\(\int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx\) [915]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 31, antiderivative size = 60 \[ \int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {d \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {c+d \sin (e+f x)}{c-d}\right ) (c+d \sin (e+f x))^{1+n}}{a^2 (c-d)^2 f (1+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx=\frac {d \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,-\frac {c+d \sin (e+f x)}{-c+d}\right ) (c+d \sin (e+f x))^{1+n}}{a^2 (-c+d)^2 f (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3042, 3312, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3312

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {d (c+d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {a c+a d \sin (e+f x)}{a (c-d)}\right )}{a^2 f (n+1) (c-d)^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 78
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3312
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( 
c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f)   Su 
bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(cos(f*x+e)*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="f 
ricas")
 

Output:

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

Sympy [F(-1)]

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

integrate(cos(f*x+e)*(c+d*sin(f*x+e))**n/(a+a*sin(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(f*x+e)*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="m 
axima")
 

Output:

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

Giac [F]

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

integrate(cos(f*x+e)*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos (e+f x) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - (sin(e + f*x)*d + c)**n*c + int(((sin(e + f*x)*d + c)**n*cos(e + f*x)* 
sin(e + f*x))/(sin(e + f*x)**3*c*d - sin(e + f*x)**3*d**2*n + sin(e + f*x) 
**2*c**2 - sin(e + f*x)**2*c*d*n + 2*sin(e + f*x)**2*c*d - 2*sin(e + f*x)* 
*2*d**2*n + 2*sin(e + f*x)*c**2 - 2*sin(e + f*x)*c*d*n + sin(e + f*x)*c*d 
- sin(e + f*x)*d**2*n + c**2 - c*d*n),x)*sin(e + f*x)*c**2*d*f*n - int(((s 
in(e + f*x)*d + c)**n*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)**3*c*d - si 
n(e + f*x)**3*d**2*n + sin(e + f*x)**2*c**2 - sin(e + f*x)**2*c*d*n + 2*si 
n(e + f*x)**2*c*d - 2*sin(e + f*x)**2*d**2*n + 2*sin(e + f*x)*c**2 - 2*sin 
(e + f*x)*c*d*n + sin(e + f*x)*c*d - sin(e + f*x)*d**2*n + c**2 - c*d*n),x 
)*sin(e + f*x)*c*d**2*f*n**2 - int(((sin(e + f*x)*d + c)**n*cos(e + f*x)*s 
in(e + f*x))/(sin(e + f*x)**3*c*d - sin(e + f*x)**3*d**2*n + sin(e + f*x)* 
*2*c**2 - sin(e + f*x)**2*c*d*n + 2*sin(e + f*x)**2*c*d - 2*sin(e + f*x)** 
2*d**2*n + 2*sin(e + f*x)*c**2 - 2*sin(e + f*x)*c*d*n + sin(e + f*x)*c*d - 
 sin(e + f*x)*d**2*n + c**2 - c*d*n),x)*sin(e + f*x)*c*d**2*f*n + int(((si 
n(e + f*x)*d + c)**n*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)**3*c*d - sin 
(e + f*x)**3*d**2*n + sin(e + f*x)**2*c**2 - sin(e + f*x)**2*c*d*n + 2*sin 
(e + f*x)**2*c*d - 2*sin(e + f*x)**2*d**2*n + 2*sin(e + f*x)*c**2 - 2*sin( 
e + f*x)*c*d*n + sin(e + f*x)*c*d - sin(e + f*x)*d**2*n + c**2 - c*d*n),x) 
*sin(e + f*x)*d**3*f*n**2 + int(((sin(e + f*x)*d + c)**n*cos(e + f*x)*sin( 
e + f*x))/(sin(e + f*x)**3*c*d - sin(e + f*x)**3*d**2*n + sin(e + f*x)*...