Integrand size = 17, antiderivative size = 85 \[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\frac {\cos ^{-1+m}(e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} \csc ^{-1+n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n)} \] Output:
cos(f*x+e)^(-1+m)*(cos(f*x+e)^2)^(1/2-1/2*m)*csc(f*x+e)^(-1+n)*hypergeom([ 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 = 2.77 (sec) , antiderivative size = 312, normalized size of antiderivative = 3.67 \[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=-\frac {2 (-3+n) \operatorname {AppellF1}\left (\frac {1}{2}-\frac {n}{2},-m,1+m-n,\frac {3}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \cos ^m(e+f x) \csc ^n(e+f x) \sin \left (\frac {1}{2} (e+f x)\right )}{f (-1+n) \left ((-3+n) \operatorname {AppellF1}\left (\frac {1}{2}-\frac {n}{2},-m,1+m-n,\frac {3}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (m \operatorname {AppellF1}\left (\frac {3}{2}-\frac {n}{2},1-m,1+m-n,\frac {5}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(1+m-n) \operatorname {AppellF1}\left (\frac {3}{2}-\frac {n}{2},-m,2+m-n,\frac {5}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[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*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 = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, 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.
\(\displaystyle \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (e+f x+\frac {\pi }{2}\right )^m \left (-\sec \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle \sin ^n(e+f x) \csc ^n(e+f x) \int \cos ^m(e+f x) \sin ^{-n}(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sin ^n(e+f x) \csc ^n(e+f x) \int \cos (e+f x)^m \sin (e+f x)^{-n}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {\cos ^{m-1}(e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} \csc ^{n-1}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n)}\) |
Input:
Int[Cos[e + f*x]^m*Csc[e + f*x]^n,x]
Output:
(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]
\[\int \cos \left (f x +e \right )^{m} \csc \left (f x +e \right )^{n}d x\]
Input:
int(cos(f*x+e)^m*csc(f*x+e)^n,x)
Output:
int(cos(f*x+e)^m*csc(f*x+e)^n,x)
\[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\int { \cos \left (f x + e\right )^{m} \csc \left (f x + e\right )^{n} \,d x } \] Input:
integrate(cos(f*x+e)^m*csc(f*x+e)^n,x, algorithm="fricas")
Output:
integral(cos(f*x + e)^m*csc(f*x + e)^n, x)
\[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\int \cos ^{m}{\left (e + f x \right )} \csc ^{n}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**m*csc(f*x+e)**n,x)
Output:
Integral(cos(e + f*x)**m*csc(e + f*x)**n, x)
\[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\int { \cos \left (f x + e\right )^{m} \csc \left (f x + e\right )^{n} \,d x } \] Input:
integrate(cos(f*x+e)^m*csc(f*x+e)^n,x, algorithm="maxima")
Output:
integrate(cos(f*x + e)^m*csc(f*x + e)^n, x)
\[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\int { \cos \left (f x + e\right )^{m} \csc \left (f x + e\right )^{n} \,d x } \] Input:
integrate(cos(f*x+e)^m*csc(f*x+e)^n,x, algorithm="giac")
Output:
integrate(cos(f*x + e)^m*csc(f*x + e)^n, x)
Timed out. \[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\int {\cos \left (e+f\,x\right )}^m\,{\left (\frac {1}{\sin \left (e+f\,x\right )}\right )}^n \,d x \] Input:
int(cos(e + f*x)^m*(1/sin(e + f*x))^n,x)
Output:
int(cos(e + f*x)^m*(1/sin(e + f*x))^n, x)
\[ \int \cos ^m(e+f x) \csc ^n(e+f x) \, dx=\int \csc \left (f x +e \right )^{n} \cos \left (f x +e \right )^{m}d x \] Input:
int(cos(f*x+e)^m*csc(f*x+e)^n,x)
Output:
int(csc(e + f*x)**n*cos(e + f*x)**m,x)