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