\(\int \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}} \, dx\) [298]

Optimal result

Integrand size = 23, antiderivative size = 81 \[ \int \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}} \, dx=\frac {b \sqrt [4]{\cos ^2(e+f x)} (b \csc (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right ) \sqrt {c \sec (e+f x)}}{c f (1-n)} \] Output:

b*(cos(f*x+e)^2)^(1/4)*(b*csc(f*x+e))^(-1+n)*hypergeom([1/4, 1/2-1/2*n],[3 
/2-1/2*n],sin(f*x+e)^2)*(c*sec(f*x+e))^(1/2)/c/f/(1-n)
 

Mathematica [A] (verified)

Time = 27.05 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.17 \[ \int \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}} \, dx=\frac {(b \csc (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {n}{2},\frac {5}{4}-\frac {n}{2},\frac {3}{2}-\frac {n}{2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{\frac {1}{4}-\frac {n}{2}} \tan (e+f x)}{f (1-n) \sqrt {c \sec (e+f x)}} \] Input:

Integrate[(b*Csc[e + f*x])^n/Sqrt[c*Sec[e + f*x]],x]
 

Output:

((b*Csc[e + f*x])^n*Hypergeometric2F1[1/2 - n/2, 5/4 - n/2, 3/2 - n/2, -Ta 
n[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4 - n/2)*Tan[e + f*x])/(f*(1 - n)*Sqrt[c 
*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, 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.

\(\displaystyle \int \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \csc (e+f x))^n}{\sqrt {c \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3111

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3057

\(\displaystyle \frac {b \sqrt [4]{\cos ^2(e+f x)} \sqrt {c \sec (e+f x)} (b \csc (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{c f (1-n)}\)

Input:

Int[(b*Csc[e + f*x])^n/Sqrt[c*Sec[e + f*x]],x]
 

Output:

(b*(Cos[e + f*x]^2)^(1/4)*(b*Csc[e + f*x])^(-1 + n)*Hypergeometric2F1[1/4, 
 (1 - n)/2, (3 - n)/2, Sin[e + f*x]^2]*Sqrt[c*Sec[e + f*x]])/(c*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]

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

Input:

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x)
 

Output:

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x)
 

Fricas [F]

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

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

integral(sqrt(c*sec(f*x + e))*(b*csc(f*x + e))^n/(c*sec(f*x + e)), x)
 

Sympy [F]

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

integrate((b*csc(f*x+e))**n/(c*sec(f*x+e))**(1/2),x)
 

Output:

Integral((b*csc(e + f*x))**n/sqrt(c*sec(e + f*x)), x)
 

Maxima [F]

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

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

integrate((b*csc(f*x + e))^n/sqrt(c*sec(f*x + e)), x)
 

Giac [F]

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

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x, algorithm="giac")
 

Output:

integrate((b*csc(f*x + e))^n/sqrt(c*sec(f*x + e)), x)
 

Mupad [F(-1)]

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

int((b/sin(e + f*x))^n/(c/cos(e + f*x))^(1/2),x)
 

Output:

int((b/sin(e + f*x))^n/(c/cos(e + f*x))^(1/2), x)
 

Reduce [F]

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

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(1/2),x)
 

Output:

(sqrt(c)*b**n*int((sqrt(sec(e + f*x))*csc(e + f*x)**n)/sec(e + f*x),x))/c