\(\int (b \cos (c+d x))^n (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [186]

Optimal result

Integrand size = 29, antiderivative size = 100 \[ \int (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {C (b \cos (c+d x))^n \sin (c+d x)}{d (1+n)}-\frac {(A+A n+C n) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d n (1+n) \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.11 \[ \int (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {b (b \cos (c+d x))^{-1+n} \cot (c+d x) \left (A (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\cos ^2(c+d x)\right )+C n \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d n (2+n)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 2030, 3493, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

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 3493

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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