\(\int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [974]

Optimal result

Integrand size = 31, antiderivative size = 196 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {5}{128} a^2 (9 A+2 B) x-\frac {a^2 (9 A+2 B) \cos ^7(c+d x)}{56 d}+\frac {5 a^2 (9 A+2 B) \cos (c+d x) \sin (c+d x)}{128 d}+\frac {5 a^2 (9 A+2 B) \cos ^3(c+d x) \sin (c+d x)}{192 d}+\frac {a^2 (9 A+2 B) \cos ^5(c+d x) \sin (c+d x)}{48 d}-\frac {B \cos ^7(c+d x) (a+a \sin (c+d x))^2}{9 d}-\frac {(9 A+2 B) \cos ^7(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{72 d} \] Output:

5/128*a^2*(9*A+2*B)*x-1/56*a^2*(9*A+2*B)*cos(d*x+c)^7/d+5/128*a^2*(9*A+2*B 
)*cos(d*x+c)*sin(d*x+c)/d+5/192*a^2*(9*A+2*B)*cos(d*x+c)^3*sin(d*x+c)/d+1/ 
48*a^2*(9*A+2*B)*cos(d*x+c)^5*sin(d*x+c)/d-1/9*B*cos(d*x+c)^7*(a+a*sin(d*x 
+c))^2/d-1/72*(9*A+2*B)*cos(d*x+c)^7*(a^2+a^2*sin(d*x+c))/d
 

Mathematica [A] (verified)

Time = 5.46 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.10 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {a^2 \cos (c+d x) \left (2880 A+1900 B+\frac {2520 (9 A+2 B) \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(c+d x)}}+32 (135 A+86 B) \cos (2 (c+d x))+16 (108 A+59 B) \cos (4 (c+d x))+288 A \cos (6 (c+d x))+64 B \cos (6 (c+d x))-28 B \cos (8 (c+d x))-13671 A \sin (c+d x)-2478 B \sin (c+d x)-2457 A \sin (3 (c+d x))+462 B \sin (3 (c+d x))-63 A \sin (5 (c+d x))+546 B \sin (5 (c+d x))+63 A \sin (7 (c+d x))+126 B \sin (7 (c+d x))\right )}{32256 d} \] Input:

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

Output:

-1/32256*(a^2*Cos[c + d*x]*(2880*A + 1900*B + (2520*(9*A + 2*B)*ArcSin[Sqr 
t[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c + d*x]^2] + 32*(135*A + 86*B)*Cos 
[2*(c + d*x)] + 16*(108*A + 59*B)*Cos[4*(c + d*x)] + 288*A*Cos[6*(c + d*x) 
] + 64*B*Cos[6*(c + d*x)] - 28*B*Cos[8*(c + d*x)] - 13671*A*Sin[c + d*x] - 
 2478*B*Sin[c + d*x] - 2457*A*Sin[3*(c + d*x)] + 462*B*Sin[3*(c + d*x)] - 
63*A*Sin[5*(c + d*x)] + 546*B*Sin[5*(c + d*x)] + 63*A*Sin[7*(c + d*x)] + 1 
26*B*Sin[7*(c + d*x)]))/d
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.87, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3339, 3042, 3157, 3042, 3148, 3042, 3115, 3042, 3115, 3042, 3115, 24}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^6 (a \sin (c+d x)+a)^2 (A+B \sin (c+d x))dx\)

\(\Big \downarrow \) 3339

\(\displaystyle \frac {1}{9} (9 A+2 B) \int \cos ^6(c+d x) (\sin (c+d x) a+a)^2dx-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 A+2 B) \int \cos (c+d x)^6 (\sin (c+d x) a+a)^2dx-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3157

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \int \cos ^6(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \int \cos (c+d x)^6 (\sin (c+d x) a+a)dx-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \int \cos ^6(c+d x)dx-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \int \cos ^4(c+d x)dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{9} (9 A+2 B) \left (\frac {9}{8} a \left (a \left (\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\)

