\(\int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\) [984]

Optimal result

Integrand size = 31, antiderivative size = 105 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {2 (A-B) (a+a \sin (c+d x))^6}{3 a^3 d}-\frac {4 (A-2 B) (a+a \sin (c+d x))^7}{7 a^4 d}+\frac {(A-5 B) (a+a \sin (c+d x))^8}{8 a^5 d}+\frac {B (a+a \sin (c+d x))^9}{9 a^6 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {a^3 (1+\sin (c+d x))^6 \left (111 A-19 B-6 (27 A-19 B) \sin (c+d x)+21 (3 A-7 B) \sin ^2(c+d x)+56 B \sin ^3(c+d x)\right )}{504 d} \] Input:

Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

(a^3*(1 + Sin[c + d*x])^6*(111*A - 19*B - 6*(27*A - 19*B)*Sin[c + d*x] + 2 
1*(3*A - 7*B)*Sin[c + d*x]^2 + 56*B*Sin[c + d*x]^3))/(504*d)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 95, 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.

\(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\)

\(\Big \downarrow \) 3315

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^6 d}\)

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

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

Input:

Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]
 

Output:

((2*a^3*(A - B)*(a + a*Sin[c + d*x])^6)/3 - (4*a^2*(A - 2*B)*(a + a*Sin[c 
+ d*x])^7)/7 + (a*(A - 5*B)*(a + a*Sin[c + d*x])^8)/8 + (B*(a + a*Sin[c + 
d*x])^9)/9)/(a^6*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 = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.29

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

Input:

int(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.23 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {63 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} - 336 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{6} + 8 \, {\left (7 \, B a^{3} \cos \left (d x + c\right )^{8} - {\left (27 \, A + 37 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 6 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \, {\left (3 \, A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{504 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/504*(63*(A + 3*B)*a^3*cos(d*x + c)^8 - 336*(A + B)*a^3*cos(d*x + c)^6 + 
8*(7*B*a^3*cos(d*x + c)^8 - (27*A + 37*B)*a^3*cos(d*x + c)^6 + 6*(3*A + B) 
*a^3*cos(d*x + c)^4 + 8*(3*A + B)*a^3*cos(d*x + c)^2 + 16*(3*A + B)*a^3)*s 
in(d*x + c))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (99) = 198\).

Time = 0.95 (sec) , antiderivative size = 471, normalized size of antiderivative = 4.49 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {A a^{3} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac {8 A a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {A a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac {4 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {4 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{3} \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {8 B a^{3} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {B a^{3} \sin ^{8}{\left (c + d x \right )}}{8 d} + \frac {4 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {B a^{3} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)
 

Output:

Piecewise((A*a**3*sin(c + d*x)**8/(24*d) + 8*A*a**3*sin(c + d*x)**7/(35*d) 
 + A*a**3*sin(c + d*x)**6*cos(c + d*x)**2/(6*d) + 4*A*a**3*sin(c + d*x)**5 
*cos(c + d*x)**2/(5*d) + 8*A*a**3*sin(c + d*x)**5/(15*d) + A*a**3*sin(c + 
d*x)**4*cos(c + d*x)**4/(4*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**4/d + 
 4*A*a**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*a**3*sin(c + d*x)*cos( 
c + d*x)**4/d - A*a**3*cos(c + d*x)**6/(2*d) + 8*B*a**3*sin(c + d*x)**9/(3 
15*d) + B*a**3*sin(c + d*x)**8/(8*d) + 4*B*a**3*sin(c + d*x)**7*cos(c + d* 
x)**2/(35*d) + 8*B*a**3*sin(c + d*x)**7/(35*d) + B*a**3*sin(c + d*x)**6*co 
s(c + d*x)**2/(2*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 4*B*a 
**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 3*B*a**3*sin(c + d*x)**4*cos(c 
 + d*x)**4/(4*d) + B*a**3*sin(c + d*x)**3*cos(c + d*x)**4/d - B*a**3*cos(c 
 + d*x)**6/(6*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c)**5 
, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {56 \, B a^{3} \sin \left (d x + c\right )^{9} + 63 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{8} + 72 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{7} + 84 \, {\left (A - 5 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} - 504 \, {\left (A + B\right )} a^{3} \sin \left (d x + c\right )^{5} - 126 \, {\left (5 \, A - B\right )} a^{3} \sin \left (d x + c\right )^{4} + 168 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} + 252 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} + 504 \, A a^{3} \sin \left (d x + c\right )}{504 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

1/504*(56*B*a^3*sin(d*x + c)^9 + 63*(A + 3*B)*a^3*sin(d*x + c)^8 + 72*(3*A 
 + B)*a^3*sin(d*x + c)^7 + 84*(A - 5*B)*a^3*sin(d*x + c)^6 - 504*(A + B)*a 
^3*sin(d*x + c)^5 - 126*(5*A - B)*a^3*sin(d*x + c)^4 + 168*(A + 3*B)*a^3*s 
in(d*x + c)^3 + 252*(3*A + B)*a^3*sin(d*x + c)^2 + 504*A*a^3*sin(d*x + c)) 
/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (97) = 194\).

