3.290 \(\int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=172 \[ -\frac{a^2 (2 n+1) \sin (e+f x) \sec ^{n-1}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{2 a^2 \sin (e+f x) \sec ^n(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt{\sin ^2(e+f x)}}+\frac{a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.139675, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3788, 3772, 2643, 4046} \[ -\frac{a^2 (2 n+1) \sin (e+f x) \sec ^{n-1}(e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{2 a^2 \sin (e+f x) \sec ^n(e+f x) \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(e+f x)\right )}{f n \sqrt{\sin ^2(e+f x)}}+\frac{a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + n)) - (a^2*(1 + 2*n)*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^2)*Sqrt[Sin[e + f*x]^2]) + (2*a^2*Hypergeome
tric2F1[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 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), 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]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rubi steps

\begin{align*} \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx &=\left (2 a^2\right ) \int \sec ^{1+n}(e+f x) \, dx+\int \sec ^n(e+f x) \left (a^2+a^2 \sec ^2(e+f x)\right ) \, dx\\ &=\frac{a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}+\frac{\left (a^2 (1+2 n)\right ) \int \sec ^n(e+f x) \, dx}{1+n}+\left (2 a^2 \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-1-n}(e+f x) \, dx\\ &=\frac{a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}+\frac{2 a^2 \, _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)}}+\frac{\left (a^2 (1+2 n) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-n}(e+f x) \, dx}{1+n}\\ &=\frac{a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}-\frac{a^2 (1+2 n) \, _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 \left (1-n^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{2 a^2 \, _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 [F]  time = 0.899763, size = 0, normalized size = 0. \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.121, 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 ) ^{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))^2,x)

[Out]

int(sec(f*x+e)^n*(a+a*sec(f*x+e))^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 )}^{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))^2,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^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^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}\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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \sec{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx + \int \sec ^{2}{\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))**2,x)

[Out]

a**2*(Integral(2*sec(e + f*x)*sec(e + f*x)**n, x) + Integral(sec(e + f*x)**2*sec(e + f*x)**n, x) + Integral(se
c(e + f*x)**n, x))

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

[Out]

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