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

Optimal. Leaf size=165 \[ \frac {2 a^2 \cos ^9(c+d x)}{9 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{10 d}-\frac {13 a^2 \sin (c+d x) \cos ^7(c+d x)}{80 d}+\frac {13 a^2 \sin (c+d x) \cos ^5(c+d x)}{480 d}+\frac {13 a^2 \sin (c+d x) \cos ^3(c+d x)}{384 d}+\frac {13 a^2 \sin (c+d x) \cos (c+d x)}{256 d}+\frac {13 a^2 x}{256} \]

[Out]

13/256*a^2*x-2/7*a^2*cos(d*x+c)^7/d+2/9*a^2*cos(d*x+c)^9/d+13/256*a^2*cos(d*x+c)*sin(d*x+c)/d+13/384*a^2*cos(d
*x+c)^3*sin(d*x+c)/d+13/480*a^2*cos(d*x+c)^5*sin(d*x+c)/d-13/80*a^2*cos(d*x+c)^7*sin(d*x+c)/d-1/10*a^2*cos(d*x
+c)^7*sin(d*x+c)^3/d

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Rubi [A]  time = 0.30, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 16, 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} \[ \frac {2 a^2 \cos ^9(c+d x)}{9 d}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \sin ^3(c+d x) \cos ^7(c+d x)}{10 d}-\frac {13 a^2 \sin (c+d x) \cos ^7(c+d x)}{80 d}+\frac {13 a^2 \sin (c+d x) \cos ^5(c+d x)}{480 d}+\frac {13 a^2 \sin (c+d x) \cos ^3(c+d x)}{384 d}+\frac {13 a^2 \sin (c+d x) \cos (c+d x)}{256 d}+\frac {13 a^2 x}{256} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(13*a^2*x)/256 - (2*a^2*Cos[c + d*x]^7)/(7*d) + (2*a^2*Cos[c + d*x]^9)/(9*d) + (13*a^2*Cos[c + d*x]*Sin[c + d*
x])/(256*d) + (13*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(384*d) + (13*a^2*Cos[c + d*x]^5*Sin[c + d*x])/(480*d) - (1
3*a^2*Cos[c + d*x]^7*Sin[c + d*x])/(80*d) - (a^2*Cos[c + d*x]^7*Sin[c + d*x]^3)/(10*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 ^6(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^2 \, dx &=\int \left (a^2 \cos ^6(c+d x) \sin ^2(c+d x)+2 a^2 \cos ^6(c+d x) \sin ^3(c+d x)+a^2 \cos ^6(c+d x) \sin ^4(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^6(c+d x) \sin ^2(c+d x) \, dx+a^2 \int \cos ^6(c+d x) \sin ^4(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^6(c+d x) \sin ^3(c+d x) \, dx\\ &=-\frac {a^2 \cos ^7(c+d x) \sin (c+d x)}{8 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d}+\frac {1}{8} a^2 \int \cos ^6(c+d x) \, dx+\frac {1}{10} \left (3 a^2\right ) \int \cos ^6(c+d x) \sin ^2(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int x^6 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {13 a^2 \cos ^7(c+d x) \sin (c+d x)}{80 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d}+\frac {1}{80} \left (3 a^2\right ) \int \cos ^6(c+d x) \, dx+\frac {1}{48} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (x^6-x^8\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^9(c+d x)}{9 d}+\frac {5 a^2 \cos ^3(c+d x) \sin (c+d x)}{192 d}+\frac {13 a^2 \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {13 a^2 \cos ^7(c+d x) \sin (c+d x)}{80 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d}+\frac {1}{32} a^2 \int \cos ^4(c+d x) \, dx+\frac {1}{64} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^9(c+d x)}{9 d}+\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{128 d}+\frac {13 a^2 \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {13 a^2 \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {13 a^2 \cos ^7(c+d x) \sin (c+d x)}{80 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d}+\frac {1}{128} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{128} \left (5 a^2\right ) \int 1 \, dx\\ &=\frac {5 a^2 x}{128}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^9(c+d x)}{9 d}+\frac {13 a^2 \cos (c+d x) \sin (c+d x)}{256 d}+\frac {13 a^2 \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {13 a^2 \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {13 a^2 \cos ^7(c+d x) \sin (c+d x)}{80 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d}+\frac {1}{256} \left (3 a^2\right ) \int 1 \, dx\\ &=\frac {13 a^2 x}{256}-\frac {2 a^2 \cos ^7(c+d x)}{7 d}+\frac {2 a^2 \cos ^9(c+d x)}{9 d}+\frac {13 a^2 \cos (c+d x) \sin (c+d x)}{256 d}+\frac {13 a^2 \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {13 a^2 \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {13 a^2 \cos ^7(c+d x) \sin (c+d x)}{80 d}-\frac {a^2 \cos ^7(c+d x) \sin ^3(c+d x)}{10 d}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 106, normalized size = 0.64 \[ \frac {a^2 (11340 \sin (2 (c+d x))-7560 \sin (4 (c+d x))-3990 \sin (6 (c+d x))-315 \sin (8 (c+d x))+126 \sin (10 (c+d x))-30240 \cos (c+d x)-13440 \cos (3 (c+d x))+2160 \cos (7 (c+d x))+560 \cos (9 (c+d x))+12600 c+32760 d x)}{645120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12600*c + 32760*d*x - 30240*Cos[c + d*x] - 13440*Cos[3*(c + d*x)] + 2160*Cos[7*(c + d*x)] + 560*Cos[9*(c
+ d*x)] + 11340*Sin[2*(c + d*x)] - 7560*Sin[4*(c + d*x)] - 3990*Sin[6*(c + d*x)] - 315*Sin[8*(c + d*x)] + 126
*Sin[10*(c + d*x)]))/(645120*d)

