\(\int (b \csc (e+f x))^n (c \sec (e+f x))^{3/2} \, dx\) [296]

Optimal result

Integrand size = 23, antiderivative size = 81 \[ \int (b \csc (e+f x))^n (c \sec (e+f x))^{3/2} \, dx=\frac {b \cos ^2(e+f x)^{5/4} (b \csc (e+f x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right ) (c \sec (e+f x))^{5/2}}{c f (1-n)} \] Output:

b*(cos(f*x+e)^2)^(5/4)*(b*csc(f*x+e))^(-1+n)*hypergeom([5/4, 1/2-1/2*n],[3 
/2-1/2*n],sin(f*x+e)^2)*(c*sec(f*x+e))^(5/2)/c/f/(1-n)
 

Mathematica [A] (verified)

Time = 10.51 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int (b \csc (e+f x))^n (c \sec (e+f x))^{3/2} \, dx=\frac {2 \cot (e+f x) (b \csc (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\sec ^2(e+f x)\right ) (c \sec (e+f x))^{3/2} \left (-\tan ^2(e+f x)\right )^{\frac {1+n}{2}}}{f (3+2 n)} \] Input:

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

Output:

(2*Cot[e + f*x]*(b*Csc[e + f*x])^n*Hypergeometric2F1[(1 + n)/2, (3 + 2*n)/ 
4, (7 + 2*n)/4, Sec[e + f*x]^2]*(c*Sec[e + f*x])^(3/2)*(-Tan[e + f*x]^2)^( 
(1 + n)/2))/(f*(3 + 2*n))
 

Rubi [A] (verified)

Time = 0.38 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3111

\(\displaystyle \frac {b^2 (c \cos (e+f x))^{5/2} (c \sec (e+f x))^{5/2} (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int \frac {(b \sin (e+f x))^{-n}}{(c \cos (e+f x))^{3/2}}dx}{c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 (c \cos (e+f x))^{5/2} (c \sec (e+f x))^{5/2} (b \sin (e+f x))^{n-1} (b \csc (e+f x))^{n-1} \int \frac {(b \sin (e+f x))^{-n}}{(c \cos (e+f x))^{3/2}}dx}{c^2}\)

\(\Big \downarrow \) 3057

\(\displaystyle \frac {b \cos ^2(e+f x)^{5/4} (c \sec (e+f x))^{5/2} (b \csc (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {1-n}{2},\frac {3-n}{2},\sin ^2(e+f x)\right )}{c f (1-n)}\)

Input:

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

Output:

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

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

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]

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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