\(\int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [992]

Optimal result

Integrand size = 31, antiderivative size = 231 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {11}{256} a^3 (10 A+3 B) x-\frac {11 a^3 (10 A+3 B) \cos ^7(c+d x)}{560 d}+\frac {11 a^3 (10 A+3 B) \cos (c+d x) \sin (c+d x)}{256 d}+\frac {11 a^3 (10 A+3 B) \cos ^3(c+d x) \sin (c+d x)}{384 d}+\frac {11 a^3 (10 A+3 B) \cos ^5(c+d x) \sin (c+d x)}{480 d}-\frac {a (10 A+3 B) \cos ^7(c+d x) (a+a \sin (c+d x))^2}{90 d}-\frac {B \cos ^7(c+d x) (a+a \sin (c+d x))^3}{10 d}-\frac {11 (10 A+3 B) \cos ^7(c+d x) \left (a^3+a^3 \sin (c+d x)\right )}{720 d} \] Output:

11/256*a^3*(10*A+3*B)*x-11/560*a^3*(10*A+3*B)*cos(d*x+c)^7/d+11/256*a^3*(1 
0*A+3*B)*cos(d*x+c)*sin(d*x+c)/d+11/384*a^3*(10*A+3*B)*cos(d*x+c)^3*sin(d* 
x+c)/d+11/480*a^3*(10*A+3*B)*cos(d*x+c)^5*sin(d*x+c)/d-1/90*a*(10*A+3*B)*c 
os(d*x+c)^7*(a+a*sin(d*x+c))^2/d-1/10*B*cos(d*x+c)^7*(a+a*sin(d*x+c))^3/d- 
11/720*(10*A+3*B)*cos(d*x+c)^7*(a^3+a^3*sin(d*x+c))/d
 

Mathematica [A] (verified)

Time = 6.10 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.49 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {B \cos ^7(c+d x) (a+a \sin (c+d x))^3}{10 d}-\frac {32 \sqrt {2} a^2 (10 a A+3 a B) \cos ^7(c+d x) \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^{13/2} \left (\frac {7}{18} \left (\frac {99}{2048 \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^6}+\frac {33}{256 \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^5}+\frac {33}{128 \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^4}+\frac {99}{224 \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^3}+\frac {11}{16 \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^2}+\frac {1}{1+\frac {1}{2} (-1+\sin (c+d x))}\right )+\frac {385 \left (-1+\frac {\sqrt {2} \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right ) \sqrt {1-\sin (c+d x)}}{\sqrt {1+\frac {1}{2} (-1+\sin (c+d x))}}-\frac {1}{3} (1-\sin (c+d x))^2-\frac {2}{15} (1-\sin (c+d x))^3+\sin (c+d x)\right )}{8192 \left (1+\frac {1}{2} (-1+\sin (c+d x))\right )^6 (1-\sin (c+d x))^4}\right )}{35 d (1+\sin (c+d x))^{7/2}} \] Input:

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

Output:

-1/10*(B*Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3)/d - (32*Sqrt[2]*a^2*(10*a* 
A + 3*a*B)*Cos[c + d*x]^7*(1 + (-1 + Sin[c + d*x])/2)^(13/2)*((7*(99/(2048 
*(1 + (-1 + Sin[c + d*x])/2)^6) + 33/(256*(1 + (-1 + Sin[c + d*x])/2)^5) + 
 33/(128*(1 + (-1 + Sin[c + d*x])/2)^4) + 99/(224*(1 + (-1 + Sin[c + d*x]) 
/2)^3) + 11/(16*(1 + (-1 + Sin[c + d*x])/2)^2) + (1 + (-1 + Sin[c + d*x])/ 
2)^(-1)))/18 + (385*(-1 + (Sqrt[2]*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]* 
Sqrt[1 - Sin[c + d*x]])/Sqrt[1 + (-1 + Sin[c + d*x])/2] - (1 - Sin[c + d*x 
])^2/3 - (2*(1 - Sin[c + d*x])^3)/15 + Sin[c + d*x]))/(8192*(1 + (-1 + Sin 
[c + d*x])/2)^6*(1 - Sin[c + d*x])^4)))/(35*d*(1 + Sin[c + d*x])^(7/2))
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.89, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3339, 3042, 3157, 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.

