3.308 \(\int \sec ^n(e+f x) (a+a \sec (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=108 \[ \frac{2 a^2 (4 n+1) \tan (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},1-n,\frac{3}{2},1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^2*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(1 + 4*n)*Hypergeome
tric2F1[1/2, 1 - n, 3/2, 1 - Sec[e + f*x]]*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]])

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Rubi [A]  time = 0.144604, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3814, 21, 3806, 65} \[ \frac{2 a^2 (4 n+1) \tan (e+f x) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^n*(a + a*Sec[e + f*x])^(3/2),x]

[Out]

(2*a^2*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(1 + 4*n)*Hypergeome
tric2F1[1/2, 1 - n, 3/2, 1 - Sec[e + f*x]]*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]])

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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) (a+a \sec (e+f x))^{3/2} \, dx &=\frac{2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{(2 a) \int \frac{\sec ^n(e+f x) \left (a \left (\frac{1}{2}+2 n\right )+a \left (\frac{1}{2}+2 n\right ) \sec (e+f x)\right )}{\sqrt{a+a \sec (e+f x)}} \, dx}{1+2 n}\\ &=\frac{2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{(a (1+4 n)) \int \sec ^n(e+f x) \sqrt{a+a \sec (e+f x)} \, dx}{1+2 n}\\ &=\frac{2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^3 (1+4 n) \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^{-1+n}}{\sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{f (1+2 n) \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (1+4 n) \, _2F_1\left (\frac{1}{2},1-n;\frac{3}{2};1-\sec (e+f x)\right ) \tan (e+f x)}{f (1+2 n) \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.362795, size = 86, normalized size = 0.8 \[ \frac{a \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sec (e+f x)+1)} \sec ^n(e+f x) \left ((4 n+1) \cos ^{n+\frac{1}{2}}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},n+\frac{3}{2},\frac{3}{2},2 \sin ^2\left (\frac{1}{2} (e+f x)\right )\right )-1\right )}{f n} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^n*(a + a*Sec[e + f*x])^(3/2),x]

[Out]

(a*(-1 + (1 + 4*n)*Cos[e + f*x]^(1/2 + n)*Hypergeometric2F1[1/2, 3/2 + n, 3/2, 2*Sin[(e + f*x)/2]^2])*Sec[e +
f*x]^n*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(f*n)

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Maple [F]  time = 0.158, 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 ) ^{{\frac{3}{2}}}\, 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))^(3/2),x)

[Out]

int(sec(f*x+e)^n*(a+a*sec(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \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))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^(3/2)*sec(f*x + e)^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}} \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))^(3/2),x, algorithm="fricas")

[Out]

integral((a*sec(f*x + e) + a)^(3/2)*sec(f*x + e)^n, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(3/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))^(3/2),x, algorithm="giac")

[Out]

Timed out