\(\int (b \sec (a+b x))^n \sqrt {c \sin (a+b x)} \, dx\) [504]

Optimal result

Integrand size = 23, antiderivative size = 76 \[ \int (b \sec (a+b x))^n \sqrt {c \sin (a+b x)} \, dx=-\frac {c \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(a+b x)\right ) (b \sec (a+b x))^{-1+n} \sqrt [4]{\sin ^2(a+b x)}}{(1-n) \sqrt {c \sin (a+b x)}} \] Output:

-c*hypergeom([1/4, 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)/(1-n)/(c*sin(b*x+a))^(1/2)
 

Mathematica [A] (verified)

Time = 10.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int (b \sec (a+b x))^n \sqrt {c \sin (a+b x)} \, dx=\frac {\cos ^2(a+b x)^{\frac {1}{2} (-1+n)} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {1+n}{2},\frac {7}{4},\sin ^2(a+b x)\right ) (b \sec (a+b x))^n \sqrt {c \sin (a+b x)} \sin (2 (a+b x))}{3 b} \] Input:

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

Output:

((Cos[a + b*x]^2)^((-1 + n)/2)*Hypergeometric2F1[3/4, (1 + n)/2, 7/4, Sin[ 
a + b*x]^2]*(b*Sec[a + b*x])^n*Sqrt[c*Sin[a + b*x]]*Sin[2*(a + b*x)])/(3*b 
)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 76, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {c \sin (a+b x)} (b \sec (a+b x))^ndx\)

\(\Big \downarrow \) 3067

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3056

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

Input:

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

Output:

-((c*Hypergeometric2F1[1/4, (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))/((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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(c*sin(b*x + a))*(b*sec(b*x + a))^n, x)
 

Sympy [F]

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

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

Output:

Integral((b*sec(a + b*x))**n*sqrt(c*sin(a + b*x)), x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(c*sin(b*x + a))*(b*sec(b*x + a))^n, x)
 

Giac [F]

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

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

Output:

integrate(sqrt(c*sin(b*x + a))*(b*sec(b*x + a))^n, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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