\(\int \frac {(a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx\) [416]

Optimal result

Integrand size = 26, antiderivative size = 81 \[ \int \frac {(a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{a c f} \] Output:

2^(1/2+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))^(-1/2-m)*(a+a*sin(f*x+e))^(1+m)/a/c/f
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3215, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

\(\displaystyle \frac {\int \sec ^2(e+f x) (\sin (e+f x) a+a)^{m+1}dx}{a c}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

(2^(1/2 + m)*Hypergeometric2F1[-1/2, 1/2 - m, 1/2, (1 - Sin[e + f*x])/2]*S 
ec[e + f*x]*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x])^m)/(c*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 3215

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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