3.1.34 \(\int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [34]

3.1.34.1 Optimal result

Integrand size = 33, antiderivative size = 142 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 C \sec ^{\frac {7}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (7+2 n)}+\frac {2 (C (5+2 n)+A (7+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-3-2 n),\frac {1}{4} (1-2 n),\cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \sin (c+d x)}{d (3+2 n) (7+2 n) \sqrt {\sin ^2(c+d x)}} \]

output

2*C*sec(d*x+c)^(7/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(7+2*n)+2*(C*(5+2*n)+A* 
(7+2*n))*hypergeom([1/2, -3/4-1/2*n],[1/4-1/2*n],cos(d*x+c)^2)*sec(d*x+c)^ 
(3/2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(4*n^2+20*n+21)/(sin(d*x+c)^2)^(1/2)
 

3.1.34.2 Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00 \[ \int \sec ^{\frac {5}{2}}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \csc (c+d x) \sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^n \left (A (9+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\sec ^2(c+d x)\right )+C (5+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (9+2 n),\frac {1}{4} (13+2 n),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (5+2 n) (9+2 n)} \]

input

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

output

(2*Csc[c + d*x]*Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^n*(A*(9 + 2*n)*Hyperge 
ometric2F1[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Sec[c + d*x]^2] + C*(5 + 2*n)*Hy 
pergeometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Sec[c + d*x]^2]*Sec[c + d* 
x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(5 + 2*n)*(9 + 2*n))
 

3.1.34.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2034, 3042, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \int \sec ^{n+\frac {5}{2}}(c+d x)dx}{2 n+7}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{n+\frac {5}{2}}dx}{2 n+7}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\)

\(\Big \downarrow \) 4259

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \cos ^{-n-\frac {5}{2}}(c+d x)dx}{2 n+7}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sec ^{-n}(c+d x) (b \sec (c+d x))^n \left (\frac {(A (2 n+7)+C (2 n+5)) \cos ^{n+\frac {1}{2}}(c+d x) \sec ^{n+\frac {1}{2}}(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{-n-\frac {5}{2}}dx}{2 n+7}+\frac {2 C \sin (c+d x) \sec ^{n+\frac {7}{2}}(c+d x)}{d (2 n+7)}\right )\)

\(\Big \downarrow \) 3122

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

input

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

output

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

3.1.34.3.1 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]
 

3.1.34.4 Maple [F]

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

input

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

output

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

3.1.34.5 Fricas [F]

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

input

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

output

integral((C*sec(d*x + c)^4 + A*sec(d*x + c)^2)*(b*sec(d*x + c))^n*sqrt(sec 
(d*x + c)), x)
 

3.1.34.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.34.7 Maxima [F]

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

input

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

output

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

3.1.34.8 Giac [F]

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

input

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

output

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

3.1.34.9 Mupad [F(-1)]

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

input

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

output

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