Integrand size = 19, antiderivative size = 88 \[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=\frac {a (a \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} \csc ^{-1+n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n)} \] Output:
a*(a*cos(f*x+e))^(-1+m)*(cos(f*x+e)^2)^(1/2-1/2*m)*csc(f*x+e)^(-1+n)*hyper geom([1/2-1/2*m, 1/2-1/2*n],[3/2-1/2*n],sin(f*x+e)^2)/f/(1-n)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.16 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.57 \[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=-\frac {2 (-3+n) \operatorname {AppellF1}\left (\frac {1}{2}-\frac {n}{2},-m,1+m-n,\frac {3}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) (a \cos (e+f x))^m \csc ^n(e+f x) \sin \left (\frac {1}{2} (e+f x)\right )}{f (-1+n) \left ((-3+n) \operatorname {AppellF1}\left (\frac {1}{2}-\frac {n}{2},-m,1+m-n,\frac {3}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (m \operatorname {AppellF1}\left (\frac {3}{2}-\frac {n}{2},1-m,1+m-n,\frac {5}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(1+m-n) \operatorname {AppellF1}\left (\frac {3}{2}-\frac {n}{2},-m,2+m-n,\frac {5}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[(a*Cos[e + f*x])^m*Csc[e + f*x]^n,x]
Output:
(-2*(-3 + n)*AppellF1[1/2 - n/2, -m, 1 + m - n, 3/2 - n/2, Tan[(e + f*x)/2 ]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^3*(a*Cos[e + f*x])^m*Csc[e + f* x]^n*Sin[(e + f*x)/2])/(f*(-1 + n)*((-3 + n)*AppellF1[1/2 - n/2, -m, 1 + m - n, 3/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2] ^2 + 2*(m*AppellF1[3/2 - n/2, 1 - m, 1 + m - n, 5/2 - n/2, Tan[(e + f*x)/2 ]^2, -Tan[(e + f*x)/2]^2] + (1 + m - n)*AppellF1[3/2 - n/2, -m, 2 + m - n, 5/2 - n/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sin[(e + f*x)/2]^2))
Time = 0.34 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3067, 3042, 3057}
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.
\(\displaystyle \int \csc ^n(e+f x) (a \cos (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (-\sec \left (e+f x+\frac {\pi }{2}\right )\right )^n \left (a \sin \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle \sin ^n(e+f x) \csc ^n(e+f x) \int (a \cos (e+f x))^m \sin ^{-n}(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^n(e+f x) \csc ^n(e+f x) \int (a \cos (e+f x))^m \sin (e+f x)^{-n}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {a \cos ^2(e+f x)^{\frac {1-m}{2}} \csc ^{n-1}(e+f x) (a \cos (e+f x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n)}\) |
Input:
Int[(a*Cos[e + f*x])^m*Csc[e + f*x]^n,x]
Output:
(a*(a*Cos[e + f*x])^(-1 + m)*(Cos[e + f*x]^2)^((1 - m)/2)*Csc[e + f*x]^(-1 + n)*Hypergeometric2F1[(1 - m)/2, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^2])/ (f*(1 - n))
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m _), x_Symbol] :> Simp[b^2*(b*Cos[e + f*x])^(n - 1)*(b*Sec[e + f*x])^(n - 1) Int[(a*Sin[e + f*x])^m/(b*Cos[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n]
\[\int \left (\cos \left (f x +e \right ) a \right )^{m} \csc \left (f x +e \right )^{n}d x\]
Input:
int((cos(f*x+e)*a)^m*csc(f*x+e)^n,x)
Output:
int((cos(f*x+e)*a)^m*csc(f*x+e)^n,x)
\[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=\int { \left (a \cos \left (f x + e\right )\right )^{m} \csc \left (f x + e\right )^{n} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*csc(f*x+e)^n,x, algorithm="fricas")
Output:
integral((a*cos(f*x + e))^m*csc(f*x + e)^n, x)
\[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=\int \left (a \cos {\left (e + f x \right )}\right )^{m} \csc ^{n}{\left (e + f x \right )}\, dx \] Input:
integrate((a*cos(f*x+e))**m*csc(f*x+e)**n,x)
Output:
Integral((a*cos(e + f*x))**m*csc(e + f*x)**n, x)
\[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=\int { \left (a \cos \left (f x + e\right )\right )^{m} \csc \left (f x + e\right )^{n} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*csc(f*x+e)^n,x, algorithm="maxima")
Output:
integrate((a*cos(f*x + e))^m*csc(f*x + e)^n, x)
\[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=\int { \left (a \cos \left (f x + e\right )\right )^{m} \csc \left (f x + e\right )^{n} \,d x } \] Input:
integrate((a*cos(f*x+e))^m*csc(f*x+e)^n,x, algorithm="giac")
Output:
integrate((a*cos(f*x + e))^m*csc(f*x + e)^n, x)
Timed out. \[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=\int {\left (a\,\cos \left (e+f\,x\right )\right )}^m\,{\left (\frac {1}{\sin \left (e+f\,x\right )}\right )}^n \,d x \] Input:
int((a*cos(e + f*x))^m*(1/sin(e + f*x))^n,x)
Output:
int((a*cos(e + f*x))^m*(1/sin(e + f*x))^n, x)
\[ \int (a \cos (e+f x))^m \csc ^n(e+f x) \, dx=a^{m} \left (\int \csc \left (f x +e \right )^{n} \cos \left (f x +e \right )^{m}d x \right ) \] Input:
int((a*cos(f*x+e))^m*csc(f*x+e)^n,x)
Output:
a**m*int(csc(e + f*x)**n*cos(e + f*x)**m,x)