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

Optimal result

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

[Out]

-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*hypergeom([1/2, 3/2-1/2*n],[5/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-3+n)
*sin(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)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {16, 4132, 3857, 2722, 4131} \[ \int \cos ^3(c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {b^4 (A (2-n)+C (3-n)) \sin (c+d x) (b \sec (c+d x))^{n-4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-n}{2},\frac {6-n}{2},\cos ^2(c+d x)\right )}{d (2-n) (4-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 B \sin (c+d x) (b \sec (c+d x))^{n-3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-n}{2},\frac {5-n}{2},\cos ^2(c+d x)\right )}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {b^3 C \tan (c+d x) (b \sec (c+d x))^{n-3}}{d (2-n)} \]

[In]

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]

[Out]

-((b^4*(A*(2 - n) + C*(3 - 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*(2 - n)*(4 - n)*Sqrt[Sin[c + d*x]^2])) - (b^3*B*Hypergeometric2F1[1/2, (3 - n)/2, (5
- n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-3 + n)*Sin[c + d*x])/(d*(3 - n)*Sqrt[Sin[c + d*x]^2]) - (b^3*C*(b*S
ec[c + d*x])^(-3 + n)*Tan[c + d*x])/(d*(2 - n))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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] + Dist[(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 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[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]

Rubi steps \begin{align*} \text {integral}& = b^3 \int (b \sec (c+d x))^{-3+n} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx \\ & = b^3 \int (b \sec (c+d x))^{-3+n} \left (A+C \sec ^2(c+d x)\right ) \, dx+\left (b^2 B\right ) \int (b \sec (c+d x))^{-2+n} \, dx \\ & = -\frac {b^3 C (b \sec (c+d x))^{-3+n} \tan (c+d x)}{d (2-n)}+\left (b^3 \left (A+\frac {C (3-n)}{2-n}\right )\right ) \int (b \sec (c+d x))^{-3+n} \, dx+\left (b^2 B \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{2-n} \, dx \\ & = -\frac {B \cos ^3(c+d x) \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))^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)}+\left (b^3 \left (A+\frac {C (3-n)}{2-n}\right ) \left (\frac {\cos (c+d x)}{b}\right )^n (b \sec (c+d x))^n\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{3-n} \, dx \\ & = -\frac {B \cos ^3(c+d x) \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))^n \sin (c+d x)}{d (3-n) \sqrt {\sin ^2(c+d x)}}-\frac {\left (A+\frac {C (3-n)}{2-n}\right ) \cos ^4(c+d x) \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))^n \sin (c+d x)}{d (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)} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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]

[Out]

(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)*Hy
pergeometric2F1[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))

Maple [F]

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

[In]

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)

[Out]

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)

Fricas [F]

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

[In]

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")

[Out]

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)

Sympy [F(-1)]

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

[In]

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)

[Out]

Timed out

Maxima [F]

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

[In]

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")

[Out]

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)

Giac [F]

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

[In]

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")

[Out]

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)

Mupad [F(-1)]

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

[In]

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)

[Out]

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)