\(\int \sec ^m(c+d x) (b \sec (c+d x))^n (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [40]

Optimal result

Integrand size = 38, antiderivative size = 167 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-n),\frac {1}{2} (2-m-n),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (m+n) \sqrt {\sin ^2(c+d x)}}+\frac {C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-m-n),\frac {1}{2} (1-m-n),\cos ^2(c+d x)\right ) \sec ^{1+m}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (1+m+n) \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.77 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^n \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\csc (c+d x) \sec ^m(c+d x) (b \sec (c+d x))^n \left (B (2+m+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1+m+n),\frac {1}{2} (3+m+n),\sec ^2(c+d x)\right )+C (1+m+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2+m+n),\frac {1}{2} (4+m+n),\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (1+m+n) (2+m+n)} \] Input:

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

Output:

(Csc[c + d*x]*Sec[c + d*x]^m*(b*Sec[c + d*x])^n*(B*(2 + m + n)*Hypergeomet 
ric2F1[1/2, (1 + m + n)/2, (3 + m + n)/2, Sec[c + d*x]^2] + C*(1 + m + n)* 
Hypergeometric2F1[1/2, (2 + m + n)/2, (4 + m + n)/2, Sec[c + d*x]^2]*Sec[c 
 + d*x])*Sqrt[-Tan[c + d*x]^2])/(d*(1 + m + n)*(2 + m + n))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2034, 3042, 4535, 27, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="fricas")
 

Output:

integral((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c))^n*sec(d*x + 
c)^m, x)
 

Sympy [F]

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

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

Output:

Integral((b*sec(c + d*x))**n*(B + C*sec(c + d*x))*sec(c + d*x)*sec(c + d*x 
)**m, x)
 

Maxima [F]

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

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^n*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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