3.291 \(\int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx\)

Optimal. Leaf size=132 \[ \frac{a \sin (e+f x) \sec ^n(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt{\sin ^2(e+f x)}}-\frac{a \sin (e+f x) \sec ^{n-1}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) \sqrt{\sin ^2(e+f x)}} \]

[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, -n/2, (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])

________________________________________________________________________________________

Rubi [A]  time = 0.0865364, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3787, 3772, 2643} \[ \frac{a \sin (e+f x) \sec ^n(e+f x) \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(e+f x)\right )}{f n \sqrt{\sin ^2(e+f x)}}-\frac{a \sin (e+f x) \sec ^{n-1}(e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) \sqrt{\sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[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, -n/2, (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 3787

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]

Rule 3772

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \sec ^n(e+f x) (a+a \sec (e+f x)) \, dx &=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 \, _2F_1\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 \, _2F_1\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]  time = 0.146162, size = 106, normalized size = 0.8 \[ \frac{a \sqrt{-\tan ^2(e+f x)} \csc (e+f x) \sec ^{n-1}(e+f x) \left ((n+1) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n}{2},\frac{n+2}{2},\sec ^2(e+f x)\right )+n \sec (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\sec ^2(e+f x)\right )\right )}{f n (n+1)} \]

Antiderivative was successfully verified.

[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]  time = 0.508, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sec \left ( fx+e \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (f x + e\right ) + a\right )} \sec \left (f x + e\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)