\(\int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [1532]

Optimal result

Integrand size = 31, antiderivative size = 349 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}+\frac {\left (a^2-b^2\right )^2 \left (6 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac {3 \left (a^2-b^2\right ) \left (5 a^2 A b-A b^3-7 a^3 B+3 a b^2 B\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac {\left (20 a^3 A b-12 a A b^3-35 a^4 B+30 a^2 b^2 B-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac {\left (15 a^2 A b-3 A b^3-35 a^3 B+15 a b^2 B\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac {3 \left (2 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac {(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac {B (a+b \sin (c+d x))^{10}}{10 b^8 d} \] Output:

-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^3-35* 
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
 

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.85 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {-3 a^4 \left (a^6-9 a^4 b^2+42 a^2 b^4-210 b^6\right ) B+2520 a^2 A b^8 \sin (c+d x)+1260 a b^8 (2 A b+a B) \sin ^2(c+d x)+840 b^8 \left (-3 a^2 A+A b^2+2 a b B\right ) \sin ^3(c+d x)+630 b^8 \left (-6 a A b-3 a^2 B+b^2 B\right ) \sin ^4(c+d x)-1512 b^8 \left (-a^2 A+A b^2+2 a b B\right ) \sin ^5(c+d x)+1260 b^8 \left (2 a A b+a^2 B-b^2 B\right ) \sin ^6(c+d x)+360 b^8 \left (-a^2 A+3 A b^2+6 a b B\right ) \sin ^7(c+d x)-315 b^8 \left (2 a A b+a^2 B-3 b^2 B\right ) \sin ^8(c+d x)-280 b^9 (A b+2 a B) \sin ^9(c+d x)-252 b^{10} B \sin ^{10}(c+d x)}{2520 b^8 d} \] Input:

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

Output:

(-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)*Sin[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)/(2 
520*b^8*d)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 652, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

\(\displaystyle \frac {\int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}{b}d(b \sin (c+d x))}{b^7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^3d(b \sin (c+d x))}{b^8 d}\)

\(\Big \downarrow \) 652

\(\displaystyle \frac {\int \left (-B (a+b \sin (c+d x))^9+(7 a B-A b) (a+b \sin (c+d x))^8-3 \left (7 B a^2-2 A b a-b^2 B\right ) (a+b \sin (c+d x))^7+\left (35 B a^3-15 A b a^2-15 b^2 B a+3 A b^3\right ) (a+b \sin (c+d x))^6+\left (-35 B a^4+20 A b a^3+30 b^2 B a^2-12 A b^3 a-3 b^4 B\right ) (a+b \sin (c+d x))^5+3 \left (a^2-b^2\right ) \left (7 B a^3-5 A b a^2-3 b^2 B a+A b^3\right ) (a+b \sin (c+d x))^4-\left (a^2-b^2\right )^2 \left (7 B a^2-6 A b a-b^2 B\right ) (a+b \sin (c+d x))^3+\left (a^2-b^2\right )^3 (a B-A b) (a+b \sin (c+d x))^2\right )d(b \sin (c+d x))}{b^8 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {3}{8} \left (-7 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^8+\frac {1}{4} \left (a^2-b^2\right )^2 \left (-7 a^2 B+6 a A b+b^2 B\right ) (a+b \sin (c+d x))^4-\frac {1}{3} \left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3-\frac {1}{7} \left (-35 a^3 B+15 a^2 A b+15 a b^2 B-3 A b^3\right ) (a+b \sin (c+d x))^7-\frac {3}{5} \left (a^2-b^2\right ) \left (-7 a^3 B+5 a^2 A b+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5+\frac {1}{6} \left (-35 a^4 B+20 a^3 A b+30 a^2 b^2 B-12 a A b^3-3 b^4 B\right ) (a+b \sin (c+d x))^6-\frac {1}{9} (A b-7 a B) (a+b \sin (c+d x))^9-\frac {1}{10} B (a+b \sin (c+d x))^{10}}{b^8 d}\)

Input:

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

Output:

(-1/3*((a^2 - b^2)^3*(A*b - a*B)*(a + b*Sin[c + d*x])^3) + ((a^2 - b^2)^2* 
(6*a*A*b - 7*a^2*B + b^2*B)*(a + b*Sin[c + d*x])^4)/4 - (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 + ((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 - ((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 + (3*(2*a*A*b - 7*a^2*B + b^2*B)*(a + b*Sin[c + d*x])^8)/8 - ((A 
*b - 7*a*B)*(a + b*Sin[c + d*x])^9)/9 - (B*(a + b*Sin[c + d*x])^10)/10)/(b 
^8*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 652

