\(\int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [39]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86 \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {2 \csc (c+d x) \left (A (5+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\sec ^2(c+d x)\right )+B (3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\sec ^2(c+d x)\right )\right ) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (3+2 n) (5+2 n)} \] Input:

Integrate[Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]
 

Output:

(2*Csc[c + d*x]*(A*(5 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (3 + 2*n) 
/4, (7 + 2*n)/4, Sec[c + d*x]^2] + B*(3 + 2*n)*Hypergeometric2F1[1/2, (5 + 
 2*n)/4, (9 + 2*n)/4, Sec[c + d*x]^2])*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x]) 
^n*Sqrt[-Tan[c + d*x]^2])/(d*(3 + 2*n)*(5 + 2*n))
 

Rubi [A] (verified)

Time = 0.58 (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.

\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)) (b \sec (c+d x))^n \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \sec ^{n+\frac {3}{2}}(c+d x) (A+B \sec (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {3}{2}} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 4274

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \sec ^{n+\frac {3}{2}}(c+d x)dx+B \int \sec ^{n+\frac {5}{2}}(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {3}{2}}dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}}dx\right )\)

\(\Big \downarrow \) 4259

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {3}{2}}(c+d x)dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {5}{2}}(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (A \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {3}{2}}dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {5}{2}}dx\right )\)

\(\Big \downarrow \) 3122

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

Input:

Int[Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]
 

Output:

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

\[\int \sec \left (d x +c \right )^{\frac {3}{2}} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )d x\]

Input:

int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)
 

Output:

int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)
 

Fricas [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:

integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm= 
"fricas")
 

Output:

integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*(b*sec(d*x + c))^n*sqrt(sec(d 
*x + c)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(3/2)*(b*sec(d*x+c))**n*(A+B*sec(d*x+c)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:

integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm= 
"maxima")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(3/2), x)
 

Giac [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \] Input:

integrate(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm= 
"giac")
 

Output:

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^n*sec(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:

int((A + B/cos(c + d*x))*(b/cos(c + d*x))^n*(1/cos(c + d*x))^(3/2),x)
 

Output:

int((A + B/cos(c + d*x))*(b/cos(c + d*x))^n*(1/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=b^{n} \left (\left (\int \sec \left (d x +c \right )^{n +\frac {1}{2}} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sec \left (d x +c \right )^{n +\frac {1}{2}} \sec \left (d x +c \right )d x \right ) a \right ) \] Input:

int(sec(d*x+c)^(3/2)*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)
 

Output:

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