\(\int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [983]

Optimal result

Integrand size = 31, antiderivative size = 134 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {8 (A-B) (a+a \sin (c+d x))^7}{7 a^4 d}-\frac {(3 A-5 B) (a+a \sin (c+d x))^8}{2 a^5 d}+\frac {2 (A-3 B) (a+a \sin (c+d x))^9}{3 a^6 d}-\frac {(A-7 B) (a+a \sin (c+d x))^{10}}{10 a^7 d}-\frac {B (a+a \sin (c+d x))^{11}}{11 a^8 d} \] Output:

8/7*(A-B)*(a+a*sin(d*x+c))^7/a^4/d-1/2*(3*A-5*B)*(a+a*sin(d*x+c))^8/a^5/d+ 
2/3*(A-3*B)*(a+a*sin(d*x+c))^9/a^6/d-1/10*(A-7*B)*(a+a*sin(d*x+c))^10/a^7/ 
d-1/11*B*(a+a*sin(d*x+c))^11/a^8/d
 

Mathematica [A] (verified)

Time = 1.74 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 (1+\sin (c+d x))^7 \left (-484 A+78 B+14 (77 A-39 B) \sin (c+d x)+(-847 A+1029 B) \sin ^2(c+d x)+21 (11 A-37 B) \sin ^3(c+d x)+210 B \sin ^4(c+d x)\right )}{2310 d} \] Input:

Integrate[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

-1/2310*(a^3*(1 + Sin[c + d*x])^7*(-484*A + 78*B + 14*(77*A - 39*B)*Sin[c 
+ d*x] + (-847*A + 1029*B)*Sin[c + d*x]^2 + 21*(11*A - 37*B)*Sin[c + d*x]^ 
3 + 210*B*Sin[c + d*x]^4))/d
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 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.

\(\displaystyle \int \cos ^7(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^7 (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\)

\(\Big \downarrow \) 3315

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^6 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^8 d}\)

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

Int[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

((8*a^4*(A - B)*(a + a*Sin[c + d*x])^7)/7 - (a^3*(3*A - 5*B)*(a + a*Sin[c 
+ d*x])^8)/2 + (2*a^2*(A - 3*B)*(a + a*Sin[c + d*x])^9)/3 - (a*(A - 7*B)*( 
a + a*Sin[c + d*x])^10)/10 - (B*(a + a*Sin[c + d*x])^11)/11)/(a^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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

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 3315

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.17

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

Input:

int(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
 

Output:

-a^3/d*(1/11*B*sin(d*x+c)^11+1/10*(A+3*B)*sin(d*x+c)^10+1/3*sin(d*x+c)^9*A 
-B*sin(d*x+c)^8+1/7*(-6*B-8*A)*sin(d*x+c)^7+1/6*(6*B-6*A)*sin(d*x+c)^6+1/5 
*(8*B+6*A)*sin(d*x+c)^5+2*sin(d*x+c)^4*A-B*sin(d*x+c)^3+1/2*(-B-3*A)*sin(d 
*x+c)^2-A*sin(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.16 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {231 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{10} - 1155 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{8} + 2 \, {\left (105 \, B a^{3} \cos \left (d x + c\right )^{10} - 35 \, {\left (11 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} + 20 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 24 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 32 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 64 \, {\left (11 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2310 \, d} \] Input:

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/2310*(231*(A + 3*B)*a^3*cos(d*x + c)^10 - 1155*(A + B)*a^3*cos(d*x + c)^ 
8 + 2*(105*B*a^3*cos(d*x + c)^10 - 35*(11*A + 15*B)*a^3*cos(d*x + c)^8 + 2 
0*(11*A + 3*B)*a^3*cos(d*x + c)^6 + 24*(11*A + 3*B)*a^3*cos(d*x + c)^4 + 3 
2*(11*A + 3*B)*a^3*cos(d*x + c)^2 + 64*(11*A + 3*B)*a^3)*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 636 vs. \(2 (124) = 248\).

