3.1536 \(\int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=349 \[ \frac {3 \left (-7 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}+\frac {\left (a^2-b^2\right )^2 \left (-7 a^2 B+6 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac {\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}-\frac {\left (-35 a^3 B+15 a^2 A b+15 a b^2 B-3 A b^3\right ) (a+b \sin (c+d x))^7}{7 b^8 d}-\frac {3 \left (a^2-b^2\right ) \left (-7 a^3 B+5 a^2 A b+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac {\left (-35 a^4 B+20 a^3 A b+30 a^2 b^2 B-12 a A b^3-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac {(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac {B (a+b \sin (c+d x))^{10}}{10 b^8 d} \]

[Out]

-1/3*(a^2-b^2)^3*(A*b-B*a)*(a+b*sin(d*x+c))^3/b^8/d+1/4*(a^2-b^2)^2*(6*A*a*b-7*B*a^2+B*b^2)*(a+b*sin(d*x+c))^4
/b^8/d-3/5*(a^2-b^2)*(5*A*a^2*b-A*b^3-7*B*a^3+3*B*a*b^2)*(a+b*sin(d*x+c))^5/b^8/d+1/6*(20*A*a^3*b-12*A*a*b^3-3
5*B*a^4+30*B*a^2*b^2-3*B*b^4)*(a+b*sin(d*x+c))^6/b^8/d-1/7*(15*A*a^2*b-3*A*b^3-35*B*a^3+15*B*a*b^2)*(a+b*sin(d
*x+c))^7/b^8/d+3/8*(2*A*a*b-7*B*a^2+B*b^2)*(a+b*sin(d*x+c))^8/b^8/d-1/9*(A*b-7*B*a)*(a+b*sin(d*x+c))^9/b^8/d-1
/10*B*(a+b*sin(d*x+c))^10/b^8/d

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Rubi [A]  time = 0.39, antiderivative size = 349, 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 = {2837, 772} \[ \frac {3 \left (-7 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac {\left (15 a^2 A b-35 a^3 B+15 a b^2 B-3 A b^3\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac {\left (20 a^3 A b+30 a^2 b^2 B-35 a^4 B-12 a A b^3-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac {3 \left (a^2-b^2\right ) \left (5 a^2 A b-7 a^3 B+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac {\left (a^2-b^2\right )^2 \left (-7 a^2 B+6 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac {\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}-\frac {(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac {B (a+b \sin (c+d x))^{10}}{10 b^8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

-((a^2 - b^2)^3*(A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^8*d) + ((a^2 - b^2)^2*(6*a*A*b - 7*a^2*B + b^2*B)*(a
+ b*Sin[c + d*x])^4)/(4*b^8*d) - (3*(a^2 - b^2)*(5*a^2*A*b - A*b^3 - 7*a^3*B + 3*a*b^2*B)*(a + b*Sin[c + d*x])
^5)/(5*b^8*d) + ((20*a^3*A*b - 12*a*A*b^3 - 35*a^4*B + 30*a^2*b^2*B - 3*b^4*B)*(a + b*Sin[c + d*x])^6)/(6*b^8*
d) - ((15*a^2*A*b - 3*A*b^3 - 35*a^3*B + 15*a*b^2*B)*(a + b*Sin[c + d*x])^7)/(7*b^8*d) + (3*(2*a*A*b - 7*a^2*B
+ b^2*B)*(a + b*Sin[c + d*x])^8)/(8*b^8*d) - ((A*b - 7*a*B)*(a + b*Sin[c + d*x])^9)/(9*b^8*d) - (B*(a + b*Sin
[c + d*x])^10)/(10*b^8*d)

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a+x)^2 \left (A+\frac {B x}{b}\right ) \left (b^2-x^2\right )^3 \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {\left (-a^2+b^2\right )^3 (A b-a B) (a+x)^2}{b}+\frac {\left (-a^2+b^2\right )^2 \left (6 a A b-7 a^2 B+b^2 B\right ) (a+x)^3}{b}-\frac {3 \left (-a^2+b^2\right ) \left (-5 a^2 A b+A b^3+7 a^3 B-3 a b^2 B\right ) (a+x)^4}{b}+\frac {\left (20 a^3 A b-12 a A b^3-35 a^4 B+30 a^2 b^2 B-3 b^4 B\right ) (a+x)^5}{b}+\frac {\left (-15 a^2 A b+3 A b^3+35 a^3 B-15 a b^2 B\right ) (a+x)^6}{b}-\frac {3 \left (-2 a A b+7 a^2 B-b^2 B\right ) (a+x)^7}{b}+\frac {(-A b+7 a B) (a+x)^8}{b}-\frac {B (a+x)^9}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=-\frac {\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}+\frac {\left (a^2-b^2\right )^2 \left (6 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac {3 \left (a^2-b^2\right ) \left (5 a^2 A b-A b^3-7 a^3 B+3 a b^2 B\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac {\left (20 a^3 A b-12 a A b^3-35 a^4 B+30 a^2 b^2 B-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac {\left (15 a^2 A b-3 A b^3-35 a^3 B+15 a b^2 B\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac {3 \left (2 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac {(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac {B (a+b \sin (c+d x))^{10}}{10 b^8 d}\\ \end {align*}

