3.5.82 \(\int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [482]

3.5.82.1 Optimal result

Integrand size = 29, antiderivative size = 177 \[ \int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {b B \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m)}-\frac {(b B m+a A (1+m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {(A b+a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \sqrt {\sin ^2(c+d x)}} \]

output

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

3.5.82.2 Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.95 \[ \int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \left (a A \left (2+3 m+m^2\right ) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )+(A b+a B) m (2+m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sec ^2(c+d x)\right )+b B m (1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sec ^2(c+d x)\right )\right ) \sec ^{1+m}(c+d x) \sqrt {-\tan ^2(c+d x)}}{d m (1+m) (2+m)} \]

input

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

output

(Csc[c + d*x]*(a*A*(2 + 3*m + m^2)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, m 
/2, (2 + m)/2, Sec[c + d*x]^2] + (A*b + a*B)*m*(2 + m)*Cos[c + d*x]*Hyperg 
eometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sec[c + d*x]^2] + b*B*m*(1 + m)*Hyp 
ergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Sec[c + d*x]^2])*Sec[c + d*x]^(1 
 + m)*Sqrt[-Tan[c + d*x]^2])/(d*m*(1 + m)*(2 + m))
 

3.5.82.3 Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 4485, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4485

\(\displaystyle \frac {\int \sec ^m(c+d x) (b B m+a A (m+1)+(A b+a B) (m+1) \sec (c+d x))dx}{m+1}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle \frac {(m+1) (a B+A b) \int \sec ^{m+1}(c+d x)dx+(a A (m+1)+b B m) \int \sec ^m(c+d x)dx}{m+1}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a A (m+1)+b B m) \int \csc \left (c+d x+\frac {\pi }{2}\right )^mdx+(m+1) (a B+A b) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+1}dx}{m+1}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {(m+1) (a B+A b) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{-m-1}(c+d x)dx+(a A (m+1)+b B m) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{-m}(c+d x)dx}{m+1}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x)}{d (m+1)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

input

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

output

(b*B*Sec[c + d*x]^(1 + m)*Sin[c + d*x])/(d*(1 + m)) + (-(((b*B*m + a*A*(1 
+ m))*Hypergeometric2F1[1/2, (1 - m)/2, (3 - m)/2, Cos[c + d*x]^2]*Sec[c + 
 d*x]^(-1 + m)*Sin[c + d*x])/(d*(1 - m)*Sqrt[Sin[c + d*x]^2])) + ((A*b + a 
*B)*(1 + m)*Hypergeometric2F1[1/2, -1/2*m, (2 - m)/2, Cos[c + d*x]^2]*Sec[ 
c + d*x]^m*Sin[c + d*x])/(d*m*Sqrt[Sin[c + d*x]^2]))/(1 + m)
 

3.5.82.3.1 Defintions of rubi rules used

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]
 

rule 4485

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

3.5.82.4 Maple [F]

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

input

int(sec(d*x+c)^m*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

output

int(sec(d*x+c)^m*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

3.5.82.5 Fricas [F]

\[ \int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m} \,d x } \]

input

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

output

integral((B*b*sec(d*x + c)^2 + A*a + (B*a + A*b)*sec(d*x + c))*sec(d*x + c 
)^m, x)
 

3.5.82.6 Sympy [F]

\[ \int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \]

input

integrate(sec(d*x+c)**m*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

output

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x)**m, x)
 

3.5.82.7 Maxima [F]

\[ \int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m} \,d x } \]

input

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

output

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)*sec(d*x + c)^m, x)
 

3.5.82.8 Giac [F]

\[ \int \sec ^m(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m} \,d x } \]

input

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

output

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)*sec(d*x + c)^m, x)
 

3.5.82.9 Mupad [F(-1)]

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

input

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

output

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