\(\int \frac {(c \sin (a+b x))^m}{\sqrt {d \sec (a+b x)}} \, dx\) [486]

Optimal result

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

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

Mathematica [A] (verified)

Time = 28.76 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.16 \[ \int \frac {(c \sin (a+b x))^m}{\sqrt {d \sec (a+b x)}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},\frac {1}{4} (5+2 m),\frac {3+m}{2},-\tan ^2(a+b x)\right ) \sec ^2(a+b x)^{\frac {1}{4}+\frac {m}{2}} (c \sin (a+b x))^m \tan (a+b x)}{b (1+m) \sqrt {d \sec (a+b x)}} \] Input:

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

Output:

(Hypergeometric2F1[(1 + m)/2, (5 + 2*m)/4, (3 + m)/2, -Tan[a + b*x]^2]*(Se 
c[a + b*x]^2)^(1/4 + m/2)*(c*Sin[a + b*x])^m*Tan[a + b*x])/(b*(1 + m)*Sqrt 
[d*Sec[a + b*x]])
 

Rubi [A] (verified)

Time = 0.53 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3066

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3057

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

Input:

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

Output:

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

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]

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

Input:

int((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x)
 

Output:

int((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x)
 

Fricas [F]

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

integrate((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((c*sin(b*x+a))**m/(d*sec(b*x+a))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((c*sin(a + b*x))^m/(d/cos(a + b*x))^(1/2),x)
 

Output:

int((c*sin(a + b*x))^m/(d/cos(a + b*x))^(1/2), x)
 

Reduce [F]

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

int((c*sin(b*x+a))^m/(d*sec(b*x+a))^(1/2),x)
 

Output:

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