\(\int \frac {(b \sec (c+d x))^n (A+C \sec ^2(c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [37]

Optimal result

Integrand size = 33, antiderivative size = 141 \[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 C \sqrt {\sec (c+d x)} (b \sec (c+d x))^n \sin (c+d x)}{d (1+2 n)}-\frac {2 (A-C (1-2 n)+2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3-2 n),\frac {1}{4} (7-2 n),\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (1+2 n) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01 \[ \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx=\frac {2 \csc (c+d x) (b \sec (c+d x))^n \left (A (3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (-1+2 n),\frac {1}{4} (3+2 n),\sec ^2(c+d x)\right )+C (-1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-1+2 n) (3+2 n) \sec ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 146, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorith 
m="fricas")
 

Output:

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

Sympy [F]

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

integrate((b*sec(d*x+c))**n*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorith 
m="maxima")
 

Output:

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

Giac [F]

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

integrate((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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