Integrand size = 21, antiderivative size = 87 \[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=-\frac {(b \cot (e+f x))^{1+n} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{2} (1-m+n),\frac {3+n}{2},\cos ^2(e+f x)\right ) (a \sin (e+f x))^m \sin ^2(e+f x)^{\frac {1}{2} (1-m+n)}}{b f (1+n)} \]
-(b*cot(f*x+e))^(1+n)*hypergeom([1/2+1/2*n, 1/2-1/2*m+1/2*n],[3/2+1/2*n],c os(f*x+e)^2)*(a*sin(f*x+e))^m*(sin(f*x+e)^2)^(1/2-1/2*m+1/2*n)/b/f/(1+n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 2.69 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.32 \[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=\frac {(3+m-n) \operatorname {AppellF1}\left (\frac {1}{2} (1+m-n),-n,1+m,\frac {1}{2} (3+m-n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (b \cot (e+f x))^n \sin (e+f x) (a \sin (e+f x))^m}{f (1+m-n) \left ((3+m-n) \operatorname {AppellF1}\left (\frac {1}{2} (1+m-n),-n,1+m,\frac {1}{2} (3+m-n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left (n \operatorname {AppellF1}\left (\frac {1}{2} (3+m-n),1-n,1+m,\frac {1}{2} (5+m-n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(1+m) \operatorname {AppellF1}\left (\frac {1}{2} (3+m-n),-n,2+m,\frac {1}{2} (5+m-n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]
((3 + m - n)*AppellF1[(1 + m - n)/2, -n, 1 + m, (3 + m - n)/2, Tan[(e + f* x)/2]^2, -Tan[(e + f*x)/2]^2]*(b*Cot[e + f*x])^n*Sin[e + f*x]*(a*Sin[e + f *x])^m)/(f*(1 + m - n)*((3 + m - n)*AppellF1[(1 + m - n)/2, -n, 1 + m, (3 + m - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(n*AppellF1[(3 + m - n)/2, 1 - n, 1 + m, (5 + m - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/ 2]^2] + (1 + m)*AppellF1[(3 + m - n)/2, -n, 2 + m, (5 + m - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))
Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3083, 3042, 3097}
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.
\(\displaystyle \int (a \sin (e+f x))^m (b \cot (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \cos \left (e+f x-\frac {\pi }{2}\right )\right )^m \left (-b \tan \left (e+f x-\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 3083 |
\(\displaystyle (a \sin (e+f x))^m \left (\frac {\csc (e+f x)}{a}\right )^m \int (b \cot (e+f x))^n \left (\frac {\csc (e+f x)}{a}\right )^{-m}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a \sin (e+f x))^m \left (\frac {\csc (e+f x)}{a}\right )^m \int \left (\frac {\sec \left (e+f x-\frac {\pi }{2}\right )}{a}\right )^{-m} \left (-b \tan \left (e+f x-\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 3097 |
\(\displaystyle -\frac {(a \sin (e+f x))^m (b \cot (e+f x))^{n+1} \sin ^2(e+f x)^{\frac {1}{2} (-m+n+1)} \operatorname {Hypergeometric2F1}\left (\frac {n+1}{2},\frac {1}{2} (-m+n+1),\frac {n+3}{2},\cos ^2(e+f x)\right )}{b f (n+1)}\) |
-(((b*Cot[e + f*x])^(1 + n)*Hypergeometric2F1[(1 + n)/2, (1 - m + n)/2, (3 + n)/2, Cos[e + f*x]^2]*(a*Sin[e + f*x])^m*(Sin[e + f*x]^2)^((1 - m + n)/ 2))/(b*f*(1 + n)))
3.1.39.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(a*Cos[e + f*x])^FracPart[m]*(Sec[e + f*x]/a)^FracPar t[m] Int[(b*Tan[e + f*x])^n/(Sec[e + f*x]/a)^m, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e + f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[(n - 1)/2] && !IntegerQ[m/2]
\[\int \left (b \cot \left (f x +e \right )\right )^{n} \left (a \sin \left (f x +e \right )\right )^{m}d x\]
\[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=\int { \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=\int \left (a \sin {\left (e + f x \right )}\right )^{m} \left (b \cot {\left (e + f x \right )}\right )^{n}\, dx \]
\[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=\int { \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=\int { \left (b \cot \left (f x + e\right )\right )^{n} \left (a \sin \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (b \cot (e+f x))^n (a \sin (e+f x))^m \, dx=\int {\left (b\,\mathrm {cot}\left (e+f\,x\right )\right )}^n\,{\left (a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]