\(\int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [34]

Optimal result

Integrand size = 29, antiderivative size = 143 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-n),\frac {1-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{1+n} \sin (c+d x)}{b d (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2-n),-\frac {n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2+n} \sin (c+d x)}{b^2 d (2+n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {16, 3872, 3857, 2722} \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {A \sin (c+d x) (b \sec (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-n-1),\frac {1-n}{2},\cos ^2(c+d x)\right )}{b d (n+1) \sqrt {\sin ^2(c+d x)}}+\frac {B \sin (c+d x) (b \sec (c+d x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-n-2),-\frac {n}{2},\cos ^2(c+d x)\right )}{b^2 d (n+2) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

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

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 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (b \sec (c+d x))^{2+n} (A+B \sec (c+d x)) \, dx}{b^2} \\ & = \frac {A \int (b \sec (c+d x))^{2+n} \, dx}{b^2}+\frac {B \int (b \sec (c+d x))^{3+n} \, dx}{b^3} \\ & = \frac {\left (A \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 {\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 )^{-3-n} \, dx}{b^3} \\ & = \frac {A \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 {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2-n),-\frac {n}{2},\cos ^2(c+d x)\right ) \sec (c+d x) (b \sec (c+d x))^n \tan (c+d x)}{d (2+n) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \sec (c+d x) (b \sec (c+d x))^n \left (A (3+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sec ^2(c+d x)\right )+B (2+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (2+n) (3+n)} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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