Input:

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

Output:

-1/9*(B*Cos[c + d*x]^7*(a + a*Sin[c + d*x])^2)/d + ((9*A + 2*B)*(-1/8*(Cos 
[c + d*x]^7*(a^2 + a^2*Sin[c + d*x]))/d + (9*a*(-1/7*(a*Cos[c + d*x]^7)/d 
+ a*((Cos[c + d*x]^5*Sin[c + d*x])/(6*d) + (5*((Cos[c + d*x]^3*Sin[c + d*x 
])/(4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6)))/8))/9
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3115

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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3148

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

rule 3157

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && NeQ[m + p, 0] && Integers 
Q[2*m, 2*p]
 

rule 3339

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* 
(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S 
imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1))   Int[(g*Cos[e + f*x])^p*(a + 
 b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ 
a^2 - b^2, 0] && NeQ[m + p + 1, 0]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.25

\[\frac {A \,a^{2} \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )+a^{2} B \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )-\frac {2 A \cos \left (d x +c \right )^{7} a^{2}}{7}+2 a^{2} B \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )+A \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {a^{2} B \cos \left (d x +c \right )^{7}}{7}}{d}\]

Input:

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

Output:

1/d*(A*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)+a^2*B*(-1/9*cos(d*x+c)^7 
*sin(d*x+c)^2-2/63*cos(d*x+c)^7)-2/7*A*cos(d*x+c)^7*a^2+2*a^2*B*(-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))*s 
in(d*x+c)+5/128*d*x+5/128*c)+A*a^2*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/ 
8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)-1/7*a^2*B*cos(d*x+c)^7)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.69 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {896 \, B a^{2} \cos \left (d x + c\right )^{9} - 2304 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{7} + 315 \, {\left (9 \, A + 2 \, B\right )} a^{2} d x - 21 \, {\left (48 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{7} - 8 \, {\left (9 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{5} - 10 \, {\left (9 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} - 15 \, {\left (9 \, A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8064 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/8064*(896*B*a^2*cos(d*x + c)^9 - 2304*(A + B)*a^2*cos(d*x + c)^7 + 315*( 
9*A + 2*B)*a^2*d*x - 21*(48*(A + 2*B)*a^2*cos(d*x + c)^7 - 8*(9*A + 2*B)*a 
^2*cos(d*x + c)^5 - 10*(9*A + 2*B)*a^2*cos(d*x + c)^3 - 15*(9*A + 2*B)*a^2 
*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 719 vs. \(2 (180) = 360\).

