\(\int (b \sec (e+f x))^n \sin ^m(e+f x) \, dx\) [490]

Optimal result

Integrand size = 19, antiderivative size = 89 \[ \int (b \sec (e+f x))^n \sin ^m(e+f x) \, dx=-\frac {b \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin ^{-1+m}(e+f x) \sin ^2(e+f x)^{\frac {1-m}{2}}}{f (1-n)} \] Output:

-b*hypergeom([1/2-1/2*m, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*sec(f*x+e 
))^(-1+n)*sin(f*x+e)^(-1+m)*(sin(f*x+e)^2)^(1/2-1/2*m)/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 = 0.58 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.22 \[ \int (b \sec (e+f x))^n \sin ^m(e+f x) \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m-n,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^3\left (\frac {1}{2} (e+f x)\right ) (b \sec (e+f x))^n \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^m(e+f x)}{f (1+m) \left ((3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},n,1+m-n,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+\cos (e+f x))-4 \left ((1+m-n) \operatorname {AppellF1}\left (\frac {3+m}{2},n,2+m-n,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-n \operatorname {AppellF1}\left (\frac {3+m}{2},1+n,1+m-n,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(4*(3 + m)*AppellF1[(1 + m)/2, n, 1 + m - n, (3 + m)/2, Tan[(e + f*x)/2]^2 
, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^3*(b*Sec[e + f*x])^n*Sin[(e + f*x) 
/2]*Sin[e + f*x]^m)/(f*(1 + m)*((3 + m)*AppellF1[(1 + m)/2, n, 1 + m - n, 
(3 + m)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + Cos[e + f*x]) - 4 
*((1 + m - n)*AppellF1[(3 + m)/2, n, 2 + m - n, (5 + m)/2, Tan[(e + f*x)/2 
]^2, -Tan[(e + f*x)/2]^2] - n*AppellF1[(3 + m)/2, 1 + n, 1 + m - n, (5 + m 
)/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sin[(e + f*x)/2]^2))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3067, 3042, 3056}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3067

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

\(\Big \downarrow \) 3042

\(\displaystyle b^2 (b \cos (e+f x))^{n-1} (b \sec (e+f x))^{n-1} \int (b \cos (e+f x))^{-n} \sin (e+f x)^mdx\)

\(\Big \downarrow \) 3056

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3056

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(-b^(2*IntPart[(n - 1)/2] + 1))*(b*Sin[e + f*x])^(2*F 
racPart[(n - 1)/2])*((a*Cos[e + f*x])^(m + 1)/(a*f*(m + 1)*(Sin[e + f*x]^2) 
^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, C 
os[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && SimplerQ[n, m]
 

rule 3067

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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