\(\int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx\) [283]

Optimal result

Integrand size = 21, antiderivative size = 89 \[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=\frac {b \cos ^2(e+f x)^{\frac {1+m}{2}} (b \csc (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right ) (a \sec (e+f x))^{1+m}}{a f (1-n)} \] Output:

b*(cos(f*x+e)^2)^(1/2+1/2*m)*(b*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)*(a*sec(f*x+e))^(1+m)/a/f/(1-n)
 

Mathematica [C] (warning: unable to verify)

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

Time = 0.75 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.18 \[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=-\frac {b (-3+n) \operatorname {AppellF1}\left (\frac {1-n}{2},m,1-m-n,\frac {3-n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (b \csc (e+f x))^{-1+n} (a \sec (e+f x))^m}{f (-1+n) \left ((-3+n) \operatorname {AppellF1}\left (\frac {1-n}{2},m,1-m-n,\frac {3-n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((-1+m+n) \operatorname {AppellF1}\left (\frac {3-n}{2},m,2-m-n,\frac {5-n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+m \operatorname {AppellF1}\left (\frac {3-n}{2},1+m,1-m-n,\frac {5-n}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

Integrate[(b*Csc[e + f*x])^n*(a*Sec[e + f*x])^m,x]
 

Output:

-((b*(-3 + n)*AppellF1[(1 - n)/2, m, 1 - m - n, (3 - n)/2, Tan[(e + f*x)/2 
]^2, -Tan[(e + f*x)/2]^2]*(b*Csc[e + f*x])^(-1 + n)*(a*Sec[e + f*x])^m)/(f 
*(-1 + n)*((-3 + n)*AppellF1[(1 - n)/2, m, 1 - m - n, (3 - n)/2, Tan[(e + 
f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*((-1 + m + n)*AppellF1[(3 - n)/2, m, 2 
 - m - n, (5 - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*AppellF1 
[(3 - n)/2, 1 + m, 1 - m - n, (5 - n)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x 
)/2]^2])*Tan[(e + f*x)/2]^2)))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3111, 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.

\(\displaystyle \int (a \sec (e+f x))^m (b \csc (e+f x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sec (e+f x))^m (b \csc (e+f x))^ndx\)

\(\Big \downarrow \) 3111

\(\displaystyle \frac {b^2 (a \cos (e+f x))^{m+1} (a \sec (e+f x))^{m+1} (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int (a \cos (e+f x))^{-m} (b \sin (e+f x))^{-n}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (a \cos (e+f x))^{m+1} (a \sec (e+f x))^{m+1} (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int (a \cos (e+f x))^{-m} (b \sin (e+f x))^{-n}dx}{a^2}\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {b \cos ^2(e+f x)^{\frac {m+1}{2}} (a \sec (e+f x))^{m+1} (b \csc (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{a f (1-n)}\)

Input:

Int[(b*Csc[e + f*x])^n*(a*Sec[e + f*x])^m,x]
 

Output:

(b*(Cos[e + f*x]^2)^((1 + m)/2)*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F 
1[(1 + m)/2, (1 - n)/2, (3 - n)/2, Sin[e + f*x]^2]*(a*Sec[e + f*x])^(1 + m 
))/(a*f*(1 - n))
 

Defintions of rubi rules used

rule 3042

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

rule 3057

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]
 

rule 3111

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a^2/b^2)*(a*Csc[e + f*x])^(m - 1)*(b*Sec[e + f*x])^( 
n + 1)*(a*Sin[e + f*x])^(m - 1)*(b*Cos[e + f*x])^(n + 1)   Int[1/((a*Sin[e 
+ f*x])^m*(b*Cos[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  ! 
SimplerQ[-m, -n]
 

Maple [F]

\[\int \left (b \csc \left (f x +e \right )\right )^{n} \left (a \sec \left (f x +e \right )\right )^{m}d x\]

Input:

int((b*csc(f*x+e))^n*(a*sec(f*x+e))^m,x)
 

Output:

int((b*csc(f*x+e))^n*(a*sec(f*x+e))^m,x)
 

Fricas [F]

\[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((b*csc(f*x+e))^n*(a*sec(f*x+e))^m,x, algorithm="fricas")
 

Output:

integral((b*csc(f*x + e))^n*(a*sec(f*x + e))^m, x)
 

Sympy [F]

\[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=\int \left (a \sec {\left (e + f x \right )}\right )^{m} \left (b \csc {\left (e + f x \right )}\right )^{n}\, dx \] Input:

integrate((b*csc(f*x+e))**n*(a*sec(f*x+e))**m,x)
 

Output:

Integral((a*sec(e + f*x))**m*(b*csc(e + f*x))**n, x)
 

Maxima [F]

\[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((b*csc(f*x+e))^n*(a*sec(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((b*csc(f*x + e))^n*(a*sec(f*x + e))^m, x)
 

Giac [F]

\[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=\int { \left (b \csc \left (f x + e\right )\right )^{n} \left (a \sec \left (f x + e\right )\right )^{m} \,d x } \] Input:

integrate((b*csc(f*x+e))^n*(a*sec(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((b*csc(f*x + e))^n*(a*sec(f*x + e))^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=\int {\left (\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n \,d x \] Input:

int((a/cos(e + f*x))^m*(b/sin(e + f*x))^n,x)
 

Output:

int((a/cos(e + f*x))^m*(b/sin(e + f*x))^n, x)
 

Reduce [F]

\[ \int (b \csc (e+f x))^n (a \sec (e+f x))^m \, dx=b^{n} a^{m} \left (\int \sec \left (f x +e \right )^{m} \csc \left (f x +e \right )^{n}d x \right ) \] Input:

int((b*csc(f*x+e))^n*(a*sec(f*x+e))^m,x)
 

Output:

b**n*a**m*int(sec(e + f*x)**m*csc(e + f*x)**n,x)