\(\int \sec (c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [73]

Optimal result

Integrand size = 37, antiderivative size = 182 \[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(C (1+n)+A (2+n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n (2+n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {16, 4132, 3857, 2722, 4131} \[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(A (n+2)+C (n+1)) \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right )}{d n (n+2) \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) (b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-n-1),\frac {1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) \sqrt {\sin ^2(c+d x)}}+\frac {C \tan (c+d x) (b \sec (c+d x))^{n+1}}{b d (n+2)} \]

[In]

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

[Out]

((C*(1 + n) + A*(2 + n))*Hypergeometric2F1[1/2, -1/2*n, (2 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c +
d*x])/(d*n*(2 + n)*Sqrt[Sin[c + d*x]^2]) + (B*Hypergeometric2F1[1/2, (-1 - n)/2, (1 - n)/2, Cos[c + d*x]^2]*(b
*Sec[c + d*x])^(1 + n)*Sin[c + d*x])/(b*d*(1 + n)*Sqrt[Sin[c + d*x]^2]) + (C*(b*Sec[c + d*x])^(1 + n)*Tan[c +
d*x])/(b*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}& = \frac {\int (b \sec (c+d x))^{1+n} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx}{b} \\ & = \frac {\int (b \sec (c+d x))^{1+n} \left (A+C \sec ^2(c+d x)\right ) \, dx}{b}+\frac {B \int (b \sec (c+d x))^{2+n} \, dx}{b^2} \\ & = \frac {C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}+\frac {\left (A+\frac {C (1+n)}{2+n}\right ) \int (b \sec (c+d x))^{1+n} \, dx}{b}+\frac {\left (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}{b^2} \\ & = \frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)}+\frac {\left (\left (A+\frac {C (1+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 )^{-1-n} \, dx}{b} \\ & = \frac {\left (A+\frac {C (1+n)}{2+n}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^{1+n} \tan (c+d x)}{b d (2+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.92 \[ \int \sec (c+d x) (b \sec (c+d x))^n \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\csc (c+d x) (b \sec (c+d x))^n \left (A \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(c+d x)\right )+(1+n) \left (B (3+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(c+d x)\right )+C (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sec ^2(c+d x)\right )\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (1+n) (2+n) (3+n)} \]

[In]

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

[Out]

(Csc[c + d*x]*(b*Sec[c + d*x])^n*(A*(6 + 5*n + n^2)*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^
2] + (1 + n)*(B*(3 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sec[c + d*x]^2] + C*(2 + n)*
Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Sec[c + d*x]^2])*Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(d*(1 + n
)*(2 + n)*(3 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((b*sec(c + d*x))**n*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x), x)

Maxima [F]

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

[In]

integrate(sec(d*x+c)*(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*sec(d*x + c), x)

Giac [F]

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

[In]

integrate(sec(d*x+c)*(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*sec(d*x + c), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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