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fricas [A]  time = 0.75, size = 111, normalized size = 0.67 \[ \frac {17920 \, a^{2} \cos \left (d x + c\right )^{9} - 23040 \, a^{2} \cos \left (d x + c\right )^{7} + 4095 \, a^{2} d x + 21 \, {\left (384 \, a^{2} \cos \left (d x + c\right )^{9} - 1008 \, a^{2} \cos \left (d x + c\right )^{7} + 104 \, a^{2} \cos \left (d x + c\right )^{5} + 130 \, a^{2} \cos \left (d x + c\right )^{3} + 195 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80640 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80640*(17920*a^2*cos(d*x + c)^9 - 23040*a^2*cos(d*x + c)^7 + 4095*a^2*d*x + 21*(384*a^2*cos(d*x + c)^9 - 100
8*a^2*cos(d*x + c)^7 + 104*a^2*cos(d*x + c)^5 + 130*a^2*cos(d*x + c)^3 + 195*a^2*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.31, size = 157, normalized size = 0.95 \[ \frac {13}{256} \, a^{2} x + \frac {a^{2} \cos \left (9 \, d x + 9 \, c\right )}{1152 \, d} + \frac {3 \, a^{2} \cos \left (7 \, d x + 7 \, c\right )}{896 \, d} - \frac {a^{2} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, a^{2} \cos \left (d x + c\right )}{64 \, d} + \frac {a^{2} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {a^{2} \sin \left (8 \, d x + 8 \, c\right )}{2048 \, d} - \frac {19 \, a^{2} \sin \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac {3 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{256 \, d} + \frac {9 \, a^{2} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/256*a^2*x + 1/1152*a^2*cos(9*d*x + 9*c)/d + 3/896*a^2*cos(7*d*x + 7*c)/d - 1/48*a^2*cos(3*d*x + 3*c)/d - 3/
64*a^2*cos(d*x + c)/d + 1/5120*a^2*sin(10*d*x + 10*c)/d - 1/2048*a^2*sin(8*d*x + 8*c)/d - 19/3072*a^2*sin(6*d*
x + 6*c)/d - 3/256*a^2*sin(4*d*x + 4*c)/d + 9/512*a^2*sin(2*d*x + 2*c)/d

