\(\int (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx\) [36]

Optimal result

Integrand size = 21, antiderivative size = 137 \[ \int (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=-\frac {A b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-1+n} \sin (c+d x)}{d (1-n) \sqrt {\sin ^2(c+d x)}}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3872, 3857, 2722} \[ \int (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {B \sin (c+d x) (b \sec (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}-\frac {A b \sin (c+d x) (b \sec (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.78 \[ \int (b \sec (c+d x))^n (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \left (A (1+n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(c+d x)\right )+B n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^n \sqrt {-\tan ^2(c+d x)}}{d n (1+n)} \]

[In]

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

[Out]

(Csc[c + d*x]*(A*(1 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[c + d*x]^2] + B*n*Hypergeomet
ric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[c + d*x]^2])*(b*Sec[c + d*x])^n*Sqrt[-Tan[c + d*x]^2])/(d*n*(1 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((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)), x)

Maxima [F]

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

[In]

integrate((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, x)

Giac [F]

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

[In]

integrate((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, x)

Mupad [F(-1)]

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

[In]

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

[Out]

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