\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx\) [71]

Optimal result

Integrand size = 34, antiderivative size = 81 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {3}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {3}{2}-m} (a+a \sin (e+f x))^{2+m}}{a^2 c^2 f} \] Output:

2^(3/2+m)*hypergeom([-1/2, -1/2-m],[1/2],1/2-1/2*sin(f*x+e))*sec(f*x+e)*(1 
+sin(f*x+e))^(-3/2-m)*(a+a*sin(f*x+e))^(2+m)/a^2/c^2/f
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {3}{2}+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a (1+\sin (e+f x)))^m}{c^2 f (1-\sin (e+f x))} \] Input:

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

Output:

(2^(3/2 + m)*Cos[e + f*x]*Hypergeometric2F1[-1/2, -1/2 - m, 1/2, (1 - Sin[ 
e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a*(1 + Sin[e + f*x]))^m)/(c^2* 
f*(1 - Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3042, 3319, 3042, 3168, 80, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3319

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

\(\displaystyle \frac {\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a} \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{(a-a \sin (e+f x))^{3/2}}d\sin (e+f x)}{c^2 f}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {2^{m+\frac {1}{2}} \sec (e+f x) \sqrt {a-a \sin (e+f x)} (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m+1} \int \frac {\left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{2}}}{(a-a \sin (e+f x))^{3/2}}d\sin (e+f x)}{c^2 f}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {2^{m+\frac {3}{2}} \sec (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{a c^2 f}\)

Input:

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

Output:

(2^(3/2 + m)*Hypergeometric2F1[-1/2, -1/2 - m, 1/2, (1 - Sin[e + f*x])/2]* 
Sec[e + f*x]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^(1 + m))/( 
a*c^2*f)
 

Defintions of rubi rules used

rule 79

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

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

rule 3168

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.), x_Symbol] :> Simp[a^2*((g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin 
[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2)))   Subst[Int[(a + 
b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; Fre 
eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]
 

rule 3319

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[ 
a^m*(c^m/g^(2*m))   Int[(g*Cos[e + f*x])^(2*m + p)*(c + d*Sin[e + f*x])^(n 
- m), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] 
&& EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && LtQ[n^2, m^2])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m} \cos ^{2}{\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx}{c^{2}} \] Input:

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

Output:

Integral((a*sin(e + f*x) + a)**m*cos(e + f*x)**2/(sin(e + f*x)**2 - 2*sin( 
e + f*x) + 1), x)/c**2
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x}{c^{2}} \] Input:

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

Output:

int(((sin(e + f*x)*a + a)**m*cos(e + f*x)**2)/(sin(e + f*x)**2 - 2*sin(e + 
 f*x) + 1),x)/c**2