\(\int (d \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\) [660]

Optimal result

Integrand size = 27, antiderivative size = 95 \[ \int (d \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=-\frac {2^{1+m} \cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,\frac {3}{2},\frac {1-\sin (e+f x)}{1+\sin (e+f x)}\right ) (d \sin (e+f x))^{-m} \left (\frac {\sin (e+f x)}{1+\sin (e+f x)}\right )^{1+m} (a+a \sin (e+f x))^m}{d f} \] Output:

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

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.95 \[ \int (d \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-2 m,-m,1-m,-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (a \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right )^m (d \sin (e+f x))^{-m} \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )^{-2 m}}{d f m} \] Input:

Integrate[(d*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 3266, 3042, 3265, 3042, 3264, 142}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m-1} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m-1}dx\)

\(\Big \downarrow \) 3266

\(\displaystyle (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (d \sin (e+f x))^{-m-1} (\sin (e+f x)+1)^mdx\)

\(\Big \downarrow \) 3042

\(\displaystyle (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int (d \sin (e+f x))^{-m-1} (\sin (e+f x)+1)^mdx\)

\(\Big \downarrow \) 3265

\(\displaystyle \frac {\sin ^m(e+f x) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m} \int \sin ^{-m-1}(e+f x) (\sin (e+f x)+1)^mdx}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin ^m(e+f x) (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m (d \sin (e+f x))^{-m} \int \sin (e+f x)^{-m-1} (\sin (e+f x)+1)^mdx}{d}\)

\(\Big \downarrow \) 3264

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

\(\Big \downarrow \) 142

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

Input:

Int[(d*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]
 

Output:

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

Defintions of rubi rules used

rule 142

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e 
 - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + 
b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f 
*x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 
 0] &&  !IntegerQ[n]
 

rule 3042

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

rule 3264

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(d/b)^n*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e 
 + f*x]]*Sqrt[a - b*Sin[e + f*x]]))   Subst[Int[(a - x)^n*((2*a - x)^(m - 1 
/2)/Sqrt[x]), x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n} 
, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]
 

rule 3265

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_), x_Symbol] :> Simp[(d/b)^IntPart[n]*((d*Sin[e + f*x])^FracPart[n 
]/(b*Sin[e + f*x])^FracPart[n])   Int[(a + b*Sin[e + f*x])^m*(b*Sin[e + f*x 
])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !I 
ntegerQ[m] && GtQ[a, 0] &&  !GtQ[d/b, 0]
 

rule 3266

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Sin[e + f*x])^FracPart[m 
]/(1 + (b/a)*Sin[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Sin[e + f*x])^m*(d 
*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 
2, 0] &&  !IntegerQ[m] &&  !GtQ[a, 0]
 

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]

Input:

int((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Output:

int((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Fricas [F]

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

integrate((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="fricas")
 

Output:

integral((a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^(-m - 1), x)
 

Sympy [F]

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

integrate((d*sin(f*x+e))**(-1-m)*(a+a*sin(f*x+e))**m,x)
 

Output:

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

Maxima [F]

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

integrate((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^(-m - 1), x)
 

Giac [F]

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

integrate((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((a*sin(f*x + e) + a)^m*(d*sin(f*x + e))^(-m - 1), x)
 

Mupad [F(-1)]

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

int((a + a*sin(e + f*x))^m/(d*sin(e + f*x))^(m + 1),x)
 

Output:

int((a + a*sin(e + f*x))^m/(d*sin(e + f*x))^(m + 1), x)
 

Reduce [F]

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

int((d*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)
 

Output:

int((sin(e + f*x)*a + a)**m/(sin(e + f*x)**m*sin(e + f*x)),x)/(d**m*d)