\(\int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [66]

Optimal result

Integrand size = 41, antiderivative size = 227 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 C \sec ^{1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (5+3 m)}-\frac {3 (C (2+3 m)+A (5+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (1-3 m) (5+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-2-3 m),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \sin (c+d x)}{d (2+3 m) \sqrt {\sin ^2(c+d x)}} \] Output:

3*C*sec(d*x+c)^(1+m)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(5+3*m)-3*(C*(2+3*m 
)+A*(5+3*m))*hypergeom([1/2, 1/6-1/2*m],[7/6-1/2*m],cos(d*x+c)^2)*sec(d*x+ 
c)^(-1+m)*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(1-3*m)/(5+3*m)/(sin(d*x+c)^2) 
^(1/2)+3*B*hypergeom([1/2, -1/3-1/2*m],[2/3-1/2*m],cos(d*x+c)^2)*sec(d*x+c 
)^m*(b*sec(d*x+c))^(2/3)*sin(d*x+c)/d/(2+3*m)/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.90 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \left (A \left (40+39 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\sec ^2(c+d x)\right )+(2+3 m) \left (B (8+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\sec ^2(c+d x)\right )+C (5+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (8+3 m),\frac {7}{3}+\frac {m}{2},\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+3 m) (5+3 m) (8+3 m) \sqrt [3]{b \sec (c+d x)}} \] Input:

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

Output:

(3*b*Csc[c + d*x]*Sec[c + d*x]^m*(A*(40 + 39*m + 9*m^2)*Hypergeometric2F1[ 
1/2, (2 + 3*m)/6, (8 + 3*m)/6, Sec[c + d*x]^2] + (2 + 3*m)*(B*(8 + 3*m)*Co 
s[c + d*x]*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Sec[c + d*x]^ 
2] + C*(5 + 3*m)*Hypergeometric2F1[1/2, (8 + 3*m)/6, 7/3 + m/2, Sec[c + d* 
x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(2 + 3*m)*(5 + 3*m)*(8 + 
3*m)*(b*Sec[c + d*x])^(1/3))
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.98, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {2034, 3042, 4535, 3042, 4259, 3042, 3122, 4534, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\int \sec ^{m+\frac {2}{3}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^{m+\frac {5}{3}}(c+d x)dx\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {5}{3}}dx\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \cos ^{-m-\frac {5}{3}}(c+d x)dx\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {5}{3}}dx\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {3 B \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 4534

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\frac {(A (3 m+5)+C (3 m+2)) \int \sec ^{m+\frac {2}{3}}(c+d x)dx}{3 m+5}+\frac {3 B \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {5}{3}}(c+d x)}{d (3 m+5)}\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\frac {(A (3 m+5)+C (3 m+2)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx}{3 m+5}+\frac {3 B \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {5}{3}}(c+d x)}{d (3 m+5)}\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\frac {(A (3 m+5)+C (3 m+2)) \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \cos ^{-m-\frac {2}{3}}(c+d x)dx}{3 m+5}+\frac {3 B \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {5}{3}}(c+d x)}{d (3 m+5)}\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (\frac {(A (3 m+5)+C (3 m+2)) \cos ^{m+\frac {2}{3}}(c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {2}{3}}dx}{3 m+5}+\frac {3 B \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {5}{3}}(c+d x)}{d (3 m+5)}\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {(b \sec (c+d x))^{2/3} \left (-\frac {3 (A (3 m+5)+C (3 m+2)) \sin (c+d x) \sec ^{m-\frac {1}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (1-3 m),\frac {1}{6} (7-3 m),\cos ^2(c+d x)\right )}{d (1-3 m) (3 m+5) \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) \sec ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-3 m-2),\frac {1}{6} (4-3 m),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) \sec ^{m+\frac {5}{3}}(c+d x)}{d (3 m+5)}\right )}{\sec ^{\frac {2}{3}}(c+d x)}\)

Input:

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

Output:

((b*Sec[c + d*x])^(2/3)*((3*C*Sec[c + d*x]^(5/3 + m)*Sin[c + d*x])/(d*(5 + 
 3*m)) - (3*(C*(2 + 3*m) + A*(5 + 3*m))*Hypergeometric2F1[1/2, (1 - 3*m)/6 
, (7 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(-1/3 + m)*Sin[c + d*x])/(d*(1 
 - 3*m)*(5 + 3*m)*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric2F1[1/2, (-2 
- 3*m)/6, (4 - 3*m)/6, Cos[c + d*x]^2]*Sec[c + d*x]^(2/3 + m)*Sin[c + d*x] 
)/(d*(2 + 3*m)*Sqrt[Sin[c + d*x]^2])))/Sec[c + d*x]^(2/3)
 

Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

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 4534

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

rule 4535

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)*se 
c(d*x + c)^m, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:

int((b/cos(c + d*x))^(2/3)*(1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2),x)
 

Output:

int((b/cos(c + d*x))^(2/3)*(1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2), x)
 

Reduce [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{2/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=b^{\frac {2}{3}} \left (\left (\int \sec \left (d x +c \right )^{m +\frac {2}{3}}d x \right ) a +\left (\int \sec \left (d x +c \right )^{m +\frac {2}{3}} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sec \left (d x +c \right )^{m +\frac {2}{3}} \sec \left (d x +c \right )d x \right ) b \right ) \] Input:

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

Output:

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