\(\displaystyle \int \cos ^6(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^6 (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\)

\(\Big \downarrow \) 3339

\(\displaystyle \frac {1}{10} (10 A+3 B) \int \cos ^6(c+d x) (\sin (c+d x) a+a)^3dx-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} (10 A+3 B) \int \cos (c+d x)^6 (\sin (c+d x) a+a)^3dx-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3157

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \int \cos ^6(c+d x) (\sin (c+d x) a+a)^2dx-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \int \cos (c+d x)^6 (\sin (c+d x) a+a)^2dx-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3157

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \int \cos ^6(c+d x) (\sin (c+d x) a+a)dx-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \int \cos (c+d x)^6 (\sin (c+d x) a+a)dx-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \int \cos ^6(c+d x)dx-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \int \cos ^4(c+d x)dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{10} (10 A+3 B) \left (\frac {11}{9} a \left (\frac {9}{8} a \left (a \left (\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )-\frac {a \cos ^7(c+d x)}{7 d}\right )-\frac {\cos ^7(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{8 d}\right )-\frac {a \cos ^7(c+d x) (a \sin (c+d x)+a)^2}{9 d}\right )-\frac {B \cos ^7(c+d x) (a \sin (c+d x)+a)^3}{10 d}\)

Input:

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

Output:

-1/10*(B*Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3)/d + ((10*A + 3*B)*(-1/9*(a 
*Cos[c + d*x]^7*(a + a*Sin[c + d*x])^2)/d + (11*a*(-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))/10
 

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 = 0.43 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.57

\[\frac {a^{3} A \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )+a^{3} B \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{7}}{10}-\frac {3 \cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{80}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{160}+\frac {3 d x}{256}+\frac {3 c}{256}\right )+3 a^{3} A \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )+3 a^{3} B \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )^{2}}{9}-\frac {2 \cos \left (d x +c \right )^{7}}{63}\right )-\frac {3 a^{3} A \cos \left (d x +c \right )^{7}}{7}+3 a^{3} B \left (-\frac {\cos \left (d x +c \right )^{7} \sin \left (d x +c \right )}{8}+\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{48}+\frac {5 d x}{128}+\frac {5 c}{128}\right )+a^{3} A \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {a^{3} B \cos \left (d x +c \right )^{7}}{7}}{d}\]

Input:

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

Output:

1/d*(a^3*A*(-1/9*cos(d*x+c)^7*sin(d*x+c)^2-2/63*cos(d*x+c)^7)+a^3*B*(-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)+3*a^3*A* 
(-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)+3*a^3*B*(-1/9*cos(d*x+c)^7*sin(d*x+ 
c)^2-2/63*cos(d*x+c)^7)-3/7*a^3*A*cos(d*x+c)^7+3*a^3*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))*sin(d*x+c) 
+5/128*d*x+5/128*c)+a^3*A*(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^3*B*cos(d*x+c)^7)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.67 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {8960 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{9} - 46080 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{7} + 3465 \, {\left (10 \, A + 3 \, B\right )} a^{3} d x + 21 \, {\left (384 \, B a^{3} \cos \left (d x + c\right )^{9} - 48 \, {\left (30 \, A + 41 \, B\right )} a^{3} \cos \left (d x + c\right )^{7} + 88 \, {\left (10 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{5} + 110 \, {\left (10 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 165 \, {\left (10 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{80640 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/80640*(8960*(A + 3*B)*a^3*cos(d*x + c)^9 - 46080*(A + B)*a^3*cos(d*x + c 
)^7 + 3465*(10*A + 3*B)*a^3*d*x + 21*(384*B*a^3*cos(d*x + c)^9 - 48*(30*A 
+ 41*B)*a^3*cos(d*x + c)^7 + 88*(10*A + 3*B)*a^3*cos(d*x + c)^5 + 110*(10* 
A + 3*B)*a^3*cos(d*x + c)^3 + 165*(10*A + 3*B)*a^3*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. 1042 vs. \(2 (219) = 438\).

