\(\int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [966]

Optimal result

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

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

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {a^2 (1+\sin (c+d x))^6 \left (-325 A+61 B+6 (115 A-61 B) \sin (c+d x)+(-525 A+651 B) \sin ^2(c+d x)+28 (5 A-17 B) \sin ^3(c+d x)+126 B \sin ^4(c+d x)\right )}{1260 d} \] Input:

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

Output:

-1/1260*(a^2*(1 + Sin[c + d*x])^6*(-325*A + 61*B + 6*(115*A - 61*B)*Sin[c 
+ d*x] + (-525*A + 651*B)*Sin[c + d*x]^2 + 28*(5*A - 17*B)*Sin[c + d*x]^3 
+ 126*B*Sin[c + d*x]^4))/d
 

Rubi [A] (verified)

Time = 0.40 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3315

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^5 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^5 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^8 d}\)

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((4*a^4*(A - B)*(a + a*Sin[c + d*x])^6)/3 - (4*a^3*(3*A - 5*B)*(a + a*Sin[ 
c + d*x])^7)/7 + (3*a^2*(A - 3*B)*(a + a*Sin[c + d*x])^8)/4 - (a*(A - 7*B) 
*(a + a*Sin[c + d*x])^9)/9 - (B*(a + a*Sin[c + d*x])^10)/10)/(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 = 152, normalized size of antiderivative = 1.13

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {126 \, B a^{2} \cos \left (d x + c\right )^{10} - 315 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{8} - 4 \, {\left (35 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{8} - 10 \, {\left (5 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{6} - 12 \, {\left (5 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 16 \, {\left (5 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 32 \, {\left (5 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \] Input:

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

Output:

1/1260*(126*B*a^2*cos(d*x + c)^10 - 315*(A + B)*a^2*cos(d*x + c)^8 - 4*(35 
*(A + 2*B)*a^2*cos(d*x + c)^8 - 10*(5*A + B)*a^2*cos(d*x + c)^6 - 12*(5*A 
+ B)*a^2*cos(d*x + c)^4 - 16*(5*A + B)*a^2*cos(d*x + c)^2 - 32*(5*A + B)*a 
^2)*sin(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 440 vs. \(2 (126) = 252\).

Time = 1.36 (sec) , antiderivative size = 440, normalized size of antiderivative = 3.28 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {16 A a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 A a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {16 A a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {2 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{8}{\left (c + d x \right )}}{4 d} + \frac {B a^{2} \sin ^{10}{\left (c + d x \right )}}{40 d} + \frac {32 B a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {B a^{2} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} + \frac {16 B a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {B a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {4 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{8}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{2} \cos ^{7}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {126 \, B a^{2} \sin \left (d x + c\right )^{10} + 140 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{9} + 315 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right )^{8} - 360 \, {\left (A + 3 \, B\right )} a^{2} \sin \left (d x + c\right )^{7} - 1260 \, A a^{2} \sin \left (d x + c\right )^{6} + 1512 \, B a^{2} \sin \left (d x + c\right )^{5} + 630 \, {\left (3 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{4} + 840 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right )^{3} - 630 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} - 1260 \, A a^{2} \sin \left (d x + c\right )}{1260 \, d} \] Input:

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

Output:

