3.1.33 \(\int \cos ^3(c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [33]

3.1.33.1 Optimal result

Integrand size = 31, antiderivative size = 132 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^4 (A (2-n)+C (3-n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-4+n} \sin (c+d x)}{d (2-n) (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)} \]

output

-b^4*(A*(2-n)+C*(3-n))*hypergeom([1/2, 2-1/2*n],[3-1/2*n],cos(d*x+c)^2)*(b 
*sec(d*x+c))^(-4+n)*sin(d*x+c)/d/(n^2-6*n+8)/(sin(d*x+c)^2)^(1/2)-b^3*C*(b 
*sec(d*x+c))^(-3+n)*tan(d*x+c)/d/(2-n)
 

3.1.33.2 Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {b \cot (c+d x) \left (A (-1+n) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\sec ^2(c+d x)\right )+C (-3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{-1+n} \sqrt {-\tan ^2(c+d x)}}{d (-3+n) (-1+n)} \]

input

Integrate[Cos[c + d*x]^3*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]
 

output

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

3.1.33.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, 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.

\(\displaystyle \int \cos ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) (b \sec (c+d x))^n \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b^3 \int \left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{n-3} \left (C \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A\right )dx\)

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (\left (A+\frac {C (3-n)}{2-n}\right ) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-3}dx-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b^3 \left (-\frac {b \left (A+\frac {C (3-n)}{2-n}\right ) \sin (c+d x) (b \sec (c+d x))^{n-4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right )}{d (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

input

Int[Cos[c + d*x]^3*(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]
 

output

b^3*(-((b*(A + (C*(3 - n))/(2 - n))*Hypergeometric2F1[1/2, (4 - n)/2, (6 - 
 n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-4 + n)*Sin[c + d*x])/(d*(4 - n)* 
Sqrt[Sin[c + d*x]^2])) - (C*(b*Sec[c + d*x])^(-3 + n)*Tan[c + d*x])/(d*(2 
- n)))
 

3.1.33.3.1 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 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]
 

3.1.33.4 Maple [F]

\[\int \cos \left (d x +c \right )^{3} \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]

input

int(cos(d*x+c)^3*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)
 

output

int(cos(d*x+c)^3*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)
 

3.1.33.5 Fricas [F]

\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{3} \,d x } \]

input

integrate(cos(d*x+c)^3*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="f 
ricas")
 

output

integral((C*cos(d*x + c)^3*sec(d*x + c)^2 + A*cos(d*x + c)^3)*(b*sec(d*x + 
 c))^n, x)
 

3.1.33.6 Sympy [F]

\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \]

input

integrate(cos(d*x+c)**3*(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)**3, x)
 

3.1.33.7 Maxima [F]

\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{3} \,d x } \]

input

integrate(cos(d*x+c)^3*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="m 
axima")
 

output

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*cos(d*x + c)^3, x)
 

3.1.33.8 Giac [F]

\[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{3} \,d x } \]

input

integrate(cos(d*x+c)^3*(b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="g 
iac")
 

output

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n*cos(d*x + c)^3, x)
 

3.1.33.9 Mupad [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^3\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

input

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n,x)
 

output

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n, x)