Time = 1.45 (sec) , antiderivative size = 1042, normalized size of antiderivative = 4.51 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((15*A*a**3*x*sin(c + d*x)**8/128 + 15*A*a**3*x*sin(c + d*x)**6*c 
os(c + d*x)**2/32 + 5*A*a**3*x*sin(c + d*x)**6/16 + 45*A*a**3*x*sin(c + d* 
x)**4*cos(c + d*x)**4/64 + 15*A*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 
+ 15*A*a**3*x*sin(c + d*x)**2*cos(c + d*x)**6/32 + 15*A*a**3*x*sin(c + d*x 
)**2*cos(c + d*x)**4/16 + 15*A*a**3*x*cos(c + d*x)**8/128 + 5*A*a**3*x*cos 
(c + d*x)**6/16 + 15*A*a**3*sin(c + d*x)**7*cos(c + d*x)/(128*d) + 55*A*a* 
*3*sin(c + d*x)**5*cos(c + d*x)**3/(128*d) + 5*A*a**3*sin(c + d*x)**5*cos( 
c + d*x)/(16*d) + 73*A*a**3*sin(c + d*x)**3*cos(c + d*x)**5/(128*d) + 5*A* 
a**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) - A*a**3*sin(c + d*x)**2*cos(c 
+ d*x)**7/(7*d) - 15*A*a**3*sin(c + d*x)*cos(c + d*x)**7/(128*d) + 11*A*a* 
*3*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 2*A*a**3*cos(c + d*x)**9/(63*d) - 
 3*A*a**3*cos(c + d*x)**7/(7*d) + 3*B*a**3*x*sin(c + d*x)**10/256 + 15*B*a 
**3*x*sin(c + d*x)**8*cos(c + d*x)**2/256 + 15*B*a**3*x*sin(c + d*x)**8/12 
8 + 15*B*a**3*x*sin(c + d*x)**6*cos(c + d*x)**4/128 + 15*B*a**3*x*sin(c + 
d*x)**6*cos(c + d*x)**2/32 + 15*B*a**3*x*sin(c + d*x)**4*cos(c + d*x)**6/1 
28 + 45*B*a**3*x*sin(c + d*x)**4*cos(c + d*x)**4/64 + 15*B*a**3*x*sin(c + 
d*x)**2*cos(c + d*x)**8/256 + 15*B*a**3*x*sin(c + d*x)**2*cos(c + d*x)**6/ 
32 + 3*B*a**3*x*cos(c + d*x)**10/256 + 15*B*a**3*x*cos(c + d*x)**8/128 + 3 
*B*a**3*sin(c + d*x)**9*cos(c + d*x)/(256*d) + 7*B*a**3*sin(c + d*x)**7*co 
s(c + d*x)**3/(128*d) + 15*B*a**3*sin(c + d*x)**7*cos(c + d*x)/(128*d) ...
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.23 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {276480 \, A a^{3} \cos \left (d x + c\right )^{7} + 92160 \, B a^{3} \cos \left (d x + c\right )^{7} - 10240 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} A a^{3} - 630 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} A a^{3} + 3360 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 30720 \, {\left (7 \, \cos \left (d x + c\right )^{9} - 9 \, \cos \left (d x + c\right )^{7}\right )} B a^{3} - 63 \, {\left (32 \, \sin \left (2 \, d x + 2 \, c\right )^{5} + 120 \, d x + 120 \, c + 5 \, \sin \left (8 \, d x + 8 \, c\right ) - 40 