\(\int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx\) [387]

Optimal result

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

(cos(f*x+e)^2)^(1/2+1/2*n)*(a*csc(f*x+e))^m*hypergeom([1/2+1/2*n, 1/2-1/2* 
m+1/2*n],[3/2-1/2*m+1/2*n],sin(f*x+e)^2)*(b*tan(f*x+e))^(1+n)/b/f/(1-m+n)
 

Mathematica [C] (warning: unable to verify)

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

Time = 1.92 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.22 \[ \int (a \csc (e+f x))^m (b \tan (e+f x))^n \, dx=-\frac {a (-3+m-n) \operatorname {AppellF1}\left (\frac {1}{2} (1-m+n),n,1-m,\frac {1}{2} (3-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (a \csc (e+f x))^{-1+m} (b \tan (e+f x))^n}{f (-1+m-n) \left ((-3+m-n) \operatorname {AppellF1}\left (\frac {1}{2} (1-m+n),n,1-m,\frac {1}{2} (3-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left ((-1+m) \operatorname {AppellF1}\left (\frac {1}{2} (3-m+n),n,2-m,\frac {1}{2} (5-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+n \operatorname {AppellF1}\left (\frac {1}{2} (3-m+n),1+n,1-m,\frac {1}{2} (5-m+n),\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3098, 3042, 3082, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3098

\(\displaystyle \left (\frac {\sin (e+f x)}{a}\right )^m (a \csc (e+f x))^m \int \left (\frac {\sin (e+f x)}{a}\right )^{-m} (b \tan (e+f x))^ndx\)

\(\Big \downarrow \) 3042

\(\displaystyle \left (\frac {\sin (e+f x)}{a}\right )^m (a \csc (e+f x))^m \int \left (\frac {\sin (e+f x)}{a}\right )^{-m} (b \tan (e+f x))^ndx\)

\(\Big \downarrow \) 3082

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3057

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

Input:

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

Output:

((Cos[e + f*x]^2)^((1 + n)/2)*(a*Csc[e + f*x])^m*Hypergeometric2F1[(1 + n) 
/2, (1 - m + n)/2, (3 - m + n)/2, Sin[e + f*x]^2]*(b*Tan[e + f*x])^(1 + n) 
)/(b*f*(1 - m + 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 3082

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[a*Cos[e + f*x]^(n + 1)*((b*Tan[e + f*x])^(n + 1)/(b* 
(a*Sin[e + f*x])^(n + 1)))   Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x 
], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[n]
 

rule 3098

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(a*Csc[e + f*x])^FracPart[m]*(Sin[e + f*x]/a)^FracPar 
t[m]   Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b, e, 
 f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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