\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [571]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^4 d (1+n)}-\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^4 d}+\frac {4 \sin ^{1+n}(c+d x)}{d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^{1+n}(c+d x) (5+4 n+\sin (c+d x)-4 (1+n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) (1+\sin (c+d x)))}{a^4 d (1+n) (1+\sin (c+d x))} \] Input:

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^4,x]
 

Output:

(Sin[c + d*x]^(1 + n)*(5 + 4*n + Sin[c + d*x] - 4*(1 + n)*Hypergeometric2F 
1[1, 1 + n, 2 + n, -Sin[c + d*x]]*(1 + Sin[c + d*x])))/(a^4*d*(1 + n)*(1 + 
 Sin[c + d*x]))
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3315, 100, 90, 74}

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.

\(\displaystyle \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^5 \sin (c+d x)^n}{(a \sin (c+d x)+a)^4}dx\)

\(\Big \downarrow \) 3315

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 74

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

Input:

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^4,x]
 

Output:

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

Defintions of rubi rules used

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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]

\[\int \frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{n}}{\left (a +a \sin \left (d x +c \right )\right )^{4}}d x\]

Input:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)
 

Output:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)
 

Fricas [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="frica 
s")
 

Output:

integral(sin(d*x + c)^n*cos(d*x + c)^5/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x 
 + c)^2 + 8*a^4 - 4*(a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*sin(d*x+c)**n/(a+a*sin(d*x+c))**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="maxim 
a")
 

Output:

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^4, x)
 

Giac [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \] Input:

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="giac" 
)
 

Output:

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \] Input:

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^4,x)
 

Output:

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^4, x)
 

Reduce [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int \frac {\cos \left (d x +c \right )^{5} \sin \left (d x +c \right )^{n}}{\left (\sin \left (d x +c \right ) a +a \right )^{4}}d x \] Input:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)
 

Output:

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)