3.11.73 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx\) [1073]

3.11.73.1 Optimal result

Integrand size = 35, antiderivative size = 35 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\frac {\sqrt {2} C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{a+b \sec (c+d x)} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}}}+\frac {\sqrt {2} (b B-a C) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \left (\frac {a+b \sec (c+d x)}{a+b}\right )^{2/3} \tan (c+d x)}{b d \sqrt {1+\sec (c+d x)} (a+b \sec (c+d x))^{2/3}}+A \text {Int}\left (\frac {1}{(a+b \sec (c+d x))^{2/3}},x\right ) \]

output

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

3.11.73.2 Mathematica [N/A]

Not integrable

Time = 125.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx \]

input

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

output

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

3.11.73.3 Rubi [N/A]

Not integrable

Time = 0.88 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {3042, 4550, 3042, 4321, 156, 155, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4550

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4321

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

\(\Big \downarrow \) 4412

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4273

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

\(\Big \downarrow \) 4321

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

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

input

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

output

$Aborted
 

3.11.73.3.1 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 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]
 

rule 4550

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

3.11.73.4 Maple [N/A] (verified)

Not integrable

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94

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

input

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

output

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

3.11.73.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.11.73.6 Sympy [N/A]

Not integrable

Time = 1.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

input

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

output

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

3.11.73.7 Maxima [N/A]

Not integrable

Time = 24.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]

input

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

output

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

3.11.73.8 Giac [N/A]

Not integrable

Time = 2.88 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]

input

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

output

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

3.11.73.9 Mupad [N/A]

Not integrable

Time = 20.91 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \]

input

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

output

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