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Mathematica [A]  time = 1.50, size = 295, normalized size = 0.85 \[ \frac {2520 a^2 A b^8 \sin (c+d x)-315 b^8 \left (a^2 B+2 a A b-3 b^2 B\right ) \sin ^8(c+d x)+360 b^8 \left (a^2 (-A)+6 a b B+3 A b^2\right ) \sin ^7(c+d x)+1260 b^8 \left (a^2 B+2 a A b-b^2 B\right ) \sin ^6(c+d x)-1512 b^8 \left (a^2 (-A)+2 a b B+A b^2\right ) \sin ^5(c+d x)+630 b^8 \left (-3 a^2 B-6 a A b+b^2 B\right ) \sin ^4(c+d x)+840 b^8 \left (-3 a^2 A+2 a b B+A b^2\right ) \sin ^3(c+d x)-3 a^4 B \left (a^6-9 a^4 b^2+42 a^2 b^4-210 b^6\right )-280 b^9 (2 a B+A b) \sin ^9(c+d x)+1260 a b^8 (a B+2 A b) \sin ^2(c+d x)-252 b^{10} B \sin ^{10}(c+d x)}{2520 b^8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(-3*a^4*(a^6 - 9*a^4*b^2 + 42*a^2*b^4 - 210*b^6)*B + 2520*a^2*A*b^8*Sin[c + d*x] + 1260*a*b^8*(2*A*b + a*B)*Si
n[c + d*x]^2 + 840*b^8*(-3*a^2*A + A*b^2 + 2*a*b*B)*Sin[c + d*x]^3 + 630*b^8*(-6*a*A*b - 3*a^2*B + b^2*B)*Sin[
c + d*x]^4 - 1512*b^8*(-(a^2*A) + A*b^2 + 2*a*b*B)*Sin[c + d*x]^5 + 1260*b^8*(2*a*A*b + a^2*B - b^2*B)*Sin[c +
d*x]^6 + 360*b^8*(-(a^2*A) + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x]^7 - 315*b^8*(2*a*A*b + a^2*B - 3*b^2*B)*Sin[c +
d*x]^8 - 280*b^9*(A*b + 2*a*B)*Sin[c + d*x]^9 - 252*b^10*B*Sin[c + d*x]^10)/(2520*b^8*d)

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fricas [A]  time = 0.51, size = 174, normalized size = 0.50 \[ \frac {252 \, B b^{2} \cos \left (d x + c\right )^{10} - 315 \, {\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{8} - 8 \, {\left (35 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{6} - 6 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 144 \, A a^{2} - 32 \, B a b - 16 \, A b^{2} - 8 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2520*(252*B*b^2*cos(d*x + c)^10 - 315*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^8 - 8*(35*(2*B*a*b + A*b^2)*cos
(d*x + c)^8 - 5*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^6 - 6*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^4 - 14
4*A*a^2 - 32*B*a*b - 16*A*b^2 - 8*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

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giac [A]  time = 0.50, size = 279, normalized size = 0.80 \[ \frac {B b^{2} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {7 \, A a^{2} \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (8 \, B a^{2} + 16 \, A a b - B b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {{\left (7 \, B a^{2} + 14 \, A a b + B b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {7 \, {\left (4 \, B a^{2} + 8 \, A a b + B b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {{\left (2 \, B a b + A b^{2}\right )} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {{\left (4 \, A a^{2} - 10 \, B a b - 5 \, A b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {{\left (7 \, A a^{2} - 4 \, B a b - 2 \, A b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {7 \, {\left (10 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/5120*B*b^2*cos(10*d*x + 10*c)/d + 7/64*A*a^2*sin(3*d*x + 3*c)/d - 1/1024*(B*a^2 + 2*A*a*b - B*b^2)*cos(8*d*x
+ 8*c)/d - 1/1024*(8*B*a^2 + 16*A*a*b - B*b^2)*cos(6*d*x + 6*c)/d - 1/256*(7*B*a^2 + 14*A*a*b + B*b^2)*cos(4*
d*x + 4*c)/d - 7/512*(4*B*a^2 + 8*A*a*b + B*b^2)*cos(2*d*x + 2*c)/d - 1/2304*(2*B*a*b + A*b^2)*sin(9*d*x + 9*c
)/d + 1/1792*(4*A*a^2 - 10*B*a*b - 5*A*b^2)*sin(7*d*x + 7*c)/d + 1/320*(7*A*a^2 - 4*B*a*b - 2*A*b^2)*sin(5*d*x
+ 5*c)/d + 7/128*(10*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c)/d

