\(\int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [65]

Optimal result

Integrand size = 41, antiderivative size = 230 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b C \sec ^{2+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {3 b (C (4+3 m)+A (7+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1-3 m),\frac {1}{6} (5-3 m),\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (1+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-4-3 m),\frac {1}{6} (2-3 m),\cos ^2(c+d x)\right ) \sec ^{1+m}(c+d x) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d (4+3 m) \sqrt {\sin ^2(c+d x)}} \] Output:

3*b*C*sec(d*x+c)^(2+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(7+3*m)+3*b*(C*(4 
+3*m)+A*(7+3*m))*hypergeom([1/2, -1/6-1/2*m],[5/6-1/2*m],cos(d*x+c)^2)*sec 
(d*x+c)^m*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(1+3*m)/(7+3*m)/(sin(d*x+c)^2) 
^(1/2)+3*b*B*hypergeom([1/2, -2/3-1/2*m],[1/3-1/2*m],cos(d*x+c)^2)*sec(d*x 
+c)^(1+m)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(4+3*m)/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.89 \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b \csc (c+d x) \sec ^m(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A \left (70+51 m+9 m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+(4+3 m) \left (C (7+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\sec ^2(c+d x)\right )+B (10+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (4+3 m) (7+3 m) (10+3 m)} \] Input:

Integrate[Sec[c + d*x]^m*(b*Sec[c + d*x])^(4/3)*(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*(b*Sec[c + d*x])^(1/3)*(A*(70 + 51*m + 9* 
m^2)*Hypergeometric2F1[1/2, (4 + 3*m)/6, 5/3 + m/2, Sec[c + d*x]^2] + (4 + 
 3*m)*(C*(7 + 3*m)*Hypergeometric2F1[1/2, 5/3 + m/2, 8/3 + m/2, Sec[c + d* 
x]^2] + B*(10 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 
+ 3*m)/6, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(4 + 
3*m)*(7 + 3*m)*(10 + 3*m))
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.97, 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.

\(\displaystyle \int (b \sec (c+d x))^{4/3} \sec ^m(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \int \sec ^{m+\frac {4}{3}}(c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 4535

\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\int \sec ^{m+\frac {4}{3}}(c+d x) \left (C \sec ^2(c+d x)+A\right )dx+B \int \sec ^{m+\frac {7}{3}}(c+d x)dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {7}{3}}dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 4259

\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \cos ^{-m-\frac {7}{3}}(c+d x)dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b \sqrt [3]{b \sec (c+d x)} \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{m+\frac {1}{3}}(c+d x) \sec ^{m+\frac {1}{3}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-m-\frac {7}{3}}dx\right )}{\sqrt [3]{\sec (c+d x)}}\)

\(\Big \downarrow \) 3122

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

Int[Sec[c + d*x]^m*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + 
d*x]^2),x]
 

Output:

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

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]

\[\int \sec \left (d x +c \right )^{m} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]

Input:

int(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(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))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Fricas [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 

Output:

integral((C*b*sec(d*x + c)^3 + B*b*sec(d*x + c)^2 + A*b*sec(d*x + c))*(b*s 
ec(d*x + c))^(1/3)*sec(d*x + c)^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**m*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c) 
**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(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))^(4/3)*s 
ec(d*x + c)^m, x)
 

Giac [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{m} \,d x } \] Input:

integrate(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(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))^(4/3)*s 
ec(d*x + c)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \] Input:

int((b/cos(c + d*x))^(4/3)*(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))^(4/3)*(1/cos(c + d*x))^m*(A + B/cos(c + d*x) + C/cos( 
c + d*x)^2), x)
 

Reduce [F]

\[ \int \sec ^m(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=b^{\frac {4}{3}} \left (\left (\int \sec \left (d x +c \right )^{m +\frac {1}{3}} \sec \left (d x +c \right )^{3}d x \right ) c +\left (\int \sec \left (d x +c \right )^{m +\frac {1}{3}} \sec \left (d x +c \right )^{2}d x \right ) b +\left (\int \sec \left (d x +c \right )^{m +\frac {1}{3}} \sec \left (d x +c \right )d x \right ) a \right ) \] Input:

int(sec(d*x+c)^m*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
                                                                                    
                                                                                    
 

Output:

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