Integrand size = 33, antiderivative size = 66 \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (1,n,1+n,-\frac {1}{2} i (i+\cot (c+d x))\right ) \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n}{2 d n} \]
-1/2*I*hypergeom([1, n],[1+n],-1/2*I*(I+cot(d*x+c)))*(a*cos(d*x+c)+I*a*sin (d*x+c))^n/d/n/(sin(d*x+c)^n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 5.00 (sec) , antiderivative size = 367, normalized size of antiderivative = 5.56 \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\operatorname {AppellF1}\left (1-n,-2 n,1,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {Hypergeometric2F1}\left (1-2 n,1-n,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a (\cos (c+d x)+i \sin (c+d x)))^n \sin ^{-n}(c+d x)}{d (-1+n) \left (2 \operatorname {AppellF1}\left (1-n,-2 n,1,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\left (-2 n \operatorname {AppellF1}\left (2-n,1-2 n,1,3-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right ) (-1+\cos (c+d x)+i \sin (c+d x))-\operatorname {AppellF1}\left (2-n,-2 n,2,3-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right ) (-1+\cos (c+d x)+i \sin (c+d x))+(-2+n) (1+\cos (c+d x)) \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )^{2 n}\right ) \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{-2+n}\right )} \]
(-4*Cos[(c + d*x)/2]*(AppellF1[1 - n, -2*n, 1, 2 - n, (-I)*Tan[(c + d*x)/2 ], I*Tan[(c + d*x)/2]] + Hypergeometric2F1[1 - 2*n, 1 - n, 2 - n, (-I)*Tan [(c + d*x)/2]])*Sin[(c + d*x)/2]*(a*(Cos[c + d*x] + I*Sin[c + d*x]))^n)/(d *(-1 + n)*Sin[c + d*x]^n*(2*AppellF1[1 - n, -2*n, 1, 2 - n, (-I)*Tan[(c + d*x)/2], I*Tan[(c + d*x)/2]] + ((-2*n*AppellF1[2 - n, 1 - 2*n, 1, 3 - n, ( -I)*Tan[(c + d*x)/2], I*Tan[(c + d*x)/2]]*(-1 + Cos[c + d*x] + I*Sin[c + d *x]) - AppellF1[2 - n, -2*n, 2, 3 - n, (-I)*Tan[(c + d*x)/2], I*Tan[(c + d *x)/2]]*(-1 + Cos[c + d*x] + I*Sin[c + d*x]) + (-2 + n)*(1 + Cos[c + d*x]) *(1 + I*Tan[(c + d*x)/2])^(2*n))*(1 - I*Tan[(c + d*x)/2]))/(-2 + n)))
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {3042, 3562}
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.
\(\displaystyle \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^{-n} (a \cos (c+d x)+i a \sin (c+d x))^ndx\) |
\(\Big \downarrow \) 3562 |
\(\displaystyle -\frac {i \sin ^{-n}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n,n+1,-\frac {1}{2} i (\cot (c+d x)+i)\right ) (a \cos (c+d x)+i a \sin (c+d x))^n}{2 d n}\) |
((-1/2*I)*Hypergeometric2F1[1, n, 1 + n, (-1/2*I)*(I + Cot[c + d*x])]*(a*C os[c + d*x] + I*a*Sin[c + d*x])^n)/(d*n*Sin[c + d*x]^n)
3.1.29.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a*Cos[c + d*x] + b*Sin[c + d*x])^n/(2*b*d*n*Sin[c + d*x]^n))*Hypergeometric2F1[1, n, n + 1, (b + a* Cot[c + d*x])/(2*b)], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[m + n, 0] && E qQ[a^2 + b^2, 0] && !IntegerQ[n]
\[\int \left (\cos \left (d x +c \right ) a +i a \sin \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{-n}d x\]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \]
integral(e^(I*d*n*x + I*c*n + n*log(a))/(1/2*(-I*e^(2*I*d*x + 2*I*c) + I)* e^(-I*d*x - I*c))^n, x)
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \left (a \left (i \sin {\left (c + d x \right )} + \cos {\left (c + d x \right )}\right )\right )^{n} \sin ^{- n}{\left (c + d x \right )}\, dx \]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \]
Timed out. \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \frac {{\left (a\,\cos \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\sin \left (c+d\,x\right )}^n} \,d x \]