3.3.23 \(\int (b \cos (c+d x))^n (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [223]

3.3.23.1 Optimal result

Integrand size = 38, antiderivative size = 139 \[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {b^2 B (b \cos (c+d x))^{-2+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2-n) \sqrt {\sin ^2(c+d x)}}+\frac {b C (b \cos (c+d x))^{-1+n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

output

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

3.3.23.2 Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.85 \[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {(b \cos (c+d x))^n \csc (c+d x) \left (B (-1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\cos ^2(c+d x)\right )+C (-2+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\cos ^2(c+d x)\right )\right ) \sec ^2(c+d x) \sqrt {\sin ^2(c+d x)}}{d (-2+n) (-1+n)} \]

input

Integrate[(b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d 
*x]^4,x]
 

output

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

3.3.23.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 2030, 3489, 3042, 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.

\(\displaystyle \int \sec ^4(c+d x) (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3489

\(\displaystyle b^3 \int (b \cos (c+d x))^{n-3} (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle b^3 \left (B \int (b \cos (c+d x))^{n-3}dx+\frac {C \int (b \cos (c+d x))^{n-2}dx}{b}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

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

input

Int[(b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4, 
x]
 

output

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

3.3.23.3.1 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]
 

rule 3489

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + 
(C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b   Int[(b*Sin[e + f* 
x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
 

3.3.23.4 Maple [F]

\[\int \left (\cos \left (d x +c \right ) b \right )^{n} \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{4}\left (d x +c \right )\right )d x\]

input

int((cos(d*x+c)*b)^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

output

int((cos(d*x+c)*b)^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

3.3.23.5 Fricas [F]

\[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \]

input

integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, a 
lgorithm="fricas")
 

output

integral((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n*sec(d*x + 
c)^4, x)
 

3.3.23.6 Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]

input

integrate((b*cos(d*x+c))**n*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x 
)
 

output

Timed out
 

3.3.23.7 Maxima [F]

\[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \]

input

integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, a 
lgorithm="maxima")
 

output

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n*sec(d*x + 
 c)^4, x)
 

3.3.23.8 Giac [F]

\[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4} \,d x } \]

input

integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, a 
lgorithm="giac")
 

output

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n*sec(d*x + 
 c)^4, x)
 

3.3.23.9 Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^4} \,d x \]

input

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

output

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