3.8.88 \(\int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx\) [788]

3.8.88.1 Optimal result

Integrand size = 21, antiderivative size = 273 \[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\frac {(a+b \sec (e+f x))^{1+m} \tan (e+f x)}{b f (2+m)}-\frac {\sqrt {2} a (a+b) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b^2 f (2+m) \sqrt {1+\sec (e+f x)}}+\frac {\sqrt {2} \left (a^2+b^2 (1+m)\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b^2 f (2+m) \sqrt {1+\sec (e+f x)}} \]

output

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

3.8.88.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(8899\) vs. \(2(273)=546\).

Time = 27.25 (sec) , antiderivative size = 8899, normalized size of antiderivative = 32.60 \[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\text {Result too large to show} \]

input

Integrate[Sec[e + f*x]^3*(a + b*Sec[e + f*x])^m,x]
 

output

Result too large to show
 

3.8.88.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4327, 3042, 4495, 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.

\(\displaystyle \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^3 \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 4327

\(\displaystyle \frac {\int \sec (e+f x) (b (m+1)-a \sec (e+f x)) (a+b \sec (e+f x))^mdx}{b (m+2)}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (b (m+1)-a \csc \left (e+f x+\frac {\pi }{2}\right )\right ) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx}{b (m+2)}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 4495

\(\displaystyle \frac {\frac {\left (a^2+b^2 (m+1)\right ) \int \sec (e+f x) (a+b \sec (e+f x))^mdx}{b}-\frac {a \int \sec (e+f x) (a+b \sec (e+f x))^{m+1}dx}{b}}{b (m+2)}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (a^2+b^2 (m+1)\right ) \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx}{b}-\frac {a \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{m+1}dx}{b}}{b (m+2)}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 4321

\(\displaystyle \frac {\frac {a \tan (e+f x) \int \frac {(a+b \sec (e+f x))^{m+1}}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}-\frac {\left (a^2+b^2 (m+1)\right ) \tan (e+f x) \int \frac {(a+b \sec (e+f x))^m}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}}{b (m+2)}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\frac {a (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (e+f x)}{a+b}\right )^{m+1}}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}-\frac {\left (a^2+b^2 (m+1)\right ) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (e+f x)}{a+b}\right )^m}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}}{b (m+2)}+\frac {\tan (e+f x) (a+b \sec (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 155

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

input

Int[Sec[e + f*x]^3*(a + b*Sec[e + f*x])^m,x]
 

output

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

3.8.88.3.1 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]
 

rule 4327

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 
x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 
))), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( 
m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - 
b^2, 0] &&  !LtQ[m, -1]
 

rule 4495

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b   Int[ 
Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] + Simp[B/b   Int[Csc[e + f*x]*( 
a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && N 
eQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
 

3.8.88.4 Maple [F]

\[\int \sec \left (f x +e \right )^{3} \left (a +b \sec \left (f x +e \right )\right )^{m}d x\]

input

int(sec(f*x+e)^3*(a+b*sec(f*x+e))^m,x)
 

output

int(sec(f*x+e)^3*(a+b*sec(f*x+e))^m,x)
 

3.8.88.5 Fricas [F]

\[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \]

input

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e))^m,x, algorithm="fricas")
 

output

integral((b*sec(f*x + e) + a)^m*sec(f*x + e)^3, x)
 

3.8.88.6 Sympy [F]

\[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\int \left (a + b \sec {\left (e + f x \right )}\right )^{m} \sec ^{3}{\left (e + f x \right )}\, dx \]

input

integrate(sec(f*x+e)**3*(a+b*sec(f*x+e))**m,x)
 

output

Integral((a + b*sec(e + f*x))**m*sec(e + f*x)**3, x)
 

3.8.88.7 Maxima [F]

\[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \]

input

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e))^m,x, algorithm="maxima")
 

output

integrate((b*sec(f*x + e) + a)^m*sec(f*x + e)^3, x)
 

3.8.88.8 Giac [F]

\[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \]

input

integrate(sec(f*x+e)^3*(a+b*sec(f*x+e))^m,x, algorithm="giac")
 

output

integrate((b*sec(f*x + e) + a)^m*sec(f*x + e)^3, x)
 

3.8.88.9 Mupad [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) (a+b \sec (e+f x))^m \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^m}{{\cos \left (e+f\,x\right )}^3} \,d x \]

input

int((a + b/cos(e + f*x))^m/cos(e + f*x)^3,x)
 

output

int((a + b/cos(e + f*x))^m/cos(e + f*x)^3, x)