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maple [A]  time = 0.28, size = 184, normalized size = 1.12 \[ \frac {a^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{10}-\frac {3 \left (\cos ^{7}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{80}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+2 a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{7}\left (d x +c \right )\right )}{9}-\frac {2 \left (\cos ^{7}\left (d x +c \right )\right )}{63}\right )+a^{2} \left (-\frac {\left (\cos ^{7}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{8}+\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+2*a^2*(-1/9*sin(d*x+c)^2*cos(d*x+c)^7-2/63*cos(d*x+c)^7)+a^2*(-1/
8*cos(d*x+c)^7*sin(d*x+c)+1/48*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/128*d*x+5/128*c))

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maxima [A]  time = 0.34, size = 128, normalized size = 0.78 \[ \frac {20480 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} a^{2} + 63 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} + 210 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{645120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/645120*(20480*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*a^2 + 63*(32*sin(2*d*x + 2*c)^5 + 120*d*x + 120*c + 5*si
n(8*d*x + 8*c) - 40*sin(4*d*x + 4*c))*a^2 + 210*(64*sin(2*d*x + 2*c)^3 + 120*d*x + 120*c - 3*sin(8*d*x + 8*c)
- 24*sin(4*d*x + 4*c))*a^2)/d

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mupad [B]  time = 12.12, size = 469, normalized size = 2.84 \[ \frac {13\,a^2\,x}{256}-\frac {\frac {13\,a^2\,\left (c+d\,x\right )}{256}-\frac {647\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{384}-\frac {2311\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{480}+\frac {457\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{32}-\frac {2169\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{64}+\frac {2169\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{64}-\frac {457\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{32}+\frac {2311\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{15}}{480}+\frac {647\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{17}}{384}-\frac {13\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{19}}{128}-\frac {a^2\,\left (4095\,c+4095\,d\,x-10240\right )}{80640}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {65\,a^2\,\left (c+d\,x\right )}{128}-\frac {a^2\,\left (40950\,c+40950\,d\,x-102400\right )}{80640}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {585\,a^2\,\left (c+d\,x\right )}{256}-\frac {a^2\,\left (184275\,c+184275\,d\,x+184320\right )}{80640}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{16}\,\left (\frac {585\,a^2\,\left (c+d\,x\right )}{256}-\frac {a^2\,\left (184275\,c+184275\,d\,x-645120\right )}{80640}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}\,\left (\frac {195\,a^2\,\left (c+d\,x\right )}{32}-\frac {a^2\,\left (491400\,c+491400\,d\,x+430080\right )}{80640}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {195\,a^2\,\left (c+d\,x\right )}{32}-\frac {a^2\,\left (491400\,c+491400\,d\,x-1658880\right )}{80640}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {819\,a^2\,\left (c+d\,x\right )}{64}-\frac {a^2\,\left (1031940\,c+1031940\,d\,x-1290240\right )}{80640}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,\left (\frac {1365\,a^2\,\left (c+d\,x\right )}{128}-\frac {a^2\,\left (859950\,c+859950\,d\,x-2150400\right )}{80640}\right )+\frac {13\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{128}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*a^2*x)/256 - ((13*a^2*(c + d*x))/256 - (647*a^2*tan(c/2 + (d*x)/2)^3)/384 - (2311*a^2*tan(c/2 + (d*x)/2)^5
)/480 + (457*a^2*tan(c/2 + (d*x)/2)^7)/32 - (2169*a^2*tan(c/2 + (d*x)/2)^9)/64 + (2169*a^2*tan(c/2 + (d*x)/2)^
11)/64 - (457*a^2*tan(c/2 + (d*x)/2)^13)/32 + (2311*a^2*tan(c/2 + (d*x)/2)^15)/480 + (647*a^2*tan(c/2 + (d*x)/
2)^17)/384 - (13*a^2*tan(c/2 + (d*x)/2)^19)/128 - (a^2*(4095*c + 4095*d*x - 10240))/80640 + tan(c/2 + (d*x)/2)
^2*((65*a^2*(c + d*x))/128 - (a^2*(40950*c + 40950*d*x - 102400))/80640) + tan(c/2 + (d*x)/2)^4*((585*a^2*(c +
d*x))/256 - (a^2*(184275*c + 184275*d*x + 184320))/80640) + tan(c/2 + (d*x)/2)^16*((585*a^2*(c + d*x))/256 -
(a^2*(184275*c + 184275*d*x - 645120))/80640) + tan(c/2 + (d*x)/2)^14*((195*a^2*(c + d*x))/32 - (a^2*(491400*c
+ 491400*d*x + 430080))/80640) + tan(c/2 + (d*x)/2)^6*((195*a^2*(c + d*x))/32 - (a^2*(491400*c + 491400*d*x -
1658880))/80640) + tan(c/2 + (d*x)/2)^10*((819*a^2*(c + d*x))/64 - (a^2*(1031940*c + 1031940*d*x - 1290240))/
80640) + tan(c/2 + (d*x)/2)^12*((1365*a^2*(c + d*x))/128 - (a^2*(859950*c + 859950*d*x - 2150400))/80640) + (1
3*a^2*tan(c/2 + (d*x)/2))/128)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^10)