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3} - 630 \, {\left (64 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 120 \, d x + 120 \, c - 3 \, \sin \left (8 \, d x + 8 \, c\right ) - 24 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a^{3}}{645120 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/645120*(276480*A*a^3*cos(d*x + c)^7 + 92160*B*a^3*cos(d*x + c)^7 - 1024 
0*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*A*a^3 - 630*(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^3 + 336 
0*(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^3 - 30720*(7*cos(d*x + c)^9 - 9*cos(d*x + c)^7)*B*a^3 - 63*( 
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))*B*a^3 - 630*(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^3)/d
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.18 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B a^{3} \sin \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {11}{256} \, {\left (10 \, A a^{3} + 3 \, B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {{\left (9 \, A a^{3} - 5 \, B a^{3}\right )} \cos \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (3 \, A a^{3} + B a^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{64 \, d} - \frac {{\left (29 \, A a^{3} + 15 \, B a^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (33 \, A a^{3} + 19 \, B a^{3}\right )} \cos \left (d x + c\right )}{128 \, d} - \frac {{\left (6 \, A a^{3} + 5 \, B a^{3}\right )} \sin \left (8 \, d x + 8 \, c\right )}{2048 \, d} - \frac {{\left (32 \, A a^{3} + 51 \, B a^{3}\right )} \sin \left (6 \, d x + 6 \, c\right )}{3072 \, d} + \frac {{\left (6 \, A a^{3} - 7 \, B a^{3}\right )} \sin \left (4 \, d x + 4 \, c\right )}{256 \, d} + \frac {{\left (144 \, A a^{3} + 25 \, B a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{512 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/5120*B*a^3*sin(10*d*x + 10*c)/d + 11/256*(10*A*a^3 + 3*B*a^3)*x + 1/2304 
*(A*a^3 + 3*B*a^3)*cos(9*d*x + 9*c)/d - 1/1792*(9*A*a^3 - 5*B*a^3)*cos(7*d 
*x + 7*c)/d - 1/64*(3*A*a^3 + B*a^3)*cos(5*d*x + 5*c)/d - 1/192*(29*A*a^3 
+ 15*B*a^3)*cos(3*d*x + 3*c)/d - 1/128*(33*A*a^3 + 19*B*a^3)*cos(d*x + c)/ 
d - 1/2048*(6*A*a^3 + 5*B*a^3)*sin(8*d*x + 8*c)/d - 1/3072*(32*A*a^3 + 51* 
B*a^3)*sin(6*d*x + 6*c)/d + 1/256*(6*A*a^3 - 7*B*a^3)*sin(4*d*x + 4*c)/d + 
 1/512*(144*A*a^3 + 25*B*a^3)*sin(2*d*x + 2*c)/d
 