-1/1260*(126*B*a^2*sin(d*x + c)^10 + 140*(A + 2*B)*a^2*sin(d*x + c)^9 + 31 
5*(A - B)*a^2*sin(d*x + c)^8 - 360*(A + 3*B)*a^2*sin(d*x + c)^7 - 1260*A*a 
^2*sin(d*x + c)^6 + 1512*B*a^2*sin(d*x + c)^5 + 630*(3*A + B)*a^2*sin(d*x 
+ c)^4 + 840*(A - B)*a^2*sin(d*x + c)^3 - 630*(2*A + B)*a^2*sin(d*x + c)^2 
 - 1260*A*a^2*sin(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.70 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {126 \, B a^{2} \sin \left (d x + c\right )^{10} + 140 \, A a^{2} \sin \left (d x + c\right )^{9} + 280 \, B a^{2} \sin \left (d x + c\right )^{9} + 315 \, A a^{2} \sin \left (d x + c\right )^{8} - 315 \, B a^{2} \sin \left (d x + c\right )^{8} - 360 \, A a^{2} \sin \left (d x + c\right )^{7} - 1080 \, B a^{2} \sin \left (d x + c\right )^{7} - 1260 \, A a^{2} \sin \left (d x + c\right )^{6} + 1512 \, B a^{2} \sin \left (d x + c\right )^{5} + 1890 \, A a^{2} \sin \left (d x + c\right )^{4} + 630 \, B a^{2} \sin \left (d x + c\right )^{4} + 840 \, A a^{2} \sin \left (d x + c\right )^{3} - 840 \, B a^{2} \sin \left (d x + c\right )^{3} - 1260 \, A a^{2} \sin \left (d x + c\right )^{2} - 630 \, B a^{2} \sin \left (d x + c\right )^{2} - 1260 \, A a^{2} \sin \left (d x + c\right )}{1260 \, d} \] Input:

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

Output:

-1/1260*(126*B*a^2*sin(d*x + c)^10 + 140*A*a^2*sin(d*x + c)^9 + 280*B*a^2* 
sin(d*x + c)^9 + 315*A*a^2*sin(d*x + c)^8 - 315*B*a^2*sin(d*x + c)^8 - 360 
*A*a^2*sin(d*x + c)^7 - 1080*B*a^2*sin(d*x + c)^7 - 1260*A*a^2*sin(d*x + c 
)^6 + 1512*B*a^2*sin(d*x + c)^5 + 1890*A*a^2*sin(d*x + c)^4 + 630*B*a^2*si 
n(d*x + c)^4 + 840*A*a^2*sin(d*x + c)^3 - 840*B*a^2*sin(d*x + c)^3 - 1260* 
A*a^2*sin(d*x + c)^2 - 630*B*a^2*sin(d*x + c)^2 - 1260*A*a^2*sin(d*x + c)) 
/d
 

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.25 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A-B\right )}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}-A\,a^2\,{\sin \left (c+d\,x\right )}^6+\frac {a^2\,{\sin \left (c+d\,x\right )}^4\,\left (3\,A+B\right )}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^8\,\left (A-B\right )}{4}-\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^7\,\left (A+3\,B\right )}{7}+\frac {a^2\,{\sin \left (c+d\,x\right )}^9\,\left (A+2\,B\right )}{9}+\frac {6\,B\,a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^{10}}{10}-A\,a^2\,\sin \left (c+d\,x\right )}{d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.34 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a^{2} \left (-126 \sin \left (d x +c \right )^{9} b -140 \sin \left (d x +c \right )^{8} a -280 \sin \left (d x +c \right )^{8} b -315 \sin \left (d x +c \right )^{7} a +315 \sin \left (d x +c \right )^{7} b +360 \sin \left (d x +c \right )^{6} a +1080 \sin \left (d x +c \right )^{6} b +1260 \sin \left (d x +c \right )^{5} a -1512 \sin \left (d x +c \right )^{4} b -1890 \sin \left (d x +c \right )^{3} a -630 \sin \left (d x +c \right )^{3} b -840 \sin \left (d x +c \right )^{2} a +840 \sin \left (d x +c \right )^{2} b +1260 \sin \left (d x +c \right ) a +630 \sin \left (d x +c \right ) b +1260 a \right )}{1260 d} \] Input:

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

Output:

(sin(c + d*x)*a**2*( - 126*sin(c + d*x)**9*b - 140*sin(c + d*x)**8*a - 280 
*sin(c + d*x)**8*b - 315*sin(c + d*x)**7*a + 315*sin(c + d*x)**7*b + 360*s 
in(c + d*x)**6*a + 1080*sin(c + d*x)**6*b + 1260*sin(c + d*x)**5*a - 1512* 
sin(c + d*x)**4*b - 1890*sin(c + d*x)**3*a - 630*sin(c + d*x)**3*b - 840*s 
in(c + d*x)**2*a + 840*sin(c + d*x)**2*b + 1260*sin(c + d*x)*a + 630*sin(c 
 + d*x)*b + 1260*a))/(1260*d)