\(\int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx\) [137]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \cos (e+f x) (1-\sin (e+f x))^{-\frac {1}{2}-m} (a-a \sin (e+f x))^m}{f} \] Output:

2^(1/2+m)*AppellF1(1/2,-n,1/2-m,3/2,1+sin(f*x+e),1/2+1/2*sin(f*x+e))*cos(f 
*x+e)*(1-sin(f*x+e))^(-1/2-m)*(a-a*sin(f*x+e))^m/f
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 3266, 3042, 3264, 150}

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.

\(\displaystyle \int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (-\sin (e+f x))^n (a-a \sin (e+f x))^mdx\)

\(\Big \downarrow \) 3266

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3264

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2^{m+\frac {1}{2}} \cos (e+f x) (1-\sin (e+f x))^{-m-\frac {1}{2}} (a-a \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},\sin (e+f x)+1,\frac {1}{2} (\sin (e+f x)+1)\right )}{f}\)

Input:

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

Output:

(2^(1/2 + m)*AppellF1[1/2, -n, 1/2 - m, 3/2, 1 + Sin[e + f*x], (1 + Sin[e 
+ f*x])/2]*Cos[e + f*x]*(1 - Sin[e + f*x])^(-1/2 - m)*(a - a*Sin[e + f*x]) 
^m)/f
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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