Integrand size = 19, antiderivative size = 88 \[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=\frac {b \cos ^{-1+m}(e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} (b \csc (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n)} \] Output:
b*cos(f*x+e)^(-1+m)*(cos(f*x+e)^2)^(1/2-1/2*m)*(b*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 = 1.22 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.57 \[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=-\frac {2 (-3+n) \operatorname {AppellF1}\left (\frac {1}{2}-\frac {n}{2},-m,1+m-n,\frac {3}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) \cos ^m(e+f x) (b \csc (e+f x))^n \sin \left (\frac {1}{2} (e+f x)\right )}{f (-1+n) \left ((-3+n) \operatorname {AppellF1}\left (\frac {1}{2}-\frac {n}{2},-m,1+m-n,\frac {3}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right )+2 \left (m \operatorname {AppellF1}\left (\frac {3}{2}-\frac {n}{2},1-m,1+m-n,\frac {5}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+(1+m-n) \operatorname {AppellF1}\left (\frac {3}{2}-\frac {n}{2},-m,2+m-n,\frac {5}{2}-\frac {n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[Cos[e + f*x]^m*(b*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*(b*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 = 0.36 (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.
\(\displaystyle \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (e+f x+\frac {\pi }{2}\right )^m \left (-b \sec \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 3067 |
\(\displaystyle b^2 (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int \cos ^m(e+f x) (b \sin (e+f x))^{-n}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int \cos (e+f x)^m (b \sin (e+f x))^{-n}dx\) |
\(\Big \downarrow \) 3057 |
\(\displaystyle \frac {b \cos ^{m-1}(e+f x) \cos ^2(e+f x)^{\frac {1-m}{2}} (b \csc (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{f (1-n)}\) |
Input:
Int[Cos[e + f*x]^m*(b*Csc[e + f*x])^n,x]
Output:
(b*Cos[e + f*x]^(-1 + m)*(Cos[e + f*x]^2)^((1 - m)/2)*(b*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]
\[\int \cos \left (f x +e \right )^{m} \left (b \csc \left (f x +e \right )\right )^{n}d x\]
Input:
int(cos(f*x+e)^m*(b*csc(f*x+e))^n,x)
Output:
int(cos(f*x+e)^m*(b*csc(f*x+e))^n,x)
\[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{m} \,d x } \] Input:
integrate(cos(f*x+e)^m*(b*csc(f*x+e))^n,x, algorithm="fricas")
Output:
integral((b*csc(f*x + e))^n*cos(f*x + e)^m, x)
\[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=\int \left (b \csc {\left (e + f x \right )}\right )^{n} \cos ^{m}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**m*(b*csc(f*x+e))**n,x)
Output:
Integral((b*csc(e + f*x))**n*cos(e + f*x)**m, x)
\[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{m} \,d x } \] Input:
integrate(cos(f*x+e)^m*(b*csc(f*x+e))^n,x, algorithm="maxima")
Output:
integrate((b*csc(f*x + e))^n*cos(f*x + e)^m, x)
\[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \cos \left (f x + e\right )^{m} \,d x } \] Input:
integrate(cos(f*x+e)^m*(b*csc(f*x+e))^n,x, algorithm="giac")
Output:
integrate((b*csc(f*x + e))^n*cos(f*x + e)^m, x)
Timed out. \[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=\int {\cos \left (e+f\,x\right )}^m\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n \,d x \] Input:
int(cos(e + f*x)^m*(b/sin(e + f*x))^n,x)
Output:
int(cos(e + f*x)^m*(b/sin(e + f*x))^n, x)
\[ \int \cos ^m(e+f x) (b \csc (e+f x))^n \, dx=b^{n} \left (\int \csc \left (f x +e \right )^{n} \cos \left (f x +e \right )^{m}d x \right ) \] Input:
int(cos(f*x+e)^m*(b*csc(f*x+e))^n,x)
Output:
b**n*int(csc(e + f*x)**n*cos(e + f*x)**m,x)