\(\int \sec ^2(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [28]

Optimal result

Integrand size = 31, antiderivative size = 120 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(C (2+n)+A (3+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) (3+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{2+n} \tan (c+d x)}{b^2 d (3+n)} \] Output:

(C*(2+n)+A*(3+n))*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)/(3+n)/(sin(d*x+c)^2)^(1/2)+C*(b*se 
c(d*x+c))^(2+n)*tan(d*x+c)/b^2/d/(3+n)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (4+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(c+d x)\right )+C (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+n) (4+n)} \] Input:

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

Output:

(Csc[c + d*x]*Sec[c + d*x]*(b*Sec[c + d*x])^n*(A*(4 + n)*Hypergeometric2F1 
[1/2, (2 + n)/2, (4 + n)/2, Sec[c + d*x]^2] + C*(2 + n)*Hypergeometric2F1[ 
1/2, (4 + n)/2, (6 + n)/2, Sec[c + d*x]^2]*Sec[c + d*x]^2)*Sqrt[-Tan[c + d 
*x]^2])/(d*(2 + n)*(4 + n))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2030, 3042, 4534, 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.

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

((b*(A + (C*(2 + n))/(3 + n))*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[S 
in[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(2 + n)*Tan[c + d*x])/(d*(3 + n)))/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 4534

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="f 
ricas")
 

Output:

integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*(b*sec(d*x + c))^n, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="m 
axima")
 

Output:

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

Giac [F]

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

integrate(sec(d*x+c)^2*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="g 
iac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sec ^2(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=b^{n} \left (\left (\int \sec \left (d x +c \right )^{n} \sec \left (d x +c \right )^{4}d x \right ) c +\left (\int \sec \left (d x +c \right )^{n} \sec \left (d x +c \right )^{2}d x \right ) a \right ) \] Input:

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

Output:

b**n*(int(sec(c + d*x)**n*sec(c + d*x)**4,x)*c + int(sec(c + d*x)**n*sec(c 
 + d*x)**2,x)*a)