\(\int (b \cos (c+d x))^n (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [187]

Optimal result

Integrand size = 31, antiderivative size = 112 \[ \int (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}-\frac {b (C (1-n)-A n) (b \cos (c+d x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) n \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int (b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {b (b \cos (c+d x))^{-1+n} \csc (c+d x) \left (A (1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right )+C (-1+n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-1+n) (1+n)} \] Input:

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

Output:

-((b*(b*Cos[c + d*x])^(-1 + n)*Csc[c + d*x]*(A*(1 + n)*Hypergeometric2F1[1 
/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2] + C*(-1 + n)*Cos[c + d*x]^2*Hyp 
ergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[c + d*x]^2])*Sqrt[Sin[c + d* 
x]^2])/(d*(-1 + n)*(1 + n)))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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