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^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]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.70

\[-\frac {\frac {B \,b^{2} \sin \left (d x +c \right )^{10}}{10}+\frac {\left (b^{2} A +2 a b B \right ) \sin \left (d x +c \right )^{9}}{9}+\frac {\left (\left (a^{2}-3 b^{2}\right ) B +2 A a b \right ) \sin \left (d x +c \right )^{8}}{8}+\frac {\left (-6 a b B +\left (a^{2}-3 b^{2}\right ) A \right ) \sin \left (d x +c \right )^{7}}{7}+\frac {\left (\left (-3 a^{2}+3 b^{2}\right ) B -6 A a b \right ) \sin \left (d x +c \right )^{6}}{6}+\frac {\left (6 a b B +\left (-3 a^{2}+3 b^{2}\right ) A \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\left (\left (3 a^{2}-b^{2}\right ) B +6 A a b \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (-2 a b B +\left (3 a^{2}-b^{2}\right ) A \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-2 A a b -a^{2} B \right ) \sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right ) A \,a^{2}}{d}\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.50 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {252 \, B b^{2} \cos \left (d x + c\right )^{10} - 315 \, {\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{8} - 8 \, {\left (35 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{6} - 6 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 144 \, A a^{2} - 32 \, B a b - 16 \, A b^{2} - 8 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \] Input:

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

Output:

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 - 144* 
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
 

Sympy [A] (verification not implemented)

Time = 75.18 (sec) , antiderivative size = 440, normalized size of antiderivative = 1.26 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {16 A a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A a b \cos ^{8}{\left (c + d x \right )}}{4 d} + \frac {16 A b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 A b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 A b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {32 B a b \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {16 B a b \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {4 B a b \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin ^{10}{\left (c + d x \right )}}{40 d} + \frac {B b^{2} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} + \frac {B b^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {B b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{7}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

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*s 
in(c + d*x)*cos(c + d*x)**6/d - A*a*b*cos(c + d*x)**8/(4*d) + 16*A*b**2*si 
n(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*co 
s(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*si 
n(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*si 
n(c + d*x)**4*cos(c + d*x)**6/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*s 
in(c))**2*cos(c)**7, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.68 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {252 \, B b^{2} \sin \left (d x + c\right )^{10} + 280 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{9} + 315 \, {\left (B a^{2} + 2 \, A a b - 3 \, B b^{2}\right )} \sin \left (d x + c\right )^{8} + 360 \, {\left (A a^{2} - 6 \, B a b - 3 \, A b^{2}\right )} \sin \left (d x + c\right )^{7} - 1260 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{6} - 1512 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 630 \, {\left (3 \, B a^{2} + 6 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 2520 \, A a^{2} \sin \left (d x + c\right ) + 840 \, {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 1260 \, {\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{2520 \, d} \] Input:

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

Output:

-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)*sin(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
 

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.95 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {252 \, B b^{2} \sin \left (d x + c\right )^{10} + 560 \, B a b \sin \left (d x + c\right )^{9} + 280 \, A b^{2} \sin \left (d x + c\right )^{9} + 315 \, B a^{2} \sin \left (d x + c\right )^{8} + 630 \, A a b \sin \left (d x + c\right )^{8} - 945 \, B b^{2} \sin \left (d x + c\right )^{8} + 360 \, A a^{2} \sin \left (d x + c\right )^{7} - 2160 \, B a b \sin \left (d x + c\right )^{7} - 1080 \, A b^{2} \sin \left (d x + c\right )^{7} - 1260 \, B a^{2} \sin \left (d x + c\right )^{6} - 2520 \, A a b \sin \left (d x + c\right )^{6} + 1260 \, B b^{2} \sin \left (d x + c\right )^{6} - 1512 \, A a^{2} \sin \left (d x + c\right )^{5} + 3024 \, B a b \sin \left (d x + c\right )^{5} + 1512 \, A b^{2} \sin \left (d x + c\right )^{5} + 1890 \, B a^{2} \sin \left (d x + c\right )^{4} + 3780 \, A a b \sin \left (d x + c\right )^{4} - 630 \, B b^{2} \sin \left (d x + c\right )^{4} + 2520 \, A a^{2} \sin \left (d x + c\right )^{3} - 1680 \, B a b \sin \left (d x + c\right )^{3} - 840 \, A b^{2} \sin \left (d x + c\right )^{3} - 1260 \, B a^{2} \sin \left (d x + c\right )^{2} - 2520 \, A a b \sin \left (d x + c\right )^{2} - 2520 \, A a^{2} \sin \left (d x + c\right )}{2520 \, d} \] Input:

