\(\int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx\) [299]

Optimal result

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

b*(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/f/(1-n)/(cos(f*x+e)^2)^(1/4)/(c*sec(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 11.10 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.42 \[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=-\frac {2 \cos (2 (e+f x)) \cot (e+f x) (b \csc (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\sec ^2(e+f x)\right ) \sqrt {c \sec (e+f x)} \left (-\tan ^2(e+f x)\right )^{\frac {1+n}{2}}}{c^2 f (-3+2 n) \left (-2+\sec ^2(e+f x)\right )} \] Input:

Integrate[(b*Csc[e + f*x])^n/(c*Sec[e + f*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.39 (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.

\(\displaystyle \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3111

\(\displaystyle \frac {b^2 (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int (c \cos (e+f x))^{3/2} (b \sin (e+f x))^{-n}dx}{c^2 \sqrt {c \cos (e+f x)} \sqrt {c \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int (c \cos (e+f x))^{3/2} (b \sin (e+f x))^{-n}dx}{c^2 \sqrt {c \cos (e+f x)} \sqrt {c \sec (e+f x)}}\)

\(\Big \downarrow \) 3057

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

Input:

Int[(b*Csc[e + f*x])^n/(c*Sec[e + f*x])^(3/2),x]
 

Output:

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

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]

\[\int \frac {\left (b \csc \left (f x +e \right )\right )^{n}}{\left (c \sec \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]

Input:

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x)
 

Output:

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x)
 

Fricas [F]

\[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=\int \frac {\left (b \csc {\left (e + f x \right )}\right )^{n}}{\left (c \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((b*csc(f*x+e))**n/(c*sec(f*x+e))**(3/2),x)
 

Output:

Integral((b*csc(e + f*x))**n/(c*sec(e + f*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*csc(f*x + e))^n/(c*sec(f*x + e))^(3/2), x)
 

Giac [F]

\[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=\int { \frac {\left (b \csc \left (f x + e\right )\right )^{n}}{\left (c \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*csc(f*x + e))^n/(c*sec(f*x + e))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (\frac {b}{\sin \left (e+f\,x\right )}\right )}^n}{{\left (\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((b/sin(e + f*x))^n/(c/cos(e + f*x))^(3/2),x)
 

Output:

int((b/sin(e + f*x))^n/(c/cos(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(b \csc (e+f x))^n}{(c \sec (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, b^{n} \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \csc \left (f x +e \right )^{n}}{\sec \left (f x +e \right )^{2}}d x \right )}{c^{2}} \] Input:

int((b*csc(f*x+e))^n/(c*sec(f*x+e))^(3/2),x)
 

Output:

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