\(\int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, dx\) [219]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, dx=-\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.01 \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, dx=\frac {\csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} \left (\frac {1}{2}+n\right ),\frac {1}{2} \left (\frac {5}{2}+n\right ),\sec ^2(c+d x)\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d \left (\frac {1}{2}+n\right ) \sqrt {\sec (c+d x)}} \] Input:

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

Output:

(Csc[c + d*x]*Hypergeometric2F1[1/2, (1/2 + n)/2, (5/2 + n)/2, Sec[c + d*x 
]^2]*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*(1/2 + n)*Sqrt[Sec[c + d 
*x]])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2034, 3042, 4259, 3042, 3122}

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.

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

\(\Big \downarrow \) 2034

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n+\frac {1}{2}}(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {1}{2}}dx\)

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\sec (c+d x)} \cos ^{n+\frac {1}{2}}(c+d x) (b \sec (c+d x))^n \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {1}{2}}dx\)

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 4259

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^(n - 1)*((Sin[c + d*x]/b)^(n - 1)   Int[1/(Sin[c + d*x]/b)^n, x]), x] /; 
FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \, dx=b^{n} \left (\int \sec \left (d x +c \right )^{n +\frac {1}{2}}d x \right ) \] Input:

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

Output:

b**n*int(sec(c + d*x)**((2*n + 1)/2),x)