Integrand size = 23, antiderivative size = 90 \[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},1+\sin (e+f x),\frac {1}{2} (1+\sin (e+f x))\right ) \cos (e+f x) (-\sin (e+f x))^{-n} (d \sin (e+f x))^n}{f \sqrt {1-\sin (e+f x)}} \]
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)*(d*sin(f*x+e))^n/f/((-sin(f*x+e))^n)/(1-sin(f*x+e))^(1/2)
\[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx \]
Time = 0.35 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3265, 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.
\(\displaystyle \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (1-\sin (e+f x))^m (d \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3265 |
\(\displaystyle (-\sin (e+f x))^{-n} (d \sin (e+f x))^n \int (1-\sin (e+f x))^m (-\sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (-\sin (e+f x))^{-n} (d \sin (e+f x))^n \int (1-\sin (e+f x))^m (-\sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3264 |
\(\displaystyle \frac {\cos (e+f x) (-\sin (e+f x))^{-n} (d \sin (e+f x))^n \int \frac {(1-\sin (e+f x))^{m-\frac {1}{2}} (-\sin (e+f x))^n}{\sqrt {\sin (e+f x)+1}}d(\sin (e+f x)+1)}{f \sqrt {1-\sin (e+f x)} \sqrt {\sin (e+f x)+1}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2^{m+\frac {1}{2}} \cos (e+f x) (-\sin (e+f x))^{-n} (d \sin (e+f x))^n \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},\sin (e+f x)+1,\frac {1}{2} (\sin (e+f x)+1)\right )}{f \sqrt {1-\sin (e+f x)}}\) |
(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]*(d*Sin[e + f*x])^n)/(f*Sqrt[1 - Sin[e + f*x]]*(-Si n[e + f*x])^n)
3.2.35.3.1 Defintions of rubi rules used
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])
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]
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]
\[\int \left (1-\sin \left (f x +e \right )\right )^{m} \left (d \sin \left (f x +e \right )\right )^{n}d x\]
\[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{n} {\left (-\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
\[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\int \left (d \sin {\left (e + f x \right )}\right )^{n} \left (1 - \sin {\left (e + f x \right )}\right )^{m}\, dx \]
\[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{n} {\left (-\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
\[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\int { \left (d \sin \left (f x + e\right )\right )^{n} {\left (-\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
Timed out. \[ \int (1-\sin (e+f x))^m (d \sin (e+f x))^n \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (1-\sin \left (e+f\,x\right )\right )}^m \,d x \]