\(\int \sec (e+f x) (a+b \sec (e+f x))^m \, dx\) [790]

Optimal result

Integrand size = 19, antiderivative size = 103 \[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{f \sqrt {1+\sec (e+f x)}} \] Output:

2^(1/2)*AppellF1(1/2,-m,1/2,3/2,b*(1-sec(f*x+e))/(a+b),1/2-1/2*sec(f*x+e)) 
*(a+b*sec(f*x+e))^m*tan(f*x+e)/f/(1+sec(f*x+e))^(1/2)/(((a+b*sec(f*x+e))/( 
a+b))^m)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2828\) vs. \(2(103)=206\).

Time = 14.36 (sec) , antiderivative size = 2828, normalized size of antiderivative = 27.46 \[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\text {Result too large to show} \] Input:

Integrate[Sec[e + f*x]*(a + b*Sec[e + f*x])^m,x]
 

Output:

(-6*(a + b)*AppellF1[1/2, 1 + m, -m, 3/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan 
[(e + f*x)/2]^2)/(a + b)]*(b + a*Cos[e + f*x])^m*Sec[e + f*x]^(1 + m)*(a + 
 b*Sec[e + f*x])^m*Tan[(e + f*x)/2])/(f*(-1 + Tan[(e + f*x)/2]^2)*(3*(a + 
b)*AppellF1[1/2, 1 + m, -m, 3/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x 
)/2]^2)/(a + b)] + 2*(-((a - b)*m*AppellF1[3/2, 1 + m, 1 - m, 5/2, Tan[(e 
+ f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)]) + (a + b)*(1 + m)*Appe 
llF1[3/2, 2 + m, -m, 5/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2) 
/(a + b)])*Tan[(e + f*x)/2]^2)*((6*(a + b)*AppellF1[1/2, 1 + m, -m, 3/2, T 
an[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)]*(b + a*Cos[e + f* 
x])^m*Sec[(e + f*x)/2]^2*Sec[e + f*x]^m*Tan[(e + f*x)/2]^2)/((-1 + Tan[(e 
+ f*x)/2]^2)^2*(3*(a + b)*AppellF1[1/2, 1 + m, -m, 3/2, Tan[(e + f*x)/2]^2 
, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)] + 2*(-((a - b)*m*AppellF1[3/2, 1 + 
 m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)]) 
 + (a + b)*(1 + m)*AppellF1[3/2, 2 + m, -m, 5/2, Tan[(e + f*x)/2]^2, ((a - 
 b)*Tan[(e + f*x)/2]^2)/(a + b)])*Tan[(e + f*x)/2]^2)) - (3*(a + b)*Appell 
F1[1/2, 1 + m, -m, 3/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/( 
a + b)]*(b + a*Cos[e + f*x])^m*Sec[(e + f*x)/2]^2*Sec[e + f*x]^m)/((-1 + T 
an[(e + f*x)/2]^2)*(3*(a + b)*AppellF1[1/2, 1 + m, -m, 3/2, Tan[(e + f*x)/ 
2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a + b)] + 2*(-((a - b)*m*AppellF1[3/2, 
 1 + m, 1 - m, 5/2, Tan[(e + f*x)/2]^2, ((a - b)*Tan[(e + f*x)/2]^2)/(a...
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 4321, 156, 155}

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.

\(\displaystyle \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 4321

\(\displaystyle -\frac {\tan (e+f x) \int \frac {(a+b \sec (e+f x))^m}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 156

\(\displaystyle -\frac {\tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (e+f x)}{a+b}\right )^m}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\sqrt {2} \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{f \sqrt {\sec (e+f x)+1}}\)

Input:

Int[Sec[e + f*x]*(a + b*Sec[e + f*x])^m,x]
 

Output:

(Sqrt[2]*AppellF1[1/2, 1/2, -m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + 
 f*x]))/(a + b)]*(a + b*Sec[e + f*x])^m*Tan[e + f*x])/(f*Sqrt[1 + Sec[e + 
f*x]]*((a + b*Sec[e + f*x])/(a + b))^m)
 

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

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

rule 3042

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

rule 4321

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

Maple [F]

\[\int \sec \left (f x +e \right ) \left (a +b \sec \left (f x +e \right )\right )^{m}d x\]

Input:

int(sec(f*x+e)*(a+b*sec(f*x+e))^m,x)
 

Output:

int(sec(f*x+e)*(a+b*sec(f*x+e))^m,x)
 

Fricas [F]

\[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \] Input:

integrate(sec(f*x+e)*(a+b*sec(f*x+e))^m,x, algorithm="fricas")
 

Output:

integral((b*sec(f*x + e) + a)^m*sec(f*x + e), x)
 

Sympy [F]

\[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\int \left (a + b \sec {\left (e + f x \right )}\right )^{m} \sec {\left (e + f x \right )}\, dx \] Input:

integrate(sec(f*x+e)*(a+b*sec(f*x+e))**m,x)
 

Output:

Integral((a + b*sec(e + f*x))**m*sec(e + f*x), x)
 

Maxima [F]

\[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \] Input:

integrate(sec(f*x+e)*(a+b*sec(f*x+e))^m,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e) + a)^m*sec(f*x + e), x)
 

Giac [F]

\[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \] Input:

integrate(sec(f*x+e)*(a+b*sec(f*x+e))^m,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e) + a)^m*sec(f*x + e), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^m}{\cos \left (e+f\,x\right )} \,d x \] Input:

int((a + b/cos(e + f*x))^m/cos(e + f*x),x)
 

Output:

int((a + b/cos(e + f*x))^m/cos(e + f*x), x)
 

Reduce [F]

\[ \int \sec (e+f x) (a+b \sec (e+f x))^m \, dx=\int \left (\sec \left (f x +e \right ) b +a \right )^{m} \sec \left (f x +e \right )d x \] Input:

int(sec(f*x+e)*(a+b*sec(f*x+e))^m,x)
 

Output:

int((sec(e + f*x)*b + a)**m*sec(e + f*x),x)