\(\int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [41]

Optimal result

Integrand size = 31, antiderivative size = 163 \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (1-2 n) \sqrt {\sec (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:

-2*A*hypergeom([1/2, 3/4-1/2*n],[7/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n 
*sin(d*x+c)/d/(3-2*n)/sec(d*x+c)^(3/2)/(sin(d*x+c)^2)^(1/2)-2*B*hypergeom( 
[1/2, 1/4-1/2*n],[5/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/( 
1-2*n)/sec(d*x+c)^(1/2)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.83 \[ \int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right )+B (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1+2 n),\frac {1}{4} (5+2 n),\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d \left (-1+4 n^2\right ) \sec ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(1 + 2*n)*Hypergeometric2F1[1/2, (-1 
 + 2*n)/4, (3 + 2*n)/4, Sec[c + d*x]^2] + B*(-1 + 2*n)*Hypergeometric2F1[1 
/2, (1 + 2*n)/4, (5 + 2*n)/4, Sec[c + d*x]^2]*Sec[c + d*x])*Sqrt[-Tan[c + 
d*x]^2])/(d*(-1 + 4*n^2)*Sec[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2034, 3042, 4274, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (-\frac {2 A \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {1}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (1-2 n),\frac {1}{4} (5-2 n),\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)}}\right )\)

Input:

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

Output:

((b*Sec[c + d*x])^n*((-2*A*Hypergeometric2F1[1/2, (3 - 2*n)/4, (7 - 2*n)/4 
, Cos[c + d*x]^2]*Sec[c + d*x]^(-3/2 + n)*Sin[c + d*x])/(d*(3 - 2*n)*Sqrt[ 
Sin[c + d*x]^2]) - (2*B*Hypergeometric2F1[1/2, (1 - 2*n)/4, (5 - 2*n)/4, C 
os[c + d*x]^2]*Sec[c + d*x]^(-1/2 + n)*Sin[c + d*x])/(d*(1 - 2*n)*Sqrt[Sin 
[c + d*x]^2])))/Sec[c + d*x]^n
 

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]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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