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sympy [A]  time = 27.32, size = 529, normalized size = 3.21 \[ \begin {cases} \frac {3 a^{2} x \sin ^{10}{\left (c + d x \right )}}{256} + \frac {15 a^{2} x \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{256} + \frac {5 a^{2} x \sin ^{8}{\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 )}}{128} + \frac {5 a^{2} x \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{32} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{128} + \frac {15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\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 )}}{256} + \frac {5 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{32} + \frac {3 a^{2} x \cos ^{10}{\left (c + d x \right )}}{256} + \frac {5 a^{2} x \cos ^{8}{\left (c + d x \right )}}{128} + \frac {3 a^{2} \sin ^{9}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{256 d} + \frac {7 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{128 d} + \frac {5 a^{2} \sin ^{7}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{128 d} + \frac {a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{10 d} + \frac {55 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{384 d} - \frac {7 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} + \frac {73 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{384 d} - \frac {2 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 )}}{256 d} - \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{7}{\left (c + d x \right )}}{128 d} - \frac {4 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 ^{2}{\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)**2*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((3*a**2*x*sin(c + d*x)**10/256 + 15*a**2*x*sin(c + d*x)**8*cos(c + d*x)**2/256 + 5*a**2*x*sin(c + d*
x)**8/128 + 15*a**2*x*sin(c + d*x)**6*cos(c + d*x)**4/128 + 5*a**2*x*sin(c + d*x)**6*cos(c + d*x)**2/32 + 15*a
**2*x*sin(c + d*x)**4*cos(c + d*x)**6/128 + 15*a**2*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 15*a**2*x*sin(c + d
*x)**2*cos(c + d*x)**8/256 + 5*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 3*a**2*x*cos(c + d*x)**10/256 + 5*a
**2*x*cos(c + d*x)**8/128 + 3*a**2*sin(c + d*x)**9*cos(c + d*x)/(256*d) + 7*a**2*sin(c + d*x)**7*cos(c + d*x)*
*3/(128*d) + 5*a**2*sin(c + d*x)**7*cos(c + d*x)/(128*d) + a**2*sin(c + d*x)**5*cos(c + d*x)**5/(10*d) + 55*a*
*2*sin(c + d*x)**5*cos(c + d*x)**3/(384*d) - 7*a**2*sin(c + d*x)**3*cos(c + d*x)**7/(128*d) + 73*a**2*sin(c +
d*x)**3*cos(c + d*x)**5/(384*d) - 2*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/(256*d) - 5*a**2*sin(c + d*x)*cos(c + d*x)**7/(128*d) - 4*a**2*cos(c + d*x)**9/(63*d), Ne(d, 0)), (x*(a
*sin(c) + a)**2*sin(c)**2*cos(c)**6, True))

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