Mupad [B] (verification not implemented)

Time = 35.62 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.08 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

(11*a^3*atan((11*a^3*tan(c/2 + (d*x)/2)*(10*A + 3*B))/(128*((55*A*a^3)/64 
+ (33*B*a^3)/128)))*(10*A + 3*B))/(128*d) - (11*a^3*(10*A + 3*B)*(atan(tan 
(c/2 + (d*x)/2)) - (d*x)/2))/(128*d) - ((58*A*a^3)/63 - tan(c/2 + (d*x)/2) 
*((73*A*a^3)/64 - (33*B*a^3)/128) + (10*B*a^3)/21 + tan(c/2 + (d*x)/2)^18* 
(6*A*a^3 + 2*B*a^3) + tan(c/2 + (d*x)/2)^16*(22*A*a^3 + 18*B*a^3) + tan(c/ 
2 + (d*x)/2)^8*(84*A*a^3 + 28*B*a^3) + tan(c/2 + (d*x)/2)^14*((136*A*a^3)/ 
3 + 8*B*a^3) + tan(c/2 + (d*x)/2)^4*((136*A*a^3)/7 + (24*B*a^3)/7) + tan(c 
/2 + (d*x)/2)^10*(116*A*a^3 + 60*B*a^3) + tan(c/2 + (d*x)/2)^19*((73*A*a^3 
)/64 - (33*B*a^3)/128) + tan(c/2 + (d*x)/2)^2*((202*A*a^3)/63 + (58*B*a^3) 
/21) + tan(c/2 + (d*x)/2)^12*((328*A*a^3)/3 + 72*B*a^3) - tan(c/2 + (d*x)/ 
2)^7*((341*A*a^3)/16 - (333*B*a^3)/32) + tan(c/2 + (d*x)/2)^13*((341*A*a^3 
)/16 - (333*B*a^3)/32) + tan(c/2 + (d*x)/2)^6*((456*A*a^3)/7 + (344*B*a^3) 
/7) - tan(c/2 + (d*x)/2)^5*((449*A*a^3)/48 + (577*B*a^3)/160) + tan(c/2 + 
(d*x)/2)^15*((449*A*a^3)/48 + (577*B*a^3)/160) - tan(c/2 + (d*x)/2)^3*((21 
17*A*a^3)/192 + (705*B*a^3)/128) + tan(c/2 + (d*x)/2)^17*((2117*A*a^3)/192 
 + (705*B*a^3)/128) - tan(c/2 + (d*x)/2)^9*((699*A*a^3)/32 + (2749*B*a^3)/ 
64) + tan(c/2 + (d*x)/2)^11*((699*A*a^3)/32 + (2749*B*a^3)/64))/(d*(10*tan 
(c/2 + (d*x)/2)^2 + 45*tan(c/2 + (d*x)/2)^4 + 120*tan(c/2 + (d*x)/2)^6 + 2 
10*tan(c/2 + (d*x)/2)^8 + 252*tan(c/2 + (d*x)/2)^10 + 210*tan(c/2 + (d*x)/ 
2)^12 + 120*tan(c/2 + (d*x)/2)^14 + 45*tan(c/2 + (d*x)/2)^16 + 10*tan(c...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.42 \[ \int \cos ^6(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^{3} \left (8064 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{9} b +8960 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} a +26880 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} b +30240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} a +9072 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7} b +10240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} a -61440 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} b -72240 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a -70056 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} b -84480 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a +23040 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b +30660 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +73710 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} b +102400 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +30720 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +45990 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -10395 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -37120 \cos \left (d x +c \right ) a -19200 \cos \left (d x +c \right ) b +34650 a d x +37120 a +10395 b d x +19200 b \right )}{80640 d} \] Input:

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

Output:

(a**3*(8064*cos(c + d*x)*sin(c + d*x)**9*b + 8960*cos(c + d*x)*sin(c + d*x 
)**8*a + 26880*cos(c + d*x)*sin(c + d*x)**8*b + 30240*cos(c + d*x)*sin(c + 
 d*x)**7*a + 9072*cos(c + d*x)*sin(c + d*x)**7*b + 10240*cos(c + d*x)*sin( 
c + d*x)**6*a - 61440*cos(c + d*x)*sin(c + d*x)**6*b - 72240*cos(c + d*x)* 
sin(c + d*x)**5*a - 70056*cos(c + d*x)*sin(c + d*x)**5*b - 84480*cos(c + d 
*x)*sin(c + d*x)**4*a + 23040*cos(c + d*x)*sin(c + d*x)**4*b + 30660*cos(c 
 + d*x)*sin(c + d*x)**3*a + 73710*cos(c + d*x)*sin(c + d*x)**3*b + 102400* 
cos(c + d*x)*sin(c + d*x)**2*a + 30720*cos(c + d*x)*sin(c + d*x)**2*b + 45 
990*cos(c + d*x)*sin(c + d*x)*a - 10395*cos(c + d*x)*sin(c + d*x)*b - 3712 
0*cos(c + d*x)*a - 19200*cos(c + d*x)*b + 34650*a*d*x + 37120*a + 10395*b* 
d*x + 19200*b))/(80640*d)