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

Optimal. Leaf size=141 \[ \frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {11 a^2 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {11 a^2 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {11 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {11 a^2 x}{128} \]

[Out]

11/128*a^2*x-2/5*a^2*cos(d*x+c)^5/d+2/7*a^2*cos(d*x+c)^7/d+11/128*a^2*cos(d*x+c)*sin(d*x+c)/d+11/192*a^2*cos(d
*x+c)^3*sin(d*x+c)/d-11/48*a^2*cos(d*x+c)^5*sin(d*x+c)/d-1/8*a^2*cos(d*x+c)^5*sin(d*x+c)^3/d

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Rubi [A]  time = 0.26, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2873, 2568, 2635, 8, 2565, 14} \[ \frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x) \cos ^5(c+d x)}{8 d}-\frac {11 a^2 \sin (c+d x) \cos ^5(c+d x)}{48 d}+\frac {11 a^2 \sin (c+d x) \cos ^3(c+d x)}{192 d}+\frac {11 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {11 a^2 x}{128} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*a^2*x)/128 - (2*a^2*Cos[c + d*x]^5)/(5*d) + (2*a^2*Cos[c + d*x]^7)/(7*d) + (11*a^2*Cos[c + d*x]*Sin[c + d*
x])/(128*d) + (11*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(192*d) - (11*a^2*Cos[c + d*x]^5*Sin[c + d*x])/(48*d) - (a^
2*Cos[c + d*x]^5*Sin[c + d*x]^3)/(8*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 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 ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^4(c+d x) \sin ^2(c+d x)+2 a^2 \cos ^4(c+d x) \sin ^3(c+d x)+a^2 \cos ^4(c+d x) \sin ^4(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx+a^2 \int \cos ^4(c+d x) \sin ^4(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx\\ &=-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{6} a^2 \int \cos ^4(c+d x) \, dx+\frac {1}{8} \left (3 a^2\right ) \int \cos ^4(c+d x) \sin ^2(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {11 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{16} a^2 \int \cos ^4(c+d x) \, dx+\frac {1}{8} a^2 \int \cos ^2(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {11 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {11 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{64} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{16} a^2 \int 1 \, dx\\ &=\frac {a^2 x}{16}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {11 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {11 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {11 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}+\frac {1}{128} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac {11 a^2 x}{128}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {11 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {11 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}-\frac {11 a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {a^2 \cos ^5(c+d x) \sin ^3(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 96, normalized size = 0.68 \[ \frac {a^2 (1680 \sin (2 (c+d x))-2520 \sin (4 (c+d x))-560 \sin (6 (c+d x))+105 \sin (8 (c+d x))-10080 \cos (c+d x)-3360 \cos (3 (c+d x))+672 \cos (5 (c+d x))+480 \cos (7 (c+d x))+3360 c+9240 d x)}{107520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3360*c + 9240*d*x - 10080*Cos[c + d*x] - 3360*Cos[3*(c + d*x)] + 672*Cos[5*(c + d*x)] + 480*Cos[7*(c + d
*x)] + 1680*Sin[2*(c + d*x)] - 2520*Sin[4*(c + d*x)] - 560*Sin[6*(c + d*x)] + 105*Sin[8*(c + d*x)]))/(107520*d
)

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fricas [A]  time = 0.55, size = 98, normalized size = 0.70 \[ \frac {3840 \, a^{2} \cos \left (d x + c\right )^{7} - 5376 \, a^{2} \cos \left (d x + c\right )^{5} + 1155 \, a^{2} d x + 35 \, {\left (48 \, a^{2} \cos \left (d x + c\right )^{7} - 136 \, a^{2} \cos \left (d x + c\right )^{5} + 22 \, a^{2} \cos \left (d x + c\right )^{3} + 33 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{13440 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13440*(3840*a^2*cos(d*x + c)^7 - 5376*a^2*cos(d*x + c)^5 + 1155*a^2*d*x + 35*(48*a^2*cos(d*x + c)^7 - 136*a^
2*cos(d*x + c)^5 + 22*a^2*cos(d*x + c)^3 + 33*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.27, size = 140, normalized size = 0.99 \[ \frac {11}{128} \, a^{2} x + \frac {a^{2} \cos \left (7 \, d x + 7 \, c\right )}{224 \, d} + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{32 \, d} - \frac {3 \, a^{2} \cos \left (d x + c\right )}{32 \, d} + \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {3 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/128*a^2*x + 1/224*a^2*cos(7*d*x + 7*c)/d + 1/160*a^2*cos(5*d*x + 5*c)/d - 1/32*a^2*cos(3*d*x + 3*c)/d - 3/3
2*a^2*cos(d*x + c)/d + 1/1024*a^2*sin(8*d*x + 8*c)/d - 1/192*a^2*sin(6*d*x + 6*c)/d - 3/128*a^2*sin(4*d*x + 4*
c)/d + 1/64*a^2*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.29, size = 164, normalized size = 1.16 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{8}-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{16}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{64}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+2 a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )+a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+2*a^2*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+a^2*(-1/6*sin(d*x+c)*cos(d*x
+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c))

