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

Optimal result

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

[Out]

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*s
in(f*x+e))^m/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2866, 2864, 138} \[ \int (-\sin (e+f x))^n (a-a \sin (e+f x))^m \, dx=\frac {2^{m+\frac {1}{2}} \cos (e+f x) (1-\sin (e+f x))^{-m-\frac {1}{2}} (a-a \sin (e+f x))^m \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},\sin (e+f x)+1,\frac {1}{2} (\sin (e+f x)+1)\right )}{f} \]

[In]

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

[Out]

(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

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> 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] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 2864

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(-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 2866

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[a^Int
Part[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]

Rubi steps \begin{align*} \text {integral}& = \left ((1-\sin (e+f x))^{-m} (a-a \sin (e+f x))^m\right ) \int (1-\sin (e+f x))^m (-\sin (e+f x))^n \, dx \\ & = \frac {\left (\cos (e+f x) (1-\sin (e+f x))^{-\frac {1}{2}-m} (a-a \sin (e+f x))^m\right ) \text {Subst}\left (\int \frac {(1-x)^n (2-x)^{-\frac {1}{2}+m}}{\sqrt {x}} \, dx,x,1+\sin (e+f x)\right )}{f \sqrt {1+\sin (e+f x)}} \\ & = \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} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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