\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx\) [1233]

Optimal result

Integrand size = 29, antiderivative size = 167 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a \left (a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a b^4 d (1+n)}+\frac {\left (a^2-2 b^2\right ) \sin ^{2+n}(c+d x)}{b^3 d (2+n)}-\frac {a \sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\sin ^{4+n}(c+d x)}{b d (4+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sin ^{1+n}(c+d x) \left (-\frac {a^3-2 a b^2}{1+n}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{a (1+n)}+\frac {b \left (a^2-2 b^2\right ) \sin (c+d x)}{2+n}-\frac {a b^2 \sin ^2(c+d x)}{3+n}+\frac {b^3 \sin ^3(c+d x)}{4+n}\right )}{b^4 d} \] Input:

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

Output:

(Sin[c + d*x]^(1 + n)*(-((a^3 - 2*a*b^2)/(1 + n)) + ((a^2 - b^2)^2*Hyperge 
ometric2F1[1, 1 + n, 2 + n, -((b*Sin[c + d*x])/a)])/(a*(1 + n)) + (b*(a^2 
- 2*b^2)*Sin[c + d*x])/(2 + n) - (a*b^2*Sin[c + d*x]^2)/(3 + n) + (b^3*Sin 
[c + d*x]^3)/(4 + n)))/(b^4*d)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3042, 3316, 522, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 522

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-((a*b*(a^2 - 2*b^2)*Sin[c + d*x]^(1 + n))/(1 + n)) + (b*(a^2 - b^2)^2*Hy 
pergeometric2F1[1, 1 + n, 2 + n, -((b*Sin[c + d*x])/a)]*Sin[c + d*x]^(1 + 
n))/(a*(1 + n)) + (b^2*(a^2 - 2*b^2)*Sin[c + d*x]^(2 + n))/(2 + n) - (a*b^ 
3*Sin[c + d*x]^(3 + n))/(3 + n) + (b^4*Sin[c + d*x]^(4 + n))/(4 + n))/(b^5 
*d)
 

Defintions of rubi rules used

rule 522

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
 

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 3316

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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sin(d*x + c)^n*cos(d*x + c)^5/(b*sin(d*x + c) + a), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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