3.589 \(\int \cos ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=183 \[ -\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{5 d}-\frac {3 a^2 \sin (c+d x) \cos ^7(c+d x)}{40 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{80 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 a^2 x}{128} \]

[Out]

3/128*a^2*x-2/7*a^2*cos(d*x+c)^7/d+1/3*a^2*cos(d*x+c)^9/d-1/11*a^2*cos(d*x+c)^11/d+3/128*a^2*cos(d*x+c)*sin(d*
x+c)/d+1/64*a^2*cos(d*x+c)^3*sin(d*x+c)/d+1/80*a^2*cos(d*x+c)^5*sin(d*x+c)/d-3/40*a^2*cos(d*x+c)^7*sin(d*x+c)/
d-1/5*a^2*cos(d*x+c)^7*sin(d*x+c)^3/d

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Rubi [A]  time = 0.27, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ -\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{5 d}-\frac {3 a^2 \sin (c+d x) \cos ^7(c+d x)}{40 d}+\frac {a^2 \sin (c+d x) \cos ^5(c+d x)}{80 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{64 d}+\frac {3 a^2 \sin (c+d x) \cos (c+d x)}{128 d}+\frac {3 a^2 x}{128} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(3*a^2*x)/128 - (2*a^2*Cos[c + d*x]^7)/(7*d) + (a^2*Cos[c + d*x]^9)/(3*d) - (a^2*Cos[c + d*x]^11)/(11*d) + (3*
a^2*Cos[c + d*x]*Sin[c + d*x])/(128*d) + (a^2*Cos[c + d*x]^3*Sin[c + d*x])/(64*d) + (a^2*Cos[c + d*x]^5*Sin[c
+ d*x])/(80*d) - (3*a^2*Cos[c + d*x]^7*Sin[c + d*x])/(40*d) - (a^2*Cos[c + d*x]^7*Sin[c + d*x]^3)/(5*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 ^6(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^6(c+d x) \sin ^3(c+d x)+2 a^2 \cos ^6(c+d x) \sin ^4(c+d x)+a^2 \cos ^6(c+d x) \sin ^5(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^6(c+d x) \sin ^3(c+d x) \, dx+a^2 \int \cos ^6(c+d x) \sin ^5(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^6(c+d x) \sin ^4(c+d x) \, dx\\ &=-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d}+\frac {1}{5} \left (3 a^2\right ) \int \cos ^6(c+d x) \sin ^2(c+d x) \, dx-\frac {a^2 \operatorname {Subst}\left (\int x^6 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^2 \operatorname {Subst}\left (\int x^6 \left (1-x^2\right )^2 \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {3 a^2 \cos ^7(c+d x) \sin (c+d x)}{40 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d}+\frac {1}{40} \left (3 a^2\right ) \int \cos ^6(c+d x) \, dx-\frac {a^2 \operatorname {Subst}\left (\int \left (x^6-x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}-\frac {a^2 \operatorname {Subst}\left (\int \left (x^6-2 x^8+x^{10}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{80 d}-\frac {3 a^2 \cos ^7(c+d x) \sin (c+d x)}{40 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d}+\frac {1}{16} a^2 \int \cos ^4(c+d x) \, dx\\ &=-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{80 d}-\frac {3 a^2 \cos ^7(c+d x) \sin (c+d x)}{40 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d}+\frac {1}{64} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{80 d}-\frac {3 a^2 \cos ^7(c+d x) \sin (c+d x)}{40 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d}+\frac {1}{128} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac {3 a^2 x}{128}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {a^2 \cos ^9(c+d x)}{3 d}-\frac {a^2 \cos ^{11}(c+d x)}{11 d}+\frac {3 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{64 d}+\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{80 d}-\frac {3 a^2 \cos ^7(c+d x) \sin (c+d x)}{40 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.99, size = 126, normalized size = 0.69 \[ \frac {a^2 (4620 \sin (2 (c+d x))-9240 \sin (4 (c+d x))-2310 \sin (6 (c+d x))+1155 \sin (8 (c+d x))+462 \sin (10 (c+d x))-39270 \cos (c+d x)-16170 \cos (3 (c+d x))+1155 \cos (5 (c+d x))+2805 \cos (7 (c+d x))+385 \cos (9 (c+d x))-105 \cos (11 (c+d x))+27720 c+27720 d x)}{1182720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(27720*c + 27720*d*x - 39270*Cos[c + d*x] - 16170*Cos[3*(c + d*x)] + 1155*Cos[5*(c + d*x)] + 2805*Cos[7*(
c + d*x)] + 385*Cos[9*(c + d*x)] - 105*Cos[11*(c + d*x)] + 4620*Sin[2*(c + d*x)] - 9240*Sin[4*(c + d*x)] - 231
0*Sin[6*(c + d*x)] + 1155*Sin[8*(c + d*x)] + 462*Sin[10*(c + d*x)]))/(1182720*d)

