\(\int \frac {(b \sec (a+b x))^n}{(c \sin (a+b x))^{3/2}} \, dx\) [506]

Optimal result

Integrand size = 23, antiderivative size = 78 \[ \int \frac {(b \sec (a+b x))^n}{(c \sin (a+b x))^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(a+b x)\right ) (b \sec (a+b x))^{-1+n} \sqrt [4]{\sin ^2(a+b x)}}{c (1-n) \sqrt {c \sin (a+b x)}} \] Output:

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

Mathematica [A] (verified)

Time = 10.77 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {(b \sec (a+b x))^n}{(c \sin (a+b x))^{3/2}} \, dx=-\frac {\cos ^2(a+b x)^{\frac {1}{2} (-1+n)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1+n}{2},\frac {3}{4},\sin ^2(a+b x)\right ) (b \sec (a+b x))^n \sin (2 (a+b x))}{b (c \sin (a+b x))^{3/2}} \] Input:

Integrate[(b*Sec[a + b*x])^n/(c*Sin[a + b*x])^(3/2),x]
 

Output:

-(((Cos[a + b*x]^2)^((-1 + n)/2)*Hypergeometric2F1[-1/4, (1 + n)/2, 3/4, S 
in[a + b*x]^2]*(b*Sec[a + b*x])^n*Sin[2*(a + b*x)])/(b*(c*Sin[a + b*x])^(3 
/2)))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 78, 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, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(b \sec (a+b x))^n}{(c \sin (a+b x))^{3/2}}dx\)

\(\Big \downarrow \) 3067

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3056

\(\displaystyle -\frac {\sqrt [4]{\sin ^2(a+b x)} (b \sec (a+b x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(a+b x)\right )}{c (1-n) \sqrt {c \sin (a+b x)}}\)

Input:

Int[(b*Sec[a + b*x])^n/(c*Sin[a + b*x])^(3/2),x]
 

Output:

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

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]

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

Input:

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x)
 

Output:

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x)
 

Fricas [F]

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

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(c*sin(b*x + a))*(b*sec(b*x + a))^n/(c^2*cos(b*x + a)^2 - c^ 
2), x)
 

Sympy [F]

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

integrate((b*sec(b*x+a))**n/(c*sin(b*x+a))**(3/2),x)
 

Output:

Integral((b*sec(a + b*x))**n/(c*sin(a + b*x))**(3/2), x)
 

Maxima [F]

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

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*sec(b*x + a))^n/(c*sin(b*x + a))^(3/2), x)
 

Giac [F]

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

integrate((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x, algorithm="giac")
 

Output:

integrate((b*sec(b*x + a))^n/(c*sin(b*x + a))^(3/2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((b*sec(b*x+a))^n/(c*sin(b*x+a))^(3/2),x)
 

Output:

(sqrt(c)*b**n*int((sqrt(sin(a + b*x))*sec(a + b*x)**n)/sin(a + b*x)**2,x)) 
/c**2