\(\int \frac {(b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [83]

Optimal result

Integrand size = 41, antiderivative size = 223 \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (A (3-2 n)+C (5-2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7-2 n),\frac {1}{4} (11-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (7-2 n) \sec ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (5-2 n) \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \] Output:

-2*C*(b*sec(d*x+c))^n*sin(d*x+c)/d/(3-2*n)/sec(d*x+c)^(3/2)-2*(A*(3-2*n)+C 
*(5-2*n))*hypergeom([1/2, 7/4-1/2*n],[11/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x 
+c))^n*sin(d*x+c)/d/(3-2*n)/(7-2*n)/sec(d*x+c)^(7/2)/(sin(d*x+c)^2)^(1/2)- 
2*B*hypergeom([1/2, 5/4-1/2*n],[9/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n* 
sin(d*x+c)/d/(5-2*n)/sec(d*x+c)^(5/2)/(sin(d*x+c)^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.91 \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A \left (3-8 n+4 n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-5+2 n),\frac {1}{4} (-1+2 n),\sec ^2(c+d x)\right )+(-5+2 n) \left (B (-1+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3+2 n),\frac {1}{4} (1+2 n),\sec ^2(c+d x)\right )+C (-3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-5+2 n) (-3+2 n) (-1+2 n) \sec ^{\frac {7}{2}}(c+d x)} \] Input:

Integrate[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec 
[c + d*x]^(5/2),x]
 

Output:

(2*Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(3 - 8*n + 4*n^2)*Hypergeometric2F1[ 
1/2, (-5 + 2*n)/4, (-1 + 2*n)/4, Sec[c + d*x]^2] + (-5 + 2*n)*(B*(-1 + 2*n 
)*Cos[c + d*x]*Hypergeometric2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Sec[c + d 
*x]^2] + C*(-3 + 2*n)*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Se 
c[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(-5 + 2*n)*(-3 + 
2*n)*(-1 + 2*n)*Sec[c + d*x]^(7/2))
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {3}{2}-n}(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+B \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {3}{2}-n}dx\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {2 B \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}\right )\)

\(\Big \downarrow \) 4534

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \int \sec ^{n-\frac {5}{2}}(c+d x)dx}{3-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n-\frac {5}{2}}dx}{3-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\)

\(\Big \downarrow \) 4259

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{\frac {5}{2}-n}(c+d x)dx}{3-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (3-2 n)+C (5-2 n)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{\frac {5}{2}-n}dx}{3-2 n}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (-\frac {2 (A (3-2 n)+C (5-2 n)) \sin (c+d x) \sec ^{n-\frac {7}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (7-2 n),\frac {1}{4} (11-2 n),\cos ^2(c+d x)\right )}{d (3-2 n) (7-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \sec ^{n-\frac {5}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5-2 n),\frac {1}{4} (9-2 n),\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \sec ^{n-\frac {3}{2}}(c+d x)}{d (3-2 n)}\right )\)

Input:

Int[((b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d 
*x]^(5/2),x]
 

Output:

((b*Sec[c + d*x])^n*((-2*C*Sec[c + d*x]^(-3/2 + n)*Sin[c + d*x])/(d*(3 - 2 
*n)) - (2*(A*(3 - 2*n) + C*(5 - 2*n))*Hypergeometric2F1[1/2, (7 - 2*n)/4, 
(11 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(-7/2 + n)*Sin[c + d*x])/(d*(3 
- 2*n)*(7 - 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Hypergeometric2F1[1/2, (5 - 
2*n)/4, (9 - 2*n)/4, Cos[c + d*x]^2]*Sec[c + d*x]^(-5/2 + n)*Sin[c + d*x]) 
/(d*(5 - 2*n)*Sqrt[Sin[c + d*x]^2])))/Sec[c + d*x]^n
 

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]

\[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\sec \left (d x +c \right )^{\frac {5}{2}}}d x\]

Input:

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)
 

Output:

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)
 

Fricas [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2 
),x, algorithm="fricas")
 

Output:

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sec(d* 
x + c)^(5/2), x)
 

Sympy [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((b*sec(d*x+c))**n*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**( 
5/2),x)
 

Output:

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)/sec( 
c + d*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2 
),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sec(d 
*x + c)^(5/2), x)
 

Giac [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{n}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2 
),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^n/sec(d 
*x + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(((b/cos(c + d*x))^n*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c 
+ d*x))^(5/2),x)
 

Output:

int(((b/cos(c + d*x))^n*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c 
+ d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=b^{n} \left (\left (\int \frac {\sec \left (d x +c \right )^{n +\frac {1}{2}}}{\sec \left (d x +c \right )^{3}}d x \right ) a +\left (\int \frac {\sec \left (d x +c \right )^{n +\frac {1}{2}}}{\sec \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sec \left (d x +c \right )^{n +\frac {1}{2}}}{\sec \left (d x +c \right )}d x \right ) c \right ) \] Input:

int((b*sec(d*x+c))^n*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

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