Time = 0.22 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.17 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {56 \, B a^{3} \sin \left (d x + c\right )^{9} + 63 \, A a^{3} \sin \left (d x + c\right )^{8} + 189 \, B a^{3} \sin \left (d x + c\right )^{8} + 216 \, A a^{3} \sin \left (d x + c\right )^{7} + 72 \, B a^{3} \sin \left (d x + c\right )^{7} + 84 \, A a^{3} \sin \left (d x + c\right )^{6} - 420 \, B a^{3} \sin \left (d x + c\right )^{6} - 504 \, A a^{3} \sin \left (d x + c\right )^{5} - 504 \, B a^{3} \sin \left (d x + c\right )^{5} - 630 \, A a^{3} \sin \left (d x + c\right )^{4} + 126 \, B a^{3} \sin \left (d x + c\right )^{4} + 168 \, A a^{3} \sin \left (d x + c\right )^{3} + 504 \, B a^{3} \sin \left (d x + c\right )^{3} + 756 \, A a^{3} \sin \left (d x + c\right )^{2} + 252 \, B a^{3} \sin \left (d x + c\right )^{2} + 504 \, A a^{3} \sin \left (d x + c\right )}{504 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

1/504*(56*B*a^3*sin(d*x + c)^9 + 63*A*a^3*sin(d*x + c)^8 + 189*B*a^3*sin(d 
*x + c)^8 + 216*A*a^3*sin(d*x + c)^7 + 72*B*a^3*sin(d*x + c)^7 + 84*A*a^3* 
sin(d*x + c)^6 - 420*B*a^3*sin(d*x + c)^6 - 504*A*a^3*sin(d*x + c)^5 - 504 
*B*a^3*sin(d*x + c)^5 - 630*A*a^3*sin(d*x + c)^4 + 126*B*a^3*sin(d*x + c)^ 
4 + 168*A*a^3*sin(d*x + c)^3 + 504*B*a^3*sin(d*x + c)^3 + 756*A*a^3*sin(d* 
x + c)^2 + 252*B*a^3*sin(d*x + c)^2 + 504*A*a^3*sin(d*x + c))/d
 

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.49 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3\,\left (A+3\,B\right )}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^7\,\left (3\,A+B\right )}{7}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A-5\,B\right )}{6}+\frac {a^3\,{\sin \left (c+d\,x\right )}^8\,\left (A+3\,B\right )}{8}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (5\,A-B\right )}{4}+A\,a^3\,\sin \left (c+d\,x\right )-a^3\,{\sin \left (c+d\,x\right )}^5\,\left (A+B\right )}{d} \] Input:

int(cos(c + d*x)^5*(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^3*sin(c + d*x)^3*(A + 3*B))/3 + (a^ 
3*sin(c + d*x)^7*(3*A + B))/7 + (a^3*sin(c + d*x)^6*(A - 5*B))/6 + (a^3*si 
n(c + d*x)^8*(A + 3*B))/8 + (B*a^3*sin(c + d*x)^9)/9 - (a^3*sin(c + d*x)^4 
*(5*A - B))/4 + A*a^3*sin(c + d*x) - a^3*sin(c + d*x)^5*(A + B))/d
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.70 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a^{3} \left (56 \sin \left (d x +c \right )^{8} b +63 \sin \left (d x +c \right )^{7} a +189 \sin \left (d x +c \right )^{7} b +216 \sin \left (d x +c \right )^{6} a +72 \sin \left (d x +c \right )^{6} b +84 \sin \left (d x +c \right )^{5} a -420 \sin \left (d x +c \right )^{5} b -504 \sin \left (d x +c \right )^{4} a -504 \sin \left (d x +c \right )^{4} b -630 \sin \left (d x +c \right )^{3} a +126 \sin \left (d x +c \right )^{3} b +168 \sin \left (d x +c \right )^{2} a +504 \sin \left (d x +c \right )^{2} b +756 \sin \left (d x +c \right ) a +252 \sin \left (d x +c \right ) b +504 a \right )}{504 d} \] Input:

int(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)
 

Output:

(sin(c + d*x)*a**3*(56*sin(c + d*x)**8*b + 63*sin(c + d*x)**7*a + 189*sin( 
c + d*x)**7*b + 216*sin(c + d*x)**6*a + 72*sin(c + d*x)**6*b + 84*sin(c + 
d*x)**5*a - 420*sin(c + d*x)**5*b - 504*sin(c + d*x)**4*a - 504*sin(c + d* 
x)**4*b - 630*sin(c + d*x)**3*a + 126*sin(c + d*x)**3*b + 168*sin(c + d*x) 
**2*a + 504*sin(c + d*x)**2*b + 756*sin(c + d*x)*a + 252*sin(c + d*x)*b + 
504*a))/(504*d)