Integrand size = 33, antiderivative size = 66 \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (1,n,1+n,-\frac {1}{2} i (i+\cot (c+d x))\right ) \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n}{2 d n} \]
[Out]
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {3162} \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \sin ^{-n}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n,n+1,-\frac {1}{2} i (\cot (c+d x)+i)\right ) (a \cos (c+d x)+i a \sin (c+d x))^n}{2 d n} \]
[In]
[Out]
Rule 3162
Rubi steps \begin{align*} \text {integral}& = -\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} \\ \end{align*}
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 \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\operatorname {AppellF1}\left (1-n,-2 n,1,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {Hypergeometric2F1}\left (1-2 n,1-n,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a (\cos (c+d x)+i \sin (c+d x)))^n \sin ^{-n}(c+d x)}{d (-1+n) \left (2 \operatorname {AppellF1}\left (1-n,-2 n,1,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\left (-2 n \operatorname {AppellF1}\left (2-n,1-2 n,1,3-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right ) (-1+\cos (c+d x)+i \sin (c+d x))-\operatorname {AppellF1}\left (2-n,-2 n,2,3-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right ) (-1+\cos (c+d x)+i \sin (c+d x))+(-2+n) (1+\cos (c+d x)) \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )^{2 n}\right ) \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{-2+n}\right )} \]
[In]
[Out]
\[\int \left (\cos \left (d x +c \right ) a +i a \sin \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{-n}d x\]
[In]
[Out]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \]
[In]
[Out]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \left (a \left (i \sin {\left (c + d x \right )} + \cos {\left (c + d x \right )}\right )\right )^{n} \sin ^{- n}{\left (c + d x \right )}\, dx \]
[In]
[Out]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \]
[In]
[Out]
\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \frac {{\left (a\,\cos \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\sin \left (c+d\,x\right )}^n} \,d x \]
[In]
[Out]