\(\int \frac {(c \sin (a+b x))^m}{(d \sec (a+b x))^{3/2}} \, dx\) [487]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {(c \sin (a+b x))^m}{(d \sec (a+b x))^{3/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1+m}{2},\frac {3+m}{2},\sin ^2(a+b x)\right ) (c \sin (a+b x))^{1+m}}{b c d (1+m) \sqrt [4]{\cos ^2(a+b x)} \sqrt {d \sec (a+b x)}} \] Output:

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

Mathematica [A] (verified)

Time = 11.70 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.51 \[ \int \frac {(c \sin (a+b x))^m}{(d \sec (a+b x))^{3/2}} \, dx=\frac {2 c \cos (2 (a+b x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} (-3-2 m),\frac {1-m}{2},\frac {1}{4} (1-2 m),\sec ^2(a+b x)\right ) (c \sin (a+b x))^{-1+m} \left (-\tan ^2(a+b x)\right )^{\frac {1-m}{2}}}{b d (3+2 m) \sqrt {d \sec (a+b x)} \left (-2+\sec ^2(a+b x)\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 77, 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, 3066, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3066

\(\displaystyle \frac {\int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^mdx}{d^2 \sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (d \cos (a+b x))^{3/2} (c \sin (a+b x))^mdx}{d^2 \sqrt {d \cos (a+b x)} \sqrt {d \sec (a+b x)}}\)

\(\Big \downarrow \) 3057

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

Input:

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

Output:

(Hypergeometric2F1[-1/4, (1 + m)/2, (3 + m)/2, Sin[a + b*x]^2]*(c*Sin[a + 
b*x])^(1 + m))/(b*c*d*(1 + m)*(Cos[a + b*x]^2)^(1/4)*Sqrt[d*Sec[a + b*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 3066

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(1/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] && LtQ[n, 1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(d*sec(b*x + a))*(c*sin(b*x + a))^m/(d^2*sec(b*x + a)^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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