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fricas [A]  time = 0.80, size = 124, normalized size = 0.68 \[ -\frac {13440 \, a^{2} \cos \left (d x + c\right )^{11} - 49280 \, a^{2} \cos \left (d x + c\right )^{9} + 42240 \, a^{2} \cos \left (d x + c\right )^{7} - 3465 \, a^{2} d x - 231 \, {\left (128 \, a^{2} \cos \left (d x + c\right )^{9} - 176 \, a^{2} \cos \left (d x + c\right )^{7} + 8 \, a^{2} \cos \left (d x + c\right )^{5} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{147840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/147840*(13440*a^2*cos(d*x + c)^11 - 49280*a^2*cos(d*x + c)^9 + 42240*a^2*cos(d*x + c)^7 - 3465*a^2*d*x - 23
1*(128*a^2*cos(d*x + c)^9 - 176*a^2*cos(d*x + c)^7 + 8*a^2*cos(d*x + c)^5 + 10*a^2*cos(d*x + c)^3 + 15*a^2*cos
(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.37, size = 191, normalized size = 1.04 \[ \frac {3}{128} \, a^{2} x - \frac {a^{2} \cos \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{3072 \, d} + \frac {17 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{7168 \, d} + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{1024 \, d} - \frac {7 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{512 \, d} - \frac {17 \, a^{2} \cos \left (d x + c\right )}{512 \, d} + \frac {a^{2} \sin \left (10 \, d x + 10 \, c\right )}{2560 \, d} + \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a^{2} \sin \left (6 \, d x + 6 \, c\right )}{512 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {a^{2} \sin \left (2 \, d x + 2 \, c\right )}{256 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

3/128*a^2*x - 1/11264*a^2*cos(11*d*x + 11*c)/d + 1/3072*a^2*cos(9*d*x + 9*c)/d + 17/7168*a^2*cos(7*d*x + 7*c)/
d + 1/1024*a^2*cos(5*d*x + 5*c)/d - 7/512*a^2*cos(3*d*x + 3*c)/d - 17/512*a^2*cos(d*x + c)/d + 1/2560*a^2*sin(
10*d*x + 10*c)/d + 1/1024*a^2*sin(8*d*x + 8*c)/d - 1/512*a^2*sin(6*d*x + 6*c)/d - 1/128*a^2*sin(4*d*x + 4*c)/d
+ 1/256*a^2*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.28, size = 172, normalized size = 0.94 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{4}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{11}-\frac {4 \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{99}-\frac {8 \left (\cos ^{7}\left (d x +c \right )\right )}{693}\right )+2 a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{10}-\frac {3 \left (\cos ^{7}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{80}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/11*sin(d*x+c)^4*cos(d*x+c)^7-4/99*sin(d*x+c)^2*cos(d*x+c)^7-8/693*cos(d*x+c)^7)+2*a^2*(-1/10*sin(
d*x+c)^3*cos(d*x+c)^7-3/80*cos(d*x+c)^7*sin(d*x+c)+1/160*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d
*x+c)+3/256*d*x+3/256*c)+a^2*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7))

