\(\int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx\) [653]

Optimal result

Integrand size = 29, antiderivative size = 267 \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{d (1+n) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{d (2+n) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right ) \sin ^{3+n}(c+d x)}{d (3+n) \sqrt {\cos ^2(c+d x)}}+\frac {a^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(c+d x)\right ) \sin ^{4+n}(c+d x)}{d (4+n) \sqrt {\cos ^2(c+d x)}} \] Output:

a^3*cos(d*x+c)*hypergeom([-5/2, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*sin(d 
*x+c)^(1+n)/d/(1+n)/(cos(d*x+c)^2)^(1/2)+3*a^3*cos(d*x+c)*hypergeom([-5/2, 
 1+1/2*n],[2+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(2+n)/d/(2+n)/(cos(d*x+c)^2)^ 
(1/2)+3*a^3*cos(d*x+c)*hypergeom([-5/2, 3/2+1/2*n],[5/2+1/2*n],sin(d*x+c)^ 
2)*sin(d*x+c)^(3+n)/d/(3+n)/(cos(d*x+c)^2)^(1/2)+a^3*cos(d*x+c)*hypergeom( 
[-5/2, 2+1/2*n],[3+1/2*n],sin(d*x+c)^2)*sin(d*x+c)^(4+n)/d/(4+n)/(cos(d*x+ 
c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.70 \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sqrt {\cos ^2(c+d x)} \sec (c+d x) \sin ^{1+n}(c+d x) \left (\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )}{1+n}+\sin (c+d x) \left (\frac {3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right )}{2+n}+\sin (c+d x) \left (\frac {3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(c+d x)\right )}{3+n}+\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(c+d x)\right ) \sin (c+d x)}{4+n}\right )\right )\right )}{d} \] Input:

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

Output:

(a^3*Sqrt[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c + d*x]^(1 + n)*(Hypergeometri 
c2F1[-5/2, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2]/(1 + n) + Sin[c + d*x]*(( 
3*Hypergeometric2F1[-5/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2])/(2 + n) + 
 Sin[c + d*x]*((3*Hypergeometric2F1[-5/2, (3 + n)/2, (5 + n)/2, Sin[c + d* 
x]^2])/(3 + n) + (Hypergeometric2F1[-5/2, (4 + n)/2, (6 + n)/2, Sin[c + d* 
x]^2]*Sin[c + d*x])/(4 + n)))))/d
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3352, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^6 (a \sin (c+d x)+a)^3 \sin (c+d x)^ndx\)

\(\Big \downarrow \) 3352

\(\displaystyle \int \left (3 a^3 \cos ^6(c+d x) \sin ^{n+1}(c+d x)+3 a^3 \cos ^6(c+d x) \sin ^{n+2}(c+d x)+a^3 \cos ^6(c+d x) \sin ^{n+3}(c+d x)+a^3 \cos ^6(c+d x) \sin ^n(c+d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 \cos (c+d x) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{d (n+1) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^3 \cos (c+d x) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{d (n+2) \sqrt {\cos ^2(c+d x)}}+\frac {3 a^3 \cos (c+d x) \sin ^{n+3}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {n+3}{2},\frac {n+5}{2},\sin ^2(c+d x)\right )}{d (n+3) \sqrt {\cos ^2(c+d x)}}+\frac {a^3 \cos (c+d x) \sin ^{n+4}(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {n+4}{2},\frac {n+6}{2},\sin ^2(c+d x)\right )}{d (n+4) \sqrt {\cos ^2(c+d x)}}\)

Input:

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

Output:

(a^3*Cos[c + d*x]*Hypergeometric2F1[-5/2, (1 + n)/2, (3 + n)/2, Sin[c + d* 
x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)*Sqrt[Cos[c + d*x]^2]) + (3*a^3*Cos[ 
c + d*x]*Hypergeometric2F1[-5/2, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin 
[c + d*x]^(2 + n))/(d*(2 + n)*Sqrt[Cos[c + d*x]^2]) + (3*a^3*Cos[c + d*x]* 
Hypergeometric2F1[-5/2, (3 + n)/2, (5 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x] 
^(3 + n))/(d*(3 + n)*Sqrt[Cos[c + d*x]^2]) + (a^3*Cos[c + d*x]*Hypergeomet 
ric2F1[-5/2, (4 + n)/2, (6 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(4 + n))/( 
d*(4 + n)*Sqrt[Cos[c + d*x]^2])
 

Defintions of rubi rules used

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 3352

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig 
[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F 
reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
 

Maple [F]

\[\int \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{n} \left (a +a \sin \left (d x +c \right )\right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{6} \,d x } \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

integral(-(3*a^3*cos(d*x + c)^8 - 4*a^3*cos(d*x + c)^6 + (a^3*cos(d*x + c) 
^8 - 4*a^3*cos(d*x + c)^6)*sin(d*x + c))*sin(d*x + c)^n, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{6} \,d x } \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

integrate((a*sin(d*x + c) + a)^3*sin(d*x + c)^n*cos(d*x + c)^6, x)
 

Giac [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{3} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{6} \,d x } \] Input:

integrate(cos(d*x+c)^6*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

integrate((a*sin(d*x + c) + a)^3*sin(d*x + c)^n*cos(d*x + c)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\int {\cos \left (c+d\,x\right )}^6\,{\sin \left (c+d\,x\right )}^n\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cos ^6(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{3}d x +3 \left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )^{2}d x \right )+3 \left (\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{6} \sin \left (d x +c \right )d x \right )+\int \sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{6}d x \right ) \] Input:

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

Output:

a**3*(int(sin(c + d*x)**n*cos(c + d*x)**6*sin(c + d*x)**3,x) + 3*int(sin(c 
 + d*x)**n*cos(c + d*x)**6*sin(c + d*x)**2,x) + 3*int(sin(c + d*x)**n*cos( 
c + d*x)**6*sin(c + d*x),x) + int(sin(c + d*x)**n*cos(c + d*x)**6,x))