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maple [A]  time = 0.48, size = 229, normalized size = 0.66 \[ \frac {\frac {a^{2} A \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}-\frac {B \,a^{2} \left (\cos ^{8}\left (d x +c \right )\right )}{8}-\frac {A a b \left (\cos ^{8}\left (d x +c \right )\right )}{4}+2 B a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+A \,b^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+B \,b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{8}\left (d x +c \right )\right )}{40}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/7*a^2*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^2*cos(d*x+c)^8-1/4*A*a
*b*cos(d*x+c)^8+2*B*a*b*(-1/9*sin(d*x+c)*cos(d*x+c)^8+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))+A*b^2*(-1/9*sin(d*x+c)*cos(d*x+c)^8+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))+B*b^2*(-1/10*sin(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8))

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maxima [A]  time = 0.31, size = 238, normalized size = 0.68 \[ -\frac {252 \, B b^{2} \sin \left (d x + c\right )^{10} + 280 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{9} + 315 \, {\left (B a^{2} + 2 \, A a b - 3 \, B b^{2}\right )} \sin \left (d x + c\right )^{8} + 360 \, {\left (A a^{2} - 6 \, B a b - 3 \, A b^{2}\right )} \sin \left (d x + c\right )^{7} - 1260 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{6} - 1512 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 630 \, {\left (3 \, B a^{2} + 6 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 2520 \, A a^{2} \sin \left (d x + c\right ) + 840 \, {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 1260 \, {\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2520*(252*B*b^2*sin(d*x + c)^10 + 280*(2*B*a*b + A*b^2)*sin(d*x + c)^9 + 315*(B*a^2 + 2*A*a*b - 3*B*b^2)*si
n(d*x + c)^8 + 360*(A*a^2 - 6*B*a*b - 3*A*b^2)*sin(d*x + c)^7 - 1260*(B*a^2 + 2*A*a*b - B*b^2)*sin(d*x + c)^6
- 1512*(A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^5 + 630*(3*B*a^2 + 6*A*a*b - B*b^2)*sin(d*x + c)^4 - 2520*A*a^2*
sin(d*x + c) + 840*(3*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 - 1260*(B*a^2 + 2*A*a*b)*sin(d*x + c)^2)/d

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mupad [B]  time = 0.18, size = 236, normalized size = 0.68 \[ \frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )-{\sin \left (c+d\,x\right )}^9\,\left (\frac {A\,b^2}{9}+\frac {2\,B\,a\,b}{9}\right )+{\sin \left (c+d\,x\right )}^3\,\left (-A\,a^2+\frac {2\,B\,a\,b}{3}+\frac {A\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {3\,A\,a^2}{5}+\frac {6\,B\,a\,b}{5}+\frac {3\,A\,b^2}{5}\right )+{\sin \left (c+d\,x\right )}^7\,\left (-\frac {A\,a^2}{7}+\frac {6\,B\,a\,b}{7}+\frac {3\,A\,b^2}{7}\right )+{\sin \left (c+d\,x\right )}^6\,\left (\frac {B\,a^2}{2}+A\,a\,b-\frac {B\,b^2}{2}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,B\,a^2}{4}+\frac {3\,A\,a\,b}{2}-\frac {B\,b^2}{4}\right )-{\sin \left (c+d\,x\right )}^8\,\left (\frac {B\,a^2}{8}+\frac {A\,a\,b}{4}-\frac {3\,B\,b^2}{8}\right )-\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^{10}}{10}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 25.92, size = 440, normalized size = 1.26 \[ \begin {cases} \frac {16 A a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A a b \cos ^{8}{\left (c + d x \right )}}{4 d} + \frac {16 A b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 A b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 A b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {32 B a b \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {16 B a b \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {4 B a b \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin ^{10}{\left (c + d x \right )}}{40 d} + \frac {B b^{2} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} + \frac {B b^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {B b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a + b \sin {\relax (c )}\right )^{2} \cos ^{7}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((16*A*a**2*sin(c + d*x)**7/(35*d) + 8*A*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*A*a**2*sin(c
+ d*x)**3*cos(c + d*x)**4/d + A*a**2*sin(c + d*x)*cos(c + d*x)**6/d - A*a*b*cos(c + d*x)**8/(4*d) + 16*A*b**2*
sin(c + d*x)**9/(315*d) + 8*A*b**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 2*A*b**2*sin(c + d*x)**5*cos(c + d
*x)**4/(5*d) + A*b**2*sin(c + d*x)**3*cos(c + d*x)**6/(3*d) - B*a**2*cos(c + d*x)**8/(8*d) + 32*B*a*b*sin(c +
d*x)**9/(315*d) + 16*B*a*b*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 4*B*a*b*sin(c + d*x)**5*cos(c + d*x)**4/(5
*d) + 2*B*a*b*sin(c + d*x)**3*cos(c + d*x)**6/(3*d) + B*b**2*sin(c + d*x)**10/(40*d) + B*b**2*sin(c + d*x)**8*
cos(c + d*x)**2/(8*d) + B*b**2*sin(c + d*x)**6*cos(c + d*x)**4/(4*d) + B*b**2*sin(c + d*x)**4*cos(c + d*x)**6/
(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))**2*cos(c)**7, True))

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