\(\int \cos (c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [31]

Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {b^2 (C (1-n)-A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2-n}{2},\frac {4-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-2+n} \sin (c+d x)}{d (2-n) n \sqrt {\sin ^2(c+d x)}}+\frac {b C (b \sec (c+d x))^{-1+n} \tan (c+d x)}{d n} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02 \[ \int \cos (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (A (1+n) \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\sec ^2(c+d x)\right )+C (-1+n) \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d (-1+n) (1+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 2030, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="fri 
cas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="max 
ima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="gia 
c")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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