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maxima [A]  time = 0.41, size = 116, normalized size = 0.63 \[ -\frac {5120 \, {\left (63 \, \cos \left (d x + c\right )^{11} - 154 \, \cos \left (d x + c\right )^{9} + 99 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 56320 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{2} - 693 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{3548160 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3548160*(5120*(63*cos(d*x + c)^11 - 154*cos(d*x + c)^9 + 99*cos(d*x + c)^7)*a^2 - 56320*(7*cos(d*x + c)^9 -
9*cos(d*x + c)^7)*a^2 - 693*(32*sin(2*d*x + 2*c)^5 + 120*d*x + 120*c + 5*sin(8*d*x + 8*c) - 40*sin(4*d*x + 4*
c))*a^2)/d

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mupad [B]  time = 12.05, size = 543, normalized size = 2.97 \[ \frac {3\,a^2\,x}{128}-\frac {\frac {3\,a^2\,\left (c+d\,x\right )}{128}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {3323\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{320}+\frac {108\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{5}-\frac {841\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{32}+\frac {841\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{32}-\frac {108\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{5}+\frac {3323\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{320}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{19}}{2}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{21}}{64}-a^2\,\left (\frac {3\,c}{128}+\frac {3\,d\,x}{128}-\frac {20}{231}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {33\,a^2\,\left (c+d\,x\right )}{128}-a^2\,\left (\frac {33\,c}{128}+\frac {33\,d\,x}{128}-\frac {20}{21}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{18}\,\left (\frac {165\,a^2\,\left (c+d\,x\right )}{128}-a^2\,\left (\frac {165\,c}{128}+\frac {165\,d\,x}{128}-4\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {165\,a^2\,\left (c+d\,x\right )}{128}-a^2\,\left (\frac {165\,c}{128}+\frac {165\,d\,x}{128}-\frac {16}{21}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (\frac {495\,a^2\,\left (c+d\,x\right )}{64}-a^2\,\left (\frac {495\,c}{64}+\frac {495\,d\,x}{64}+16\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,\left (\frac {495\,a^2\,\left (c+d\,x\right )}{128}-a^2\,\left (\frac {495\,c}{128}+\frac {495\,d\,x}{128}-12\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {495\,a^2\,\left (c+d\,x\right )}{128}-a^2\,\left (\frac {495\,c}{128}+\frac {495\,d\,x}{128}-\frac {16}{7}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {495\,a^2\,\left (c+d\,x\right )}{64}-a^2\,\left (\frac {495\,c}{64}+\frac {495\,d\,x}{64}-\frac {312}{7}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {693\,a^2\,\left (c+d\,x\right )}{64}-a^2\,\left (\frac {693\,c}{64}+\frac {693\,d\,x}{64}+40\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {693\,a^2\,\left (c+d\,x\right )}{64}-a^2\,\left (\frac {693\,c}{64}+\frac {693\,d\,x}{64}-80\right )\right )+\frac {3\,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 )}^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*sin(c + d*x)^3*(a + a*sin(c + d*x))^2,x)

[Out]

