\(\int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx\) [477]

Optimal result

Integrand size = 25, antiderivative size = 25 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\frac {\sqrt {2} B \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}+A \text {Int}\left (\frac {1}{(a+b \sec (c+d x))^{2/3}},x\right ) \] Output:

2^(1/2)*B*AppellF1(1/2,2/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+ 
c))*((a+b*sec(d*x+c))/(a+b))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))^(1/2)/(a+b* 
sec(d*x+c))^(2/3)+A*Defer(Int)(1/(a+b*sec(d*x+c))^(2/3),x)
 

Mathematica [N/A]

Not integrable

Time = 24.64 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx \] Input:

Integrate[(A + B*Sec[c + d*x])/(a + b*Sec[c + d*x])^(2/3),x]
 

Output:

Integrate[(A + B*Sec[c + d*x])/(a + b*Sec[c + d*x])^(2/3), x]
 

Rubi [N/A]

Not integrable

Time = 0.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {3042, 4412, 3042, 4273, 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.

\(\displaystyle \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\)

\(\Big \downarrow \) 4412

\(\displaystyle A \int \frac {1}{(a+b \sec (c+d x))^{2/3}}dx+B \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^{2/3}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle A \int \frac {1}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+B \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\)

\(\Big \downarrow \) 4273

\(\displaystyle A \int \frac {1}{(a+b \sec (c+d x))^{2/3}}dx+B \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\)

\(\Big \downarrow \) 4321

\(\displaystyle A \int \frac {1}{(a+b \sec (c+d x))^{2/3}}dx-\frac {B \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\)

\(\Big \downarrow \) 156

\(\displaystyle A \int \frac {1}{(a+b \sec (c+d x))^{2/3}}dx-\frac {B \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \left (\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}\right )^{2/3}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}\)

\(\Big \downarrow \) 155

\(\displaystyle A \int \frac {1}{(a+b \sec (c+d x))^{2/3}}dx+\frac {\sqrt {2} B \tan (c+d x) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt {\sec (c+d x)+1} (a+b \sec (c+d x))^{2/3}}\)

Input:

Int[(A + B*Sec[c + d*x])/(a + b*Sec[c + d*x])^(2/3),x]
 

Output:

$Aborted
 

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 4273

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Unintegrable[ 
(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0 
] &&  !IntegerQ[2*n]
 

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 4412

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[c   Int[(a + b*Csc[e + f*x])^m, x], x] + Sim 
p[d   Int[(a + b*Csc[e + f*x])^m*Csc[e + f*x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[2*m]
 

Maple [N/A]

Not integrable

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

\[\int \frac {A +B \sec \left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]

Input:

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(2/3),x)
 

Output:

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(2/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(2/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [N/A]

Not integrable

Time = 1.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \] Input:

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(2/3),x)
 

Output:

Integral((A + B*sec(c + d*x))/(a + b*sec(c + d*x))**(2/3), x)
 

Maxima [N/A]

Not integrable

Time = 1.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(2/3),x, algorithm="maxima")
 

Output:

integrate((B*sec(d*x + c) + A)/(b*sec(d*x + c) + a)^(2/3), x)
 

Giac [N/A]

Not integrable

Time = 0.90 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(2/3),x, algorithm="giac")
 

Output:

integrate((B*sec(d*x + c) + A)/(b*sec(d*x + c) + a)^(2/3), x)
 

Mupad [N/A]

Not integrable

Time = 13.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \] Input:

int((A + B/cos(c + d*x))/(a + b/cos(c + d*x))^(2/3),x)
 

Output:

int((A + B/cos(c + d*x))/(a + b/cos(c + d*x))^(2/3), x)
 

Reduce [N/A]

Not integrable

Time = 0.16 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\left (\int \frac {\sec \left (d x +c \right )}{\left (\sec \left (d x +c \right ) b +a \right )^{\frac {2}{3}}}d x \right ) b +\left (\int \frac {1}{\left (\sec \left (d x +c \right ) b +a \right )^{\frac {2}{3}}}d x \right ) a \] Input:

int((A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(2/3),x)
 

Output:

int(sec(c + d*x)/(sec(c + d*x)*b + a)**(2/3),x)*b + int(1/(sec(c + d*x)*b 
+ a)**(2/3),x)*a