Time = 1.87 (sec) , antiderivative size = 636, normalized size of antiderivative = 4.75 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)**7*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
 

Output:

Piecewise((A*a**3*sin(c + d*x)**10/(40*d) + 16*A*a**3*sin(c + d*x)**9/(105 
*d) + A*a**3*sin(c + d*x)**8*cos(c + d*x)**2/(8*d) + 24*A*a**3*sin(c + d*x 
)**7*cos(c + d*x)**2/(35*d) + 16*A*a**3*sin(c + d*x)**7/(35*d) + A*a**3*si 
n(c + d*x)**6*cos(c + d*x)**4/(4*d) + 6*A*a**3*sin(c + d*x)**5*cos(c + d*x 
)**4/(5*d) + 8*A*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + A*a**3*sin(c 
 + d*x)**4*cos(c + d*x)**6/(4*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**6/ 
d + 2*A*a**3*sin(c + d*x)**3*cos(c + d*x)**4/d + A*a**3*sin(c + d*x)*cos(c 
 + d*x)**6/d - 3*A*a**3*cos(c + d*x)**8/(8*d) + 16*B*a**3*sin(c + d*x)**11 
/(1155*d) + 3*B*a**3*sin(c + d*x)**10/(40*d) + 8*B*a**3*sin(c + d*x)**9*co 
s(c + d*x)**2/(105*d) + 16*B*a**3*sin(c + d*x)**9/(105*d) + 3*B*a**3*sin(c 
 + d*x)**8*cos(c + d*x)**2/(8*d) + 6*B*a**3*sin(c + d*x)**7*cos(c + d*x)** 
4/(35*d) + 24*B*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 3*B*a**3*sin 
(c + d*x)**6*cos(c + d*x)**4/(4*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)** 
6/(5*d) + 6*B*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 3*B*a**3*sin(c 
+ d*x)**4*cos(c + d*x)**6/(4*d) + B*a**3*sin(c + d*x)**3*cos(c + d*x)**6/d 
 - B*a**3*cos(c + d*x)**8/(8*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + 
a)**3*cos(c)**7, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.36 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {210 \, B a^{3} \sin \left (d x + c\right )^{11} + 231 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{10} + 770 \, A a^{3} \sin \left (d x + c\right )^{9} - 2310 \, B a^{3} \sin \left (d x + c\right )^{8} - 660 \, {\left (4 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{7} - 2310 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{6} + 924 \, {\left (3 \, A + 4 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 4620 \, A a^{3} \sin \left (d x + c\right )^{4} - 2310 \, B a^{3} \sin \left (d x + c\right )^{3} - 1155 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 2310 \, A a^{3} \sin \left (d x + c\right )}{2310 \, d} \] Input:

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/2310*(210*B*a^3*sin(d*x + c)^11 + 231*(A + 3*B)*a^3*sin(d*x + c)^10 + 7 
70*A*a^3*sin(d*x + c)^9 - 2310*B*a^3*sin(d*x + c)^8 - 660*(4*A + 3*B)*a^3* 
sin(d*x + c)^7 - 2310*(A - B)*a^3*sin(d*x + c)^6 + 924*(3*A + 4*B)*a^3*sin 
(d*x + c)^5 + 4620*A*a^3*sin(d*x + c)^4 - 2310*B*a^3*sin(d*x + c)^3 - 1155 
*(3*A + B)*a^3*sin(d*x + c)^2 - 2310*A*a^3*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.70 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {210 \, B a^{3} \sin \left (d x + c\right )^{11} + 231 \, A a^{3} \sin \left (d x + c\right )^{10} + 693 \, B a^{3} \sin \left (d x + c\right )^{10} + 770 \, A a^{3} \sin \left (d x + c\right )^{9} - 2310 \, B a^{3} \sin \left (d x + c\right )^{8} - 2640 \, A a^{3} \sin \left (d x + c\right )^{7} - 1980 \, B a^{3} \sin \left (d x + c\right )^{7} - 2310 \, A a^{3} \sin \left (d x + c\right )^{6} + 2310 \, B a^{3} \sin \left (d x + c\right )^{6} + 2772 \, A a^{3} \sin \left (d x + c\right )^{5} + 3696 \, B a^{3} \sin \left (d x + c\right )^{5} + 4620 \, A a^{3} \sin \left (d x + c\right )^{4} - 2310 \, B a^{3} \sin \left (d x + c\right )^{3} - 3465 \, A a^{3} \sin \left (d x + c\right )^{2} - 1155 \, B a^{3} \sin \left (d x + c\right )^{2} - 2310 \, A a^{3} \sin \left (d x + c\right )}{2310 \, d} \] Input:

