\(\int \frac {\cos (e+f x) (a+a \sin (e+f x))^m}{(c+d \sin (e+f x))^2} \, dx\) [922]

Optimal result

Integrand size = 31, antiderivative size = 59 \[ \int \frac {\cos (e+f x) (a+a \sin (e+f x))^m}{(c+d \sin (e+f x))^2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d (1+\sin (e+f x))}{c-d}\right ) (a+a \sin (e+f x))^{1+m}}{a (c-d)^2 f (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (e+f x) (a+a \sin (e+f x))^m}{(c+d \sin (e+f x))^2} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (2,1+m,2+m,-\frac {d (1+\sin (e+f x))}{c-d}\right ) (a+a \sin (e+f x))^{1+m}}{a (c-d)^2 f (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3042, 3312, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3312

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((sin(e + f*x)*a + a)**m + int(((sin(e + f*x)*a + a)**m*cos(e + f*x)*sin(e 
 + f*x))/(sin(e + f*x)**3*c*d**2*m - sin(e + f*x)**3*d**3 + 2*sin(e + f*x) 
**2*c**2*d*m + sin(e + f*x)**2*c*d**2*m - 2*sin(e + f*x)**2*c*d**2 - sin(e 
 + f*x)**2*d**3 + sin(e + f*x)*c**3*m + 2*sin(e + f*x)*c**2*d*m - sin(e + 
f*x)*c**2*d - 2*sin(e + f*x)*c*d**2 + c**3*m - c**2*d),x)*sin(e + f*x)*c** 
2*d*f*m**2 - int(((sin(e + f*x)*a + a)**m*cos(e + f*x)*sin(e + f*x))/(sin( 
e + f*x)**3*c*d**2*m - sin(e + f*x)**3*d**3 + 2*sin(e + f*x)**2*c**2*d*m + 
 sin(e + f*x)**2*c*d**2*m - 2*sin(e + f*x)**2*c*d**2 - sin(e + f*x)**2*d** 
3 + sin(e + f*x)*c**3*m + 2*sin(e + f*x)*c**2*d*m - sin(e + f*x)*c**2*d - 
2*sin(e + f*x)*c*d**2 + c**3*m - c**2*d),x)*sin(e + f*x)*c*d**2*f*m**2 - i 
nt(((sin(e + f*x)*a + a)**m*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)**3*c* 
d**2*m - sin(e + f*x)**3*d**3 + 2*sin(e + f*x)**2*c**2*d*m + sin(e + f*x)* 
*2*c*d**2*m - 2*sin(e + f*x)**2*c*d**2 - sin(e + f*x)**2*d**3 + sin(e + f* 
x)*c**3*m + 2*sin(e + f*x)*c**2*d*m - sin(e + f*x)*c**2*d - 2*sin(e + f*x) 
*c*d**2 + c**3*m - c**2*d),x)*sin(e + f*x)*c*d**2*f*m + int(((sin(e + f*x) 
*a + a)**m*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)**3*c*d**2*m - sin(e + 
f*x)**3*d**3 + 2*sin(e + f*x)**2*c**2*d*m + sin(e + f*x)**2*c*d**2*m - 2*s 
in(e + f*x)**2*c*d**2 - sin(e + f*x)**2*d**3 + sin(e + f*x)*c**3*m + 2*sin 
(e + f*x)*c**2*d*m - sin(e + f*x)*c**2*d - 2*sin(e + f*x)*c*d**2 + c**3*m 
- c**2*d),x)*sin(e + f*x)*d**3*f*m + int(((sin(e + f*x)*a + a)**m*cos(e...