Time = 1.01 (sec) , antiderivative size = 719, normalized size of antiderivative = 3.67 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((5*A*a**2*x*sin(c + d*x)**8/128 + 5*A*a**2*x*sin(c + d*x)**6*cos 
(c + d*x)**2/32 + 5*A*a**2*x*sin(c + d*x)**6/16 + 15*A*a**2*x*sin(c + d*x) 
**4*cos(c + d*x)**4/64 + 15*A*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 
5*A*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 15*A*a**2*x*sin(c + d*x)** 
2*cos(c + d*x)**4/16 + 5*A*a**2*x*cos(c + d*x)**8/128 + 5*A*a**2*x*cos(c + 
 d*x)**6/16 + 5*A*a**2*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 55*A*a**2*si 
n(c + d*x)**5*cos(c + d*x)**3/(384*d) + 5*A*a**2*sin(c + d*x)**5*cos(c + d 
*x)/(16*d) + 73*A*a**2*sin(c + d*x)**3*cos(c + d*x)**5/(384*d) + 5*A*a**2* 
sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - 5*A*a**2*sin(c + d*x)*cos(c + d*x) 
**7/(128*d) + 11*A*a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 2*A*a**2*cos 
(c + d*x)**7/(7*d) + 5*B*a**2*x*sin(c + d*x)**8/64 + 5*B*a**2*x*sin(c + d* 
x)**6*cos(c + d*x)**2/16 + 15*B*a**2*x*sin(c + d*x)**4*cos(c + d*x)**4/32 
+ 5*B*a**2*x*sin(c + d*x)**2*cos(c + d*x)**6/16 + 5*B*a**2*x*cos(c + d*x)* 
*8/64 + 5*B*a**2*sin(c + d*x)**7*cos(c + d*x)/(64*d) + 55*B*a**2*sin(c + d 
*x)**5*cos(c + d*x)**3/(192*d) + 73*B*a**2*sin(c + d*x)**3*cos(c + d*x)**5 
/(192*d) - B*a**2*sin(c + d*x)**2*cos(c + d*x)**7/(7*d) - 5*B*a**2*sin(c + 
 d*x)*cos(c + d*x)**7/(64*d) - 2*B*a**2*cos(c + d*x)**9/(63*d) - B*a**2*co 
s(c + d*x)**7/(7*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c) 
**6, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.06 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {18432 \, A a^{2} \cos \left (d x + c\right )^{7} + 9216 \, B a^{2} \cos \left (d x + c\right )^{7} - 21 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{2} + 336 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 1024 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} B a^{2} - 42 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{2}}{64512 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/64512*(18432*A*a^2*cos(d*x + c)^7 + 9216*B*a^2*cos(d*x + c)^7 - 21*(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*a^2 + 336*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4 
*c) - 48*sin(2*d*x + 2*c))*A*a^2 - 1024*(7*cos(d*x + c)^9 - 9*cos(d*x + c) 
^7)*B*a^2 - 42*(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))*B*a^2)/d
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.20 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B a^{2} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {B a^{2} \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} + \frac {5}{128} \, {\left (9 \, A a^{2} + 2 \, B a^{2}\right )} x - \frac {{\left (8 \, A a^{2} + B a^{2}\right )} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (2 \, A a^{2} + B a^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {{\left (18 \, A a^{2} + 11 \, B a^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (20 \, A a^{2} + 13 \, B a^{2}\right )} \cos \left (d x + c\right )}{128 \, d} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {{\left (5 \, A a^{2} - 2 \, B a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{128 \, d} + \frac {{\left (8 \, A a^{2} + B a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/2304*B*a^2*cos(9*d*x + 9*c)/d - 1/96*B*a^2*sin(6*d*x + 6*c)/d + 5/128*(9 
*A*a^2 + 2*B*a^2)*x - 1/1792*(8*A*a^2 + B*a^2)*cos(7*d*x + 7*c)/d - 1/64*( 
2*A*a^2 + B*a^2)*cos(5*d*x + 5*c)/d - 1/192*(18*A*a^2 + 11*B*a^2)*cos(3*d* 
x + 3*c)/d - 1/128*(20*A*a^2 + 13*B*a^2)*cos(d*x + c)/d - 1/1024*(A*a^2 + 
2*B*a^2)*sin(8*d*x + 8*c)/d + 1/128*(5*A*a^2 - 2*B*a^2)*sin(4*d*x + 4*c)/d 
 + 1/32*(8*A*a^2 + B*a^2)*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 36.13 (sec) , antiderivative size = 622, normalized size of antiderivative = 3.17 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

