\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [570]

Optimal result

Integrand size = 29, antiderivative size = 85 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {\sin ^{2+n}(c+d x)}{a^3 d (2+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^{1+n}(c+d x) (-3 (2+n)+4 (2+n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x))+(1+n) \sin (c+d x))}{a^3 d (1+n) (2+n)} \] Input:

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

Output:

(Sin[c + d*x]^(1 + n)*(-3*(2 + n) + 4*(2 + n)*Hypergeometric2F1[1, 1 + n, 
2 + n, -Sin[c + d*x]] + (1 + n)*Sin[c + d*x]))/(a^3*d*(1 + n)*(2 + n))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3315, 99, 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.

\(\displaystyle \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^5 \sin (c+d x)^n}{(a \sin (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 3315

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

\(\Big \downarrow \) 99

\(\displaystyle \frac {\int \left (-3 a \sin ^n(c+d x)+\frac {4 a^2 \sin ^n(c+d x)}{\sin (c+d x) a+a}+a \sin ^{n+1}(c+d x)\right )d(a \sin (c+d x))}{a^5 d}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

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 [F]

\[\int \frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{n}}{\left (a +a \sin \left (d x +c \right )\right )^{3}}d x\]

Input:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)
 

Output:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)
 

Fricas [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 

Output:

integral(-sin(d*x + c)^n*cos(d*x + c)^5/(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a 
^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*sin(d*x+c)**n/(a+a*sin(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 

Output:

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^3, x)
 

Giac [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 

Output:

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \] Input:

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3,x)
 

Output:

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3, x)
 

Reduce [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin \left (d x +c \right )^{n} \cos \left (d x +c \right )^{5}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x}{a^{3}} \] Input:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)
 

Output:

int((sin(c + d*x)**n*cos(c + d*x)**5)/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 
 + 3*sin(c + d*x) + 1),x)/a**3