(3*a^2*x)/128 - ((3*a^2*(c + d*x))/128 + (a^2*tan(c/2 + (d*x)/2)^3)/2 - (3323*a^2*tan(c/2 + (d*x)/2)^5)/320 +
(108*a^2*tan(c/2 + (d*x)/2)^7)/5 - (841*a^2*tan(c/2 + (d*x)/2)^9)/32 + (841*a^2*tan(c/2 + (d*x)/2)^13)/32 - (1
08*a^2*tan(c/2 + (d*x)/2)^15)/5 + (3323*a^2*tan(c/2 + (d*x)/2)^17)/320 - (a^2*tan(c/2 + (d*x)/2)^19)/2 - (3*a^
2*tan(c/2 + (d*x)/2)^21)/64 - a^2*((3*c)/128 + (3*d*x)/128 - 20/231) + tan(c/2 + (d*x)/2)^2*((33*a^2*(c + d*x)
)/128 - a^2*((33*c)/128 + (33*d*x)/128 - 20/21)) + tan(c/2 + (d*x)/2)^18*((165*a^2*(c + d*x))/128 - a^2*((165*
c)/128 + (165*d*x)/128 - 4)) + tan(c/2 + (d*x)/2)^4*((165*a^2*(c + d*x))/128 - a^2*((165*c)/128 + (165*d*x)/12
8 - 16/21)) + tan(c/2 + (d*x)/2)^14*((495*a^2*(c + d*x))/64 - a^2*((495*c)/64 + (495*d*x)/64 + 16)) + tan(c/2
+ (d*x)/2)^16*((495*a^2*(c + d*x))/128 - a^2*((495*c)/128 + (495*d*x)/128 - 12)) + tan(c/2 + (d*x)/2)^6*((495*
a^2*(c + d*x))/128 - a^2*((495*c)/128 + (495*d*x)/128 - 16/7)) + tan(c/2 + (d*x)/2)^8*((495*a^2*(c + d*x))/64
- a^2*((495*c)/64 + (495*d*x)/64 - 312/7)) + tan(c/2 + (d*x)/2)^10*((693*a^2*(c + d*x))/64 - a^2*((693*c)/64 +
(693*d*x)/64 + 40)) + tan(c/2 + (d*x)/2)^12*((693*a^2*(c + d*x))/64 - a^2*((693*c)/64 + (693*d*x)/64 - 80)) +
(3*a^2*tan(c/2 + (d*x)/2))/64)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^11)

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sympy [A]  time = 40.49, size = 384, normalized size = 2.10 \[ \begin {cases} \frac {3 a^{2} x \sin ^{10}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{64} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{64} + \frac {15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 a^{2} x \cos ^{10}{\left (c + d x \right )}}{128} + \frac {3 a^{2} \sin ^{9}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {7 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{64 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {7 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{64 d} - \frac {4 a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{63 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{7 d} - \frac {3 a^{2} \sin {\left (c + d x \right )} \cos ^{9}{\left (c + d x \right )}}{128 d} - \frac {8 a^{2} \cos ^{11}{\left (c + d x \right )}}{693 d} - \frac {2 a^{2} \cos ^{9}{\left (c + d x \right )}}{63 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right )^{2} \sin ^{3}{\relax (c )} \cos ^{6}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((3*a**2*x*sin(c + d*x)**10/128 + 15*a**2*x*sin(c + d*x)**8*cos(c + d*x)**2/128 + 15*a**2*x*sin(c + d
*x)**6*cos(c + d*x)**4/64 + 15*a**2*x*sin(c + d*x)**4*cos(c + d*x)**6/64 + 15*a**2*x*sin(c + d*x)**2*cos(c + d
*x)**8/128 + 3*a**2*x*cos(c + d*x)**10/128 + 3*a**2*sin(c + d*x)**9*cos(c + d*x)/(128*d) + 7*a**2*sin(c + d*x)
**7*cos(c + d*x)**3/(64*d) + a**2*sin(c + d*x)**5*cos(c + d*x)**5/(5*d) - a**2*sin(c + d*x)**4*cos(c + d*x)**7
/(7*d) - 7*a**2*sin(c + d*x)**3*cos(c + d*x)**7/(64*d) - 4*a**2*sin(c + d*x)**2*cos(c + d*x)**9/(63*d) - a**2*
sin(c + d*x)**2*cos(c + d*x)**7/(7*d) - 3*a**2*sin(c + d*x)*cos(c + d*x)**9/(128*d) - 8*a**2*cos(c + d*x)**11/
(693*d) - 2*a**2*cos(c + d*x)**9/(63*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**3*cos(c)**6, True))

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