\(\int (b \sec (e+f x))^n \, dx\) [500]

Optimal result

Integrand size = 10, antiderivative size = 73 \[ \int (b \sec (e+f x))^n \, dx=-\frac {b \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin (e+f x)}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3857, 2722} \[ \int (b \sec (e+f x))^n \, dx=-\frac {b \sin (e+f x) (b \sec (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]

[In]

Int[(b*Sec[e + f*x])^n,x]

[Out]

-((b*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*x])^(-1 + n)*Sin[e + f*x])/(f*(
1 - n)*Sqrt[Sin[e + f*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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int (b \sec (e+f x))^n \, dx=\frac {\cot (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(e+f x)\right ) (b \sec (e+f x))^n \sqrt {-\tan ^2(e+f x)}}{f n} \]

[In]

Integrate[(b*Sec[e + f*x])^n,x]

[Out]

(Cot[e + f*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[e + f*x]^2]*(b*Sec[e + f*x])^n*Sqrt[-Tan[e + f*x]^2])
/(f*n)

Maple [F]

\[\int \left (b \sec \left (f x +e \right )\right )^{n}d x\]

[In]

int((b*sec(f*x+e))^n,x)

[Out]

int((b*sec(f*x+e))^n,x)

Fricas [F]

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

[In]

integrate((b*sec(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e))^n, x)

Sympy [F]

\[ \int (b \sec (e+f x))^n \, dx=\int \left (b \sec {\left (e + f x \right )}\right )^{n}\, dx \]

[In]

integrate((b*sec(f*x+e))**n,x)

[Out]

Integral((b*sec(e + f*x))**n, x)

Maxima [F]

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

[In]

integrate((b*sec(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^n, x)

Giac [F]

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

[In]

integrate((b*sec(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^n, x)

Mupad [F(-1)]

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

[In]

int((b/cos(e + f*x))^n,x)

[Out]

int((b/cos(e + f*x))^n, x)