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maxima [A]  time = 0.32, size = 102, normalized size = 0.72 \[ \frac {6144 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{2} + 560 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 105 \, {\left (24 \, d x + 24 \, c + \sin \left (8 \, d x + 8 \, c\right ) - 8 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{107520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/107520*(6144*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^2 + 560*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4
*d*x + 4*c))*a^2 + 105*(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 = 12.35, size = 363, normalized size = 2.57 \[ \frac {11\,a^2\,x}{128}-\frac {\frac {11\,a^2\,\left (c+d\,x\right )}{128}-\frac {259\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{192}-\frac {1103\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{192}+\frac {2261\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{192}-\frac {2261\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{192}+\frac {1103\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{192}+\frac {259\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{192}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{64}-\frac {a^2\,\left (1155\,c+1155\,d\,x-3072\right )}{13440}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {11\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (9240\,c+9240\,d\,x-24576\right )}{13440}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {77\,a^2\,\left (c+d\,x\right )}{32}-\frac {a^2\,\left (32340\,c+32340\,d\,x+21504\right )}{13440}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {77\,a^2\,\left (c+d\,x\right )}{32}-\frac {a^2\,\left (32340\,c+32340\,d\,x-107520\right )}{13440}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {385\,a^2\,\left (c+d\,x\right )}{64}-\frac {a^2\,\left (80850\,c+80850\,d\,x-107520\right )}{13440}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {77\,a^2\,\left (c+d\,x\right )}{16}-\frac {a^2\,\left (64680\,c+64680\,d\,x-172032\right )}{13440}\right )+\frac {11\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*a^2*x)/128 - ((11*a^2*(c + d*x))/128 - (259*a^2*tan(c/2 + (d*x)/2)^3)/192 - (1103*a^2*tan(c/2 + (d*x)/2)^5
)/192 + (2261*a^2*tan(c/2 + (d*x)/2)^7)/192 - (2261*a^2*tan(c/2 + (d*x)/2)^9)/192 + (1103*a^2*tan(c/2 + (d*x)/
2)^11)/192 + (259*a^2*tan(c/2 + (d*x)/2)^13)/192 - (11*a^2*tan(c/2 + (d*x)/2)^15)/64 - (a^2*(1155*c + 1155*d*x
- 3072))/13440 + tan(c/2 + (d*x)/2)^2*((11*a^2*(c + d*x))/16 - (a^2*(9240*c + 9240*d*x - 24576))/13440) + tan
(c/2 + (d*x)/2)^4*((77*a^2*(c + d*x))/32 - (a^2*(32340*c + 32340*d*x + 21504))/13440) + tan(c/2 + (d*x)/2)^12*
((77*a^2*(c + d*x))/32 - (a^2*(32340*c + 32340*d*x - 107520))/13440) + tan(c/2 + (d*x)/2)^8*((385*a^2*(c + d*x
))/64 - (a^2*(80850*c + 80850*d*x - 107520))/13440) + tan(c/2 + (d*x)/2)^6*((77*a^2*(c + d*x))/16 - (a^2*(6468
0*c + 64680*d*x - 172032))/13440) + (11*a^2*tan(c/2 + (d*x)/2))/64)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^8)

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sympy [A]  time = 14.02, size = 420, normalized size = 2.98 \[ \begin {cases} \frac {3 a^{2} x \sin ^{8}{\left (c + d x \right )}}{128} + \frac {3 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {3 a^{2} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {3 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {11 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} - \frac {11 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{128 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {2 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {3 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {4 a^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin ^{2}{\relax (c )} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**2*x*sin(c + d*x)**8/128 + 3*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + a**2*x*sin(c + d*x)**6
/16 + 9*a**2*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 3*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**2*x*sin
(c + d*x)**2*cos(c + d*x)**6/32 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**2*x*cos(c + d*x)**8/128 +
a**2*x*cos(c + d*x)**6/16 + 3*a**2*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 11*a**2*sin(c + d*x)**5*cos(c + d*x
)**3/(128*d) + a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 11*a**2*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) + a*
*2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - 2*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/(128*d) - a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 4*a**2*cos(c + d*x)**7/(35*d), Ne(d, 0)),
(x*(a*sin(c) + a)**2*sin(c)**2*cos(c)**4, True))

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