3.296 \(\int \sec ^n(e+f x) \sqrt{1+\sec (e+f x)} \, dx\)

Optimal. Leaf size=45 \[ \frac{2 \tan (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},1-\sec (e+f x)\right )}{f \sqrt{\sec (e+f x)+1}} \]

[Out]

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

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Rubi [A]  time = 0.0517975, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3806, 65} \[ \frac{2 \tan (e+f x) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1-\sec (e+f x)\right )}{f \sqrt{\sec (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^n*Sqrt[1 + Sec[e + f*x]],x]

[Out]

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

Rule 3806

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(a^2*d*
Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[(d*x)^(n - 1)/Sqrt[a - b*x], x]
, x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \sec ^n(e+f x) \sqrt{1+\sec (e+f x)} \, dx &=-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{x^{-1+n}}{\sqrt{1-x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{1-\sec (e+f x)} \sqrt{1+\sec (e+f x)}}\\ &=\frac{2 \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1-\sec (e+f x)\right ) \tan (e+f x)}{f \sqrt{1+\sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0372123, size = 45, normalized size = 1. \[ \frac{2 \tan (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},1-\sec (e+f x)\right )}{f \sqrt{\sec (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^n*Sqrt[1 + Sec[e + f*x]],x]

[Out]

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

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Maple [F]  time = 0.178, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{n}\sqrt{1+\sec \left ( fx+e \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^n*(1+sec(f*x+e))^(1/2),x)

[Out]

int(sec(f*x+e)^n*(1+sec(f*x+e))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (f x + e\right )^{n} \sqrt{\sec \left (f x + e\right ) + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n*(1+sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^n*sqrt(sec(f*x + e) + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sec \left (f x + e\right )^{n} \sqrt{\sec \left (f x + e\right ) + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n*(1+sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sec(f*x + e)^n*sqrt(sec(f*x + e) + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\sec{\left (e + f x \right )} + 1} \sec ^{n}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**n*(1+sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(sec(e + f*x) + 1)*sec(e + f*x)**n, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (f x + e\right )^{n} \sqrt{\sec \left (f x + e\right ) + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^n*(1+sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^n*sqrt(sec(f*x + e) + 1), x)