\(\int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\) [918]

Optimal result

Integrand size = 29, antiderivative size = 141 \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {A b^3 (b \cos (c+d x))^{-3+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}}+\frac {b^2 B (b \cos (c+d x))^{-2+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84 \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=-\frac {(b \cos (c+d x))^n \csc (c+d x) \left (A (-2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\cos ^2(c+d x)\right )+B (-3+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\cos ^2(c+d x)\right )\right ) \sec ^3(c+d x) \sqrt {\sin ^2(c+d x)}}{d (-3+n) (-2+n)} \] Input:

Integrate[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
 

Output:

-(((b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(-2 + n)*Hypergeometric2F1[1/2, (-3 
+ n)/2, (-1 + n)/2, Cos[c + d*x]^2] + B*(-3 + n)*Cos[c + d*x]*Hypergeometr 
ic2F1[1/2, (-2 + n)/2, n/2, Cos[c + d*x]^2])*Sec[c + d*x]^3*Sqrt[Sin[c + d 
*x]^2])/(d*(-3 + n)*(-2 + n)))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 2030, 3227, 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.

\(\displaystyle \int \sec ^4(c+d x) (A+B \cos (c+d x)) (b \cos (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^n}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^4 \int \left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{n-4} \left (A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx\)

\(\Big \downarrow \) 3227

\(\displaystyle b^4 \left (A \int (b \cos (c+d x))^{n-4}dx+\frac {B \int (b \cos (c+d x))^{n-3}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (A \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{n-4}dx+\frac {B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{n-3}dx}{b}\right )\)

\(\Big \downarrow \) 3122

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

Input:

Int[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]
 

Output:

b^4*((A*(b*Cos[c + d*x])^(-3 + n)*Hypergeometric2F1[1/2, (-3 + n)/2, (-1 + 
 n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(3 - n)*Sqrt[Sin[c + d*x]^2]) + 
(B*(b*Cos[c + d*x])^(-2 + n)*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Cos[c 
 + d*x]^2]*Sin[c + d*x])/(b^2*d*(2 - 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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

Maple [F]

\[\int \left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{4}d x\]

Input:

int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)
 

Output:

int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)
 

Fricas [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fri 
cas")
 

Output:

integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="max 
ima")
 

Output:

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^4, x)
 

Giac [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="gia 
c")
 

Output:

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:

int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^4,x)
 

Output:

int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^4, x)
 

Reduce [F]

\[ \int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=b^{n} \left (\left (\int \cos \left (d x +c \right )^{n} \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) b +\left (\int \cos \left (d x +c \right )^{n} \sec \left (d x +c \right )^{4}d x \right ) a \right ) \] Input:

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)
 

Output:

b**n*(int(cos(c + d*x)**n*cos(c + d*x)*sec(c + d*x)**4,x)*b + int(cos(c + 
d*x)**n*sec(c + d*x)**4,x)*a)