\(\int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx\) [29]

Optimal result

Integrand size = 33, antiderivative size = 66 \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (1,n,1+n,-\frac {1}{2} i (i+\cot (c+d x))\right ) \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n}{2 d n} \] Output:

-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)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 3.20 (sec) , antiderivative size = 367, normalized size of antiderivative = 5.56 \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\operatorname {AppellF1}\left (1-n,-2 n,1,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {Hypergeometric2F1}\left (1-2 n,1-n,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (a (\cos (c+d x)+i \sin (c+d x)))^n \sin ^{-n}(c+d x)}{d (-1+n) \left (2 \operatorname {AppellF1}\left (1-n,-2 n,1,2-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\left (-2 n \operatorname {AppellF1}\left (2-n,1-2 n,1,3-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right ) (-1+\cos (c+d x)+i \sin (c+d x))-\operatorname {AppellF1}\left (2-n,-2 n,2,3-n,-i \tan \left (\frac {1}{2} (c+d x)\right ),i \tan \left (\frac {1}{2} (c+d x)\right )\right ) (-1+\cos (c+d x)+i \sin (c+d x))+(-2+n) (1+\cos (c+d x)) \left (1+i \tan \left (\frac {1}{2} (c+d x)\right )\right )^{2 n}\right ) \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{-2+n}\right )} \] Input:

Integrate[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^n/Sin[c + d*x]^n,x]
 

Output:

(-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)))
 

Rubi [A] (verified)

Time = 0.25 (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

\(\displaystyle -\frac {i \sin ^{-n}(c+d x) \operatorname {Hypergeometric2F1}\left (1,n,n+1,-\frac {1}{2} i (\cot (c+d x)+i)\right ) (a \cos (c+d x)+i a \sin (c+d x))^n}{2 d n}\)

Input:

Int[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^n/Sin[c + d*x]^n,x]
 

Output:

((-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)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3562

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]
 

Maple [F]

\[\int \left (a \cos \left (d x +c \right )+i a \sin \left (d x +c \right )\right )^{n} \sin \left (d x +c \right )^{-n}d x\]

Input:

int((a*cos(d*x+c)+I*a*sin(d*x+c))^n/(sin(d*x+c)^n),x)
 

Output:

int((a*cos(d*x+c)+I*a*sin(d*x+c))^n/(sin(d*x+c)^n),x)
 

Fricas [F]

\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \] Input:

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))^n/(sin(d*x+c)^n),x, algorithm="fri 
cas")
 

Output:

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)
 

Sympy [F]

\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \left (a \left (i \sin {\left (c + d x \right )} + \cos {\left (c + d x \right )}\right )\right )^{n} \sin ^{- n}{\left (c + d x \right )}\, dx \] Input:

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))**n/(sin(d*x+c)**n),x)
 

Output:

Integral((a*(I*sin(c + d*x) + cos(c + d*x)))**n/sin(c + d*x)**n, x)
 

Maxima [F]

\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \] Input:

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))^n/(sin(d*x+c)^n),x, algorithm="max 
ima")
 

Output:

integrate((a*cos(d*x + c) + I*a*sin(d*x + c))^n*sin(d*x + c)^(-n), x)
 

Giac [F]

\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int { \frac {{\left (a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )\right )}^{n}}{\sin \left (d x + c\right )^{n}} \,d x } \] Input:

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))^n/(sin(d*x+c)^n),x, algorithm="gia 
c")
 

Output:

integrate((a*cos(d*x + c) + I*a*sin(d*x + c))^n/sin(d*x + c)^n, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \frac {{\left (a\,\cos \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\sin \left (c+d\,x\right )}^n} \,d x \] Input:

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^n/sin(c + d*x)^n,x)
 

Output:

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^n/sin(c + d*x)^n, x)
 

Reduce [F]

\[ \int \sin ^{-n}(c+d x) (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int \frac {\left (\cos \left (d x +c \right ) a +\sin \left (d x +c \right ) a i \right )^{n}}{\sin \left (d x +c \right )^{n}}d x \] Input:

int((a*cos(d*x+c)+I*a*sin(d*x+c))^n/(sin(d*x+c)^n),x)
 

Output:

int((cos(c + d*x)*a + sin(c + d*x)*a*i)**n/sin(c + d*x)**n,x)