\(\int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx\) [423]

Optimal result

Integrand size = 28, antiderivative size = 74 \[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{2 c f (1+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 10.41 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a (1+\sin (e+f x)))^m}{2 c f (1+2 m) \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 3224, 3042, 3146, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m}{(c-c \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3224

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {3}{2}}}{\cos (e+f x)^3}dx}{a c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\)

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

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 3146

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

rule 3224

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e 
+ f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*FracP 
art[m]))   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F 
reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
 && (FractionQ[m] ||  !FractionQ[n])
 

Maple [F]

\[\int \frac {\left (a +\sin \left (f x +e \right ) a \right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

Input:

int((a+sin(f*x+e)*a)^m/(c-c*sin(f*x+e))^(3/2),x)
 

Output:

int((a+sin(f*x+e)*a)^m/(c-c*sin(f*x+e))^(3/2),x)
 

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m}}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a+a*sin(f*x+e))**m/(c-c*sin(f*x+e))**(3/2),x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m/(-c*(sin(e + f*x) - 1))**(3/2), x)
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{1,[0,1,1,0,0,0,0]%%%} / %%%{16,[0,0,0,1,1,1,1]%%%} Error: 
Bad Argum
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(3/2),x)
 

Output:

int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right )}{c^{2}} \] Input:

int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x)
 

Output:

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