3.1.77 \(\int \cos ^3(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [77]

3.1.77.1 Optimal result

Integrand size = 39, antiderivative size = 208 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^4 (A (2-n)+C (3-n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-4+n} \sin (c+d x)}{d (2-n) (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-3+n} \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)} \]

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*B*hy 
pergeom([1/2, 3/2-1/2*n],[5/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-3+n)*s 
in(d*x+c)/d/(3-n)/(sin(d*x+c)^2)^(1/2)-b^3*C*(b*sec(d*x+c))^(-3+n)*tan(d*x 
+c)/d/(2-n)
 

3.1.77.2 Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.81 \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {b \cot (c+d x) \left (A \left (2-3 n+n^2\right ) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+n),\frac {1}{2} (-1+n),\sec ^2(c+d x)\right )+(-3+n) \left (B (-1+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+n),\frac {n}{2},\sec ^2(c+d x)\right )+C (-2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n),\frac {1+n}{2},\sec ^2(c+d x)\right )\right )\right ) (b \sec (c+d x))^{-1+n} \sqrt {-\tan ^2(c+d x)}}{d (-3+n) (-2+n) (-1+n)} \]

input

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

output

(b*Cot[c + d*x]*(A*(2 - 3*n + n^2)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, ( 
-3 + n)/2, (-1 + n)/2, Sec[c + d*x]^2] + (-3 + n)*(B*(-1 + n)*Cos[c + d*x] 
*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Sec[c + d*x]^2] + C*(-2 + n)*Hype 
rgeometric2F1[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)*(-2 + n)*(-1 + n))
 

3.1.77.3 Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 2030, 4535, 3042, 4259, 3042, 3122, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4259

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle b^3 \left (\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}\right )\)

\(\Big \downarrow \) 4534

\(\displaystyle b^3 \left (\left (A+\frac {C (3-n)}{2-n}\right ) \int (b \sec (c+d x))^{n-3}dx-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (\left (A+\frac {C (3-n)}{2-n}\right ) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{n-3}dx-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

\(\Big \downarrow \) 4259

\(\displaystyle b^3 \left (\left (A+\frac {C (3-n)}{2-n}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\cos (c+d x)}{b}\right )^{3-n}dx-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (\left (A+\frac {C (3-n)}{2-n}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{3-n}dx-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

\(\Big \downarrow \) 3122

\(\displaystyle b^3 \left (-\frac {b \left (A+\frac {C (3-n)}{2-n}\right ) \sin (c+d x) (b \sec (c+d x))^{n-4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right )}{d (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)}\right )\)

input

Int[Cos[c + d*x]^3*(b*Sec[c + d*x])^n*(A + B*Sec[c + d*x] + 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])) - (B*Hypergeometric2F1[1/2, (3 - n)/2, (5 - n)/2, C 
os[c + d*x]^2]*(b*Sec[c + d*x])^(-3 + n)*Sin[c + d*x])/(d*(3 - n)*Sqrt[Sin 
[c + d*x]^2]) - (C*(b*Sec[c + d*x])^(-3 + n)*Tan[c + d*x])/(d*(2 - n)))
 

3.1.77.3.1 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]
 

rule 4535

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

3.1.77.4 Maple [F]

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

input

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

output

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

3.1.77.5 Fricas [F]

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

input

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

output

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

3.1.77.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.77.7 Maxima [F]

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

input

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

output

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

3.1.77.8 Giac [F]

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

input

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

output

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

3.1.77.9 Mupad [F(-1)]

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

input

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

output

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