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

Output:

-1/2520*(252*B*b^2*sin(d*x + c)^10 + 560*B*a*b*sin(d*x + c)^9 + 280*A*b^2* 
sin(d*x + c)^9 + 315*B*a^2*sin(d*x + c)^8 + 630*A*a*b*sin(d*x + c)^8 - 945 
*B*b^2*sin(d*x + c)^8 + 360*A*a^2*sin(d*x + c)^7 - 2160*B*a*b*sin(d*x + c) 
^7 - 1080*A*b^2*sin(d*x + c)^7 - 1260*B*a^2*sin(d*x + c)^6 - 2520*A*a*b*si 
n(d*x + c)^6 + 1260*B*b^2*sin(d*x + c)^6 - 1512*A*a^2*sin(d*x + c)^5 + 302 
4*B*a*b*sin(d*x + c)^5 + 1512*A*b^2*sin(d*x + c)^5 + 1890*B*a^2*sin(d*x + 
c)^4 + 3780*A*a*b*sin(d*x + c)^4 - 630*B*b^2*sin(d*x + c)^4 + 2520*A*a^2*s 
in(d*x + c)^3 - 1680*B*a*b*sin(d*x + c)^3 - 840*A*b^2*sin(d*x + c)^3 - 126 
0*B*a^2*sin(d*x + c)^2 - 2520*A*a*b*sin(d*x + c)^2 - 2520*A*a^2*sin(d*x + 
c))/d
 

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.68 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )-{\sin \left (c+d\,x\right )}^9\,\left (\frac {A\,b^2}{9}+\frac {2\,B\,a\,b}{9}\right )+{\sin \left (c+d\,x\right )}^3\,\left (-A\,a^2+\frac {2\,B\,a\,b}{3}+\frac {A\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {3\,A\,a^2}{5}+\frac {6\,B\,a\,b}{5}+\frac {3\,A\,b^2}{5}\right )+{\sin \left (c+d\,x\right )}^7\,\left (-\frac {A\,a^2}{7}+\frac {6\,B\,a\,b}{7}+\frac {3\,A\,b^2}{7}\right )+{\sin \left (c+d\,x\right )}^6\,\left (\frac {B\,a^2}{2}+A\,a\,b-\frac {B\,b^2}{2}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,B\,a^2}{4}+\frac {3\,A\,a\,b}{2}-\frac {B\,b^2}{4}\right )-{\sin \left (c+d\,x\right )}^8\,\left (\frac {B\,a^2}{8}+\frac {A\,a\,b}{4}-\frac {3\,B\,b^2}{8}\right )-\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^{10}}{10}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \] Input:

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

Output:

(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
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.62 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) \left (-84 \sin \left (d x +c \right )^{9} b^{3}-280 \sin \left (d x +c \right )^{8} a \,b^{2}-315 \sin \left (d x +c \right )^{7} a^{2} b +315 \sin \left (d x +c \right )^{7} b^{3}-120 \sin \left (d x +c \right )^{6} a^{3}+1080 \sin \left (d x +c \right )^{6} a \,b^{2}+1260 \sin \left (d x +c \right )^{5} a^{2} b -420 \sin \left (d x +c \right )^{5} b^{3}+504 \sin \left (d x +c \right )^{4} a^{3}-1512 \sin \left (d x +c \right )^{4} a \,b^{2}-1890 \sin \left (d x +c \right )^{3} a^{2} b +210 \sin \left (d x +c \right )^{3} b^{3}-840 \sin \left (d x +c \right )^{2} a^{3}+840 \sin \left (d x +c \right )^{2} a \,b^{2}+1260 \sin \left (d x +c \right ) a^{2} b +840 a^{3}\right )}{840 d} \] Input:

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

Output:

(sin(c + d*x)*( - 84*sin(c + d*x)**9*b**3 - 280*sin(c + d*x)**8*a*b**2 - 3 
15*sin(c + d*x)**7*a**2*b + 315*sin(c + d*x)**7*b**3 - 120*sin(c + d*x)**6 
*a**3 + 1080*sin(c + d*x)**6*a*b**2 + 1260*sin(c + d*x)**5*a**2*b - 420*si 
n(c + d*x)**5*b**3 + 504*sin(c + d*x)**4*a**3 - 1512*sin(c + d*x)**4*a*b** 
2 - 1890*sin(c + d*x)**3*a**2*b + 210*sin(c + d*x)**3*b**3 - 840*sin(c + d 
*x)**2*a**3 + 840*sin(c + d*x)**2*a*b**2 + 1260*sin(c + d*x)*a**2*b + 840* 
a**3))/(840*d)