(5*a^2*atan((5*a^2*tan(c/2 + (d*x)/2)*(9*A + 2*B))/(64*((45*A*a^2)/64 + (5 
*B*a^2)/32)))*(9*A + 2*B))/(64*d) - (5*a^2*(9*A + 2*B)*(atan(tan(c/2 + (d* 
x)/2)) - (d*x)/2))/(64*d) - ((4*A*a^2)/7 - tan(c/2 + (d*x)/2)*((83*A*a^2)/ 
64 - (5*B*a^2)/32) + (22*B*a^2)/63 + tan(c/2 + (d*x)/2)^16*(4*A*a^2 + 2*B* 
a^2) + tan(c/2 + (d*x)/2)^14*(8*A*a^2 + 8*B*a^2) + tan(c/2 + (d*x)/2)^2*(( 
8*A*a^2)/7 + (8*B*a^2)/7) + tan(c/2 + (d*x)/2)^8*(32*A*a^2 + 4*B*a^2) + ta 
n(c/2 + (d*x)/2)^6*(24*A*a^2 + 24*B*a^2) + tan(c/2 + (d*x)/2)^12*(24*A*a^2 
 + (16*B*a^2)/3) + tan(c/2 + (d*x)/2)^10*(40*A*a^2 + 40*B*a^2) + tan(c/2 + 
 (d*x)/2)^4*((88*A*a^2)/7 + (32*B*a^2)/7) + tan(c/2 + (d*x)/2)^17*((83*A*a 
^2)/64 - (5*B*a^2)/32) - tan(c/2 + (d*x)/2)^5*((149*A*a^2)/32 - (83*B*a^2) 
/16) + tan(c/2 + (d*x)/2)^13*((149*A*a^2)/32 - (83*B*a^2)/16) - tan(c/2 + 
(d*x)/2)^3*((189*A*a^2)/32 + (191*B*a^2)/48) + tan(c/2 + (d*x)/2)^15*((189 
*A*a^2)/32 + (191*B*a^2)/48) - tan(c/2 + (d*x)/2)^7*((409*A*a^2)/32 + (145 
*B*a^2)/16) + tan(c/2 + (d*x)/2)^11*((409*A*a^2)/32 + (145*B*a^2)/16))/(d* 
(9*tan(c/2 + (d*x)/2)^2 + 36*tan(c/2 + (d*x)/2)^4 + 84*tan(c/2 + (d*x)/2)^ 
6 + 126*tan(c/2 + (d*x)/2)^8 + 126*tan(c/2 + (d*x)/2)^10 + 84*tan(c/2 + (d 
*x)/2)^12 + 36*tan(c/2 + (d*x)/2)^14 + 9*tan(c/2 + (d*x)/2)^16 + tan(c/2 + 
 (d*x)/2)^18 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.50 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {a^{2} \left (896 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} b +1008 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} a +2016 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} b +2304 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a -1280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} b -1512 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a -5712 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b -6912 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a -1536 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b -1890 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +4956 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +6912 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +3328 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +5229 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -630 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -2304 \cos \left (d x +c \right ) a -1408 \cos \left (d x +c \right ) b +2835 a d x +2304 a +630 b d x +1408 b \right )}{8064 d} \] Input:

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

Output:

(a**2*(896*cos(c + d*x)*sin(c + d*x)**8*b + 1008*cos(c + d*x)*sin(c + d*x) 
**7*a + 2016*cos(c + d*x)*sin(c + d*x)**7*b + 2304*cos(c + d*x)*sin(c + d* 
x)**6*a - 1280*cos(c + d*x)*sin(c + d*x)**6*b - 1512*cos(c + d*x)*sin(c + 
d*x)**5*a - 5712*cos(c + d*x)*sin(c + d*x)**5*b - 6912*cos(c + d*x)*sin(c 
+ d*x)**4*a - 1536*cos(c + d*x)*sin(c + d*x)**4*b - 1890*cos(c + d*x)*sin( 
c + d*x)**3*a + 4956*cos(c + d*x)*sin(c + d*x)**3*b + 6912*cos(c + d*x)*si 
n(c + d*x)**2*a + 3328*cos(c + d*x)*sin(c + d*x)**2*b + 5229*cos(c + d*x)* 
sin(c + d*x)*a - 630*cos(c + d*x)*sin(c + d*x)*b - 2304*cos(c + d*x)*a - 1 
408*cos(c + d*x)*b + 2835*a*d*x + 2304*a + 630*b*d*x + 1408*b))/(8064*d)