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

-1/2310*(210*B*a^3*sin(d*x + c)^11 + 231*A*a^3*sin(d*x + c)^10 + 693*B*a^3 
*sin(d*x + c)^10 + 770*A*a^3*sin(d*x + c)^9 - 2310*B*a^3*sin(d*x + c)^8 - 
2640*A*a^3*sin(d*x + c)^7 - 1980*B*a^3*sin(d*x + c)^7 - 2310*A*a^3*sin(d*x 
 + c)^6 + 2310*B*a^3*sin(d*x + c)^6 + 2772*A*a^3*sin(d*x + c)^5 + 3696*B*a 
^3*sin(d*x + c)^5 + 4620*A*a^3*sin(d*x + c)^4 - 2310*B*a^3*sin(d*x + c)^3 
- 3465*A*a^3*sin(d*x + c)^2 - 1155*B*a^3*sin(d*x + c)^2 - 2310*A*a^3*sin(d 
*x + c))/d
 

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.32 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}-\frac {A\,a^3\,{\sin \left (c+d\,x\right )}^9}{3}-2\,A\,a^3\,{\sin \left (c+d\,x\right )}^4+a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A-B\right )-\frac {a^3\,{\sin \left (c+d\,x\right )}^{10}\,\left (A+3\,B\right )}{10}+B\,a^3\,{\sin \left (c+d\,x\right )}^3+B\,a^3\,{\sin \left (c+d\,x\right )}^8-\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^{11}}{11}-\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^5\,\left (3\,A+4\,B\right )}{5}+\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^7\,\left (4\,A+3\,B\right )}{7}+A\,a^3\,\sin \left (c+d\,x\right )}{d} \] Input:

int(cos(c + d*x)^7*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)
 

Output:

((a^3*sin(c + d*x)^2*(3*A + B))/2 - (A*a^3*sin(c + d*x)^9)/3 - 2*A*a^3*sin 
(c + d*x)^4 + a^3*sin(c + d*x)^6*(A - B) - (a^3*sin(c + d*x)^10*(A + 3*B)) 
/10 + B*a^3*sin(c + d*x)^3 + B*a^3*sin(c + d*x)^8 - (B*a^3*sin(c + d*x)^11 
)/11 - (2*a^3*sin(c + d*x)^5*(3*A + 4*B))/5 + (2*a^3*sin(c + d*x)^7*(4*A + 
 3*B))/7 + A*a^3*sin(c + d*x))/d
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.34 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a^{3} \left (-210 \sin \left (d x +c \right )^{10} b -231 \sin \left (d x +c \right )^{9} a -693 \sin \left (d x +c \right )^{9} b -770 \sin \left (d x +c \right )^{8} a +2310 \sin \left (d x +c \right )^{7} b +2640 \sin \left (d x +c \right )^{6} a +1980 \sin \left (d x +c \right )^{6} b +2310 \sin \left (d x +c \right )^{5} a -2310 \sin \left (d x +c \right )^{5} b -2772 \sin \left (d x +c \right )^{4} a -3696 \sin \left (d x +c \right )^{4} b -4620 \sin \left (d x +c \right )^{3} a +2310 \sin \left (d x +c \right )^{2} b +3465 \sin \left (d x +c \right ) a +1155 \sin \left (d x +c \right ) b +2310 a \right )}{2310 d} \] Input:

int(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
 

Output:

(sin(c + d*x)*a**3*( - 210*sin(c + d*x)**10*b - 231*sin(c + d*x)**9*a - 69 
3*sin(c + d*x)**9*b - 770*sin(c + d*x)**8*a + 2310*sin(c + d*x)**7*b + 264 
0*sin(c + d*x)**6*a + 1980*sin(c + d*x)**6*b + 2310*sin(c + d*x)**5*a - 23 
10*sin(c + d*x)**5*b - 2772*sin(c + d*x)**4*a - 3696*sin(c + d*x)**4*b - 4 
620*sin(c + d*x)**3*a + 2310*sin(c + d*x)**2*b + 3465*sin(c + d*x)*a + 115 
5*sin(c + d*x)*b + 2310*a))/(2310*d)