\(\int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [34]

Optimal result

Integrand size = 29, antiderivative size = 143 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2-n),-\frac {n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2+n} \sin (c+d x)}{b^2 d (2+n) \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(c+d x)\right )+B (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+n) (3+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {2030, 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.

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

((A*b*Hypergeometric2F1[1/2, (-1 - n)/2, (1 - n)/2, Cos[c + d*x]^2]*(b*Sec 
[c + d*x])^(1 + n)*Sin[c + d*x])/(d*(1 + n)*Sqrt[Sin[c + d*x]^2]) + (B*Hyp 
ergeometric2F1[1/2, (-2 - n)/2, -1/2*n, Cos[c + d*x]^2]*(b*Sec[c + d*x])^( 
2 + n)*Sin[c + d*x])/(d*(2 + n)*Sqrt[Sin[c + d*x]^2]))/b^2
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

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

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

Output:

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x))*sec(c + d*x)**2, x)
 

Maxima [F]

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

integrate(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorithm="max 
ima")
 

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

b**n*(int(sec(c + d*x)**n*sec(c + d*x)**3,x)*b + int(sec(c + d*x)**n*sec(c 
 + d*x)**2,x)*a)