\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [1234]

Optimal result

Integrand size = 29, antiderivative size = 191 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (3 a^2-2 b^2\right ) \sin ^{1+n}(c+d x)}{b^4 d (1+n)}+\frac {\left (a^2-b^2\right ) \left (b^2 n-a^2 (4+n)\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a^2 b^4 d (1+n)}-\frac {2 a \sin ^{2+n}(c+d x)}{b^3 d (2+n)}+\frac {\sin ^{3+n}(c+d x)}{b^2 d (3+n)}+\frac {\left (a^2-b^2\right )^2 \sin ^{1+n}(c+d x)}{a b^4 d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\sin ^{1+n}(c+d x) \left (\frac {3 a^2-2 b^2}{1+n}-\frac {4 \left (a^2-b^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{1+n}+\frac {\left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,-\frac {b \sin (c+d x)}{a}\right )}{a^2 (1+n)}-\frac {2 a b \sin (c+d x)}{2+n}+\frac {b^2 \sin ^2(c+d x)}{3+n}\right )}{b^4 d} \] Input:

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + b*Sin[c + d*x])^2,x]
 

Output:

(Sin[c + d*x]^(1 + n)*((3*a^2 - 2*b^2)/(1 + n) - (4*(a^2 - b^2)*Hypergeome 
tric2F1[1, 1 + n, 2 + n, -((b*Sin[c + d*x])/a)])/(1 + n) + ((a^2 - b^2)^2* 
Hypergeometric2F1[2, 1 + n, 2 + n, -((b*Sin[c + d*x])/a)])/(a^2*(1 + n)) - 
 (2*a*b*Sin[c + d*x])/(2 + n) + (b^2*Sin[c + d*x]^2)/(3 + n)))/(b^4*d)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3316, 519, 2123, 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.

\(\displaystyle \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 2123

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + b*Sin[c + d*x])^2,x]
 

Output:

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

Defintions of rubi rules used

rule 519

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, c + d*x, x], R = 
 PolynomialRemainder[(a + b*x^2)^p, c + d*x, x]}, Simp[(-R)*(e*x)^(m + 1)*( 
(c + d*x)^(n + 1)/(c*e*(n + 1))), x] + Simp[1/(c*(n + 1))   Int[(e*x)^m*(c 
+ d*x)^(n + 1)*ExpandToSum[c*(n + 1)*Qx + R*(m + n + 2), x], x], x]] /; Fre 
eQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0] && LtQ[n, -1] &&  !IntegerQ[m]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2123

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] 
:> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c 
, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x, algorithm="frica 
s")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*sin(d*x+c)**n/(a+b*sin(d*x+c))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x, algorithm="maxim 
a")
 

Output:

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(b*sin(d*x + c) + a)^2, x)
 

Giac [F]

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

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+b*sin(d*x+c))^2,x, algorithm="giac" 
)
 

Output:

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(b*sin(d*x + c) + a)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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