\(\int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx\) [291]

Optimal result

Integrand size = 19, antiderivative size = 132 \[ \int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx=-\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) \sin (e+f x)}{f (1-n) \sqrt {\sin ^2(e+f x)}}+\frac {a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right ) \sec ^n(e+f x) \sin (e+f x)}{f n \sqrt {\sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3872, 3857, 2722} \[ \int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx=\frac {a \sin (e+f x) \sec ^n(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}-\frac {a \sin (e+f x) \sec ^{n-1}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]

[In]

Int[Sec[e + f*x]^n*(a + a*Sec[e + f*x]),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.80 \[ \int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx=\frac {a \csc (e+f x) \sec ^{-1+n}(e+f x) \left ((1+n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(e+f x)\right )+n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(e+f x)\right ) \sec (e+f x)\right ) \sqrt {-\tan ^2(e+f x)}}{f n (1+n)} \]

[In]

Integrate[Sec[e + f*x]^n*(a + a*Sec[e + f*x]),x]

[Out]

(a*Csc[e + f*x]*Sec[e + f*x]^(-1 + n)*((1 + n)*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Sec[e + f*x]^2] + n*Hype
rgeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sec[e + f*x]^2]*Sec[e + f*x])*Sqrt[-Tan[e + f*x]^2])/(f*n*(1 + n))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((a*sec(f*x + e) + a)*sec(f*x + e)^n, x)

Sympy [F]

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

[In]

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

[Out]

a*(Integral(sec(e + f*x)*sec(e + f*x)**n, x) + Integral(sec(e + f*x)**n, x))

Maxima [F]

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

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)*sec(f*x + e)^n, x)

Giac [F]

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

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)*sec(f*x + e)^n, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx=\int \left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]

[In]

int((a + a/cos(e + f*x))*(1/cos(e + f*x))^n,x)

[Out]

int((a + a/cos(e + f*x))*(1/cos(e + f*x))^n, x)