\(\int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [276]

Optimal result

Integrand size = 35, antiderivative size = 141 \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {A \sec ^{-n}(c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)}+\frac {2^{1+n} (B+A n+B n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+n,\frac {3}{2},\frac {1-\sec (c+d x)}{1+\sec (c+d x)}\right ) \sec ^{-n}(c+d x) \left (\frac {\sec (c+d x)}{1+\sec (c+d x)}\right )^{1+n} (a+a \sec (c+d x))^n \sin (c+d x)}{d (1+n)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.78 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79 \[ \int \sec ^{-1-n}(c+d x) (a+a \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\left (A+\frac {(B+A n+B n) \left (-\cot ^2\left (\frac {1}{2} (c+d x)\right )\right )^{\frac {1}{2}-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-n,-n,1-n,\csc ^2\left (\frac {1}{2} (c+d x)\right )\right )}{n (1+\cos (c+d x))}\right ) \sec ^{-n}(c+d x) (a (1+\sec (c+d x)))^n \sin (c+d x)}{d (1+n)} \] Input:

Integrate[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + B*Sec[c + d*x] 
),x]
 

Output:

((A + ((B + A*n + B*n)*(-Cot[(c + d*x)/2]^2)^(1/2 - n)*Hypergeometric2F1[1 
/2 - n, -n, 1 - n, Csc[(c + d*x)/2]^2])/(n*(1 + Cos[c + d*x])))*(a*(1 + Se 
c[c + d*x]))^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n)
 

Rubi [A] (warning: unable to verify)

Time = 0.64 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4501, 3042, 4315, 3042, 4312, 142}

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.

\(\displaystyle \int \sec ^{-n-1}(c+d x) (a \sec (c+d x)+a)^n (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

\(\displaystyle \frac {(A n+B n+B) \int \sec ^{-n}(c+d x) (\sec (c+d x) a+a)^ndx}{n+1}+\frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4315

\(\displaystyle \frac {(A n+B n+B) (\sec (c+d x)+1)^{-n} (a \sec (c+d x)+a)^n \int \sec ^{-n}(c+d x) (\sec (c+d x)+1)^ndx}{n+1}+\frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4312

\(\displaystyle \frac {(A n+B n+B) \tan (c+d x) (\sec (c+d x)+1)^{-n-\frac {1}{2}} (a \sec (c+d x)+a)^n \int \frac {\sec ^{-n-1}(c+d x) (\sec (c+d x)+1)^{n-\frac {1}{2}}}{\sqrt {1-\sec (c+d x)}}d(1-\sec (c+d x))}{d (n+1) \sqrt {1-\sec (c+d x)}}+\frac {A \sin (c+d x) \sec ^{-n}(c+d x) (a \sec (c+d x)+a)^n}{d (n+1)}\)

\(\Big \downarrow \) 142

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

Input:

Int[Sec[c + d*x]^(-1 - n)*(a + a*Sec[c + d*x])^n*(A + B*Sec[c + d*x]),x]
 

Output:

(A*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*(1 + n)*Sec[c + d*x]^n) + ((B + 
 A*n + B*n)*Hypergeometric2F1[1/2 - n, -n, 1 - n, (-2*Sec[c + d*x])/(1 - S 
ec[c + d*x])]*Sec[c + d*x]^(1 - n)*((1 + Sec[c + d*x])/(1 - Sec[c + d*x])) 
^(1/2 - n)*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/(d*n*(1 + n)*(1 + Sec[c + 
d*x]))
 

Defintions of rubi rules used

rule 142

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((b*e 
 - a*f)*(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2, (-(d*e - c*f))*((a + 
b*x)/((b*c - a*d)*(e + f*x)))])/((b*e - a*f)*((c + d*x)/((b*c - a*d)*(e + f 
*x))))^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[m + n + p + 2, 
 0] &&  !IntegerQ[n]
 

rule 3042

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

rule 4312

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-(a*(d/b))^n)*(Cot[e + f*x]/(a^(n - 2)*f*Sqrt 
[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]))   Subst[Int[(a - x)^(n - 1) 
*((2*a - x)^(m - 1/2)/Sqrt[x]), x], x, a - b*Csc[e + f*x]], x] /; FreeQ[{a, 
 b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] & 
&  !IntegerQ[n] && GtQ[a*(d/b), 0]
 

rule 4315

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m 
]/(1 + (b/a)*Csc[e + f*x])^FracPart[m])   Int[(1 + (b/a)*Csc[e + f*x])^m*(d 
*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 
2, 0] &&  !IntegerQ[m] &&  !GtQ[a, 0]
 

rule 4501

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]
 

Maple [F]

\[\int \sec \left (d x +c \right )^{-1-n} \left (a +a \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )\right )d x\]

Input:

int(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)
 

Output:

int(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)
 

Fricas [F]

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

integrate(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorit 
hm="fricas")
 

Output:

integral((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1) 
, x)
 

Sympy [F]

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

integrate(sec(d*x+c)**(-1-n)*(a+a*sec(d*x+c))**n*(A+B*sec(d*x+c)),x)
 

Output:

Integral((a*(sec(c + d*x) + 1))**n*(A + B*sec(c + d*x))*sec(c + d*x)**(-n 
- 1), x)
 

Maxima [F]

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

integrate(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorit 
hm="maxima")
 

Output:

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1 
), x)
 

Giac [F]

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

integrate(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x, algorit 
hm="giac")
 

Output:

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^n*sec(d*x + c)^(-n - 1 
), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^(-1-n)*(a+a*sec(d*x+c))^n*(A+B*sec(d*x+c)),x)
 

Output:

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