Integrand size = 21, antiderivative size = 87 \[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=-\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right ) \cos (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f} \]
-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
\[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=\int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx \]
Time = 0.33 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, 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.
\(\displaystyle \int \sin ^n(e+f x) (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^n (a \sin (e+f x)+a)^mdx\) |
\(\Big \downarrow \) 3266 |
\(\displaystyle (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int \sin ^n(e+f x) (\sin (e+f x)+1)^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (\sin (e+f x)+1)^{-m} (a \sin (e+f x)+a)^m \int \sin (e+f x)^n (\sin (e+f x)+1)^mdx\) |
\(\Big \downarrow \) 3264 |
\(\displaystyle -\frac {\cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \int \frac {\sin ^n(e+f x) (\sin (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)}}d(1-\sin (e+f x))}{f \sqrt {1-\sin (e+f x)}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {2^{m+\frac {1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {AppellF1}\left (\frac {1}{2},-n,\frac {1}{2}-m,\frac {3}{2},1-\sin (e+f x),\frac {1}{2} (1-\sin (e+f x))\right )}{f}\) |
-((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)
3.2.36.3.1 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[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]
\[\int \left (\sin ^{n}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
\[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sin \left (f x + e\right )^{n} \,d x } \]
\[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \sin ^{n}{\left (e + f x \right )}\, dx \]
\[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sin \left (f x + e\right )^{n} \,d x } \]
\[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sin \left (f x + e\right )^{n} \,d x } \]
Timed out. \[ \int \sin ^n(e+f x) (a+a \sin (e+f x))^m \, dx=\int {\sin \left (e+f\,x\right )}^n\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]