d x - 315 \, {\left (16 \, a^{2} \cos \left (d x + c\right )^{7} - 24 \, a^{2} \cos \left (d x + c\right )^{5} + 2 \, a^{2} \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{20160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/20160*(2240*a^2*cos(d*x + c)^9 - 8640*a^2*cos(d*x + c)^7 + 8064*a^2*cos(d*x + c)^5 - 945*a^2*d*x - 315*(16*
a^2*cos(d*x + c)^7 - 24*a^2*cos(d*x + c)^5 + 2*a^2*cos(d*x + c)^3 + 3*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.31, size = 123, normalized size = 0.77 \[ \frac {3}{64} \, a^{2} x - \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {5 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {9 \, a^{2} \cos \left (d x + c\right )}{128 \, d} + \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

3/64*a^2*x - 1/2304*a^2*cos(9*d*x + 9*c)/d + 5/1792*a^2*cos(7*d*x + 7*c)/d + 1/160*a^2*cos(5*d*x + 5*c)/d - 1/
48*a^2*cos(3*d*x + 3*c)/d - 9/128*a^2*cos(d*x + c)/d + 1/512*a^2*sin(8*d*x + 8*c)/d - 1/64*a^2*sin(4*d*x + 4*c
)/d

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maple [A]  time = 0.28, size = 162, normalized size = 1.02 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{9}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{63}-\frac {8 \left (\cos ^{5}\left (d x +c \right )\right )}{315}\right )+2 a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/9*sin(d*x+c)^4*cos(d*x+c)^5-4/63*sin(d*x+c)^2*cos(d*x+c)^5-8/315*cos(d*x+c)^5)+2*a^2*(-1/8*sin(d*
x+c)^3*cos(d*x+c)^5-1/16*sin(d*x+c)*cos(d*x+c)^5+1/64*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/128*d*x+3/128
*c)+a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5))

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maxima [A]  time = 0.34, size = 101, normalized size = 0.64 \[ -\frac {512 \, {\left (35 \, \cos \left (d x + c\right )^{9} - 90 \, \cos \left (d x + c\right )^{7} + 63 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 4608 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} - 315 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{161280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/161280*(512*(35*cos(d*x + c)^9 - 90*cos(d*x + c)^7 + 63*cos(d*x + c)^5)*a^2 - 4608*(5*cos(d*x + c)^7 - 7*co
s(d*x + c)^5)*a^2 - 315*(24*d*x + 24*c + sin(8*d*x + 8*c) - 8*sin(4*d*x + 4*c))*a^2)/d

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mupad [B]  time = 12.15, size = 437, normalized size = 2.75 \[ \frac {3\,a^2\,x}{64}-\frac {\frac {3\,a^2\,\left (c+d\,x\right )}{64}+\frac {13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{16}-\frac {155\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{16}+\frac {169\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{16}-\frac {169\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{16}+\frac {155\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{16}-\frac {13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{16}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{32}-\frac {a^2\,\left (945\,c+945\,d\,x-3328\right )}{20160}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {27\,a^2\,\left (c+d\,x\right )}{64}-\frac {a^2\,\left (8505\,c+8505\,d\,x-29952\right )}{20160}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {27\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (34020\,c+34020\,d\,x-39168\right )}{20160}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (\frac {27\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (34020\,c+34020\,d\,x-80640\right )}{20160}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {63\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (79380\,c+79380\,d\,x+16128\right )}{20160}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {63\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (79380\,c+79380\,d\,x-295680\right )}{20160}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {189\,a^2\,\left (c+d\,x\right )}{32}-\frac {a^2\,\left (119070\,c+119070\,d\,x+241920\right )}{20160}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {189\,a^2\,\left (c+d\,x\right )}{32}-\frac {a^2\,\left (119070\,c+119070\,d\,x-661248\right )}{20160}\right )+\frac {3\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*sin(c + d*x)^3*(a + a*sin(c + d*x))^2,x)

[Out]

(3*a^2*x)/64 - ((3*a^2*(c + d*x))/64 + (13*a^2*tan(c/2 + (d*x)/2)^3)/16 - (155*a^2*tan(c/2 + (d*x)/2)^5)/16 +
(169*a^2*tan(c/2 + (d*x)/2)^7)/16 - (169*a^2*tan(c/2 + (d*x)/2)^11)/16 + (155*a^2*tan(c/2 + (d*x)/2)^13)/16 -
(13*a^2*tan(c/2 + (d*x)/2)^15)/16 - (3*a^2*tan(c/2 + (d*x)/2)^17)/32 - (a^2*(945*c + 945*d*x - 3328))/20160 +
tan(c/2 + (d*x)/2)^2*((27*a^2*(c + d*x))/64 - (a^2*(8505*c + 8505*d*x - 29952))/20160) + tan(c/2 + (d*x)/2)^4*
((27*a^2*(c + d*x))/16 - (a^2*(34020*c + 34020*d*x - 39168))/20160) + tan(c/2 + (d*x)/2)^14*((27*a^2*(c + d*x)
)/16 - (a^2*(34020*c + 34020*d*x - 80640))/20160) + tan(c/2 + (d*x)/2)^6*((63*a^2*(c + d*x))/16 - (a^2*(79380*
c + 79380*d*x + 16128))/20160) + tan(c/2 + (d*x)/2)^12*((63*a^2*(c + d*x))/16 - (a^2*(79380*c + 79380*d*x - 29
5680))/20160) + tan(c/2 + (d*x)/2)^10*((189*a^2*(c + d*x))/32 - (a^2*(119070*c + 119070*d*x + 241920))/20160)
+ tan(c/2 + (d*x)/2)^8*((189*a^2*(c + d*x))/32 - (a^2*(119070*c + 119070*d*x - 661248))/20160) + (3*a^2*tan(c/
2 + (d*x)/2))/32)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^9)

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sympy [A]  time = 22.16, size = 335, normalized size = 2.11 \[ \begin {cases} \frac {3 a^{2} x \sin ^{8}{\left (c + d x \right )}}{64} + \frac {3 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{32} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \cos ^{8}{\left (c + d x \right )}}{64} + \frac {3 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{64 d} + \frac {11 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{64 d} - \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {11 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{64 d} - \frac {4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{35 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {3 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac {8 a^{2} \cos ^{9}{\left (c + d x \right )}}{315 d} - \frac {2 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin ^{3}{\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((3*a**2*x*sin(c + d*x)**8/64 + 3*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/16 + 9*a**2*x*sin(c + d*x)**
4*cos(c + d*x)**4/32 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/16 + 3*a**2*x*cos(c + d*x)**8/64 + 3*a**2*sin(
c + d*x)**7*cos(c + d*x)/(64*d) + 11*a**2*sin(c + d*x)**5*cos(c + d*x)**3/(64*d) - a**2*sin(c + d*x)**4*cos(c
+ d*x)**5/(5*d) - 11*a**2*sin(c + d*x)**3*cos(c + d*x)**5/(64*d) - 4*a**2*sin(c + d*x)**2*cos(c + d*x)**7/(35*
d) - a**2*sin(c + d*x)**2*cos(c + d*x)**5/(5*d) - 3*a**2*sin(c + d*x)*cos(c + d*x)**7/(64*d) - 8*a**2*cos(c +
d*x)**9/(315*d) - 2*a**2*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**3*cos(c)**4, True))

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