3.288 \(\int \sec ^n(e+f x) (a+a \sec (e+f x))^4 \, dx\)

Optimal. Leaf size=304 \[ -\frac{a^4 \left (8 n^2+24 n+3\right ) \sin (e+f x) \sec ^{n-1}(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3-n}{2},\cos ^2(e+f x)\right )}{f (1-n) (n+1) (n+3) \sqrt{\sin ^2(e+f x)}}+\frac{4 a^4 (2 n+3) \sin (e+f x) \sec ^n(e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n}{2},\frac{2-n}{2},\cos ^2(e+f x)\right )}{f n (n+2) \sqrt{\sin ^2(e+f x)}}+\frac{a^4 \left (4 n^2+21 n+30\right ) \sin (e+f x) \sec ^{n+1}(e+f x)}{f (n+1) (n+2) (n+3)}+\frac{\sin (e+f x) \left (a^2 \sec (e+f x)+a^2\right )^2 \sec ^{n+1}(e+f x)}{f (n+3)}+\frac{2 (n+4) \sin (e+f x) \left (a^4 \sec (e+f x)+a^4\right ) \sec ^{n+1}(e+f x)}{f (n+2) (n+3)} \]

[Out]

(a^4*(30 + 21*n + 4*n^2)*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + n)*(2 + n)*(3 + n)) + (Sec[e + f*x]^(1 + n
)*(a^2 + a^2*Sec[e + f*x])^2*Sin[e + f*x])/(f*(3 + n)) + (2*(4 + n)*Sec[e + f*x]^(1 + n)*(a^4 + a^4*Sec[e + f*
x])*Sin[e + f*x])/(f*(2 + n)*(3 + n)) - (a^4*(3 + 24*n + 8*n^2)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, C
os[e + f*x]^2]*Sec[e + f*x]^(-1 + n)*Sin[e + f*x])/(f*(1 - n)*(1 + n)*(3 + n)*Sqrt[Sin[e + f*x]^2]) + (4*a^4*(
3 + 2*n)*Hypergeometric2F1[1/2, -n/2, (2 - n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^n*Sin[e + f*x])/(f*n*(2 + n)*Sqr
t[Sin[e + f*x]^2])

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Rubi [A]  time = 0.484646, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3814, 4018, 3997, 3787, 3772, 2643} \[ -\frac{a^4 \left (8 n^2+24 n+3\right ) \sin (e+f x) \sec ^{n-1}(e+f x) \, _2F_1\left (\frac{1}{2},\frac{1-n}{2};\frac{3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) (n+1) (n+3) \sqrt{\sin ^2(e+f x)}}+\frac{4 a^4 (2 n+3) \sin (e+f x) \sec ^n(e+f x) \, _2F_1\left (\frac{1}{2},-\frac{n}{2};\frac{2-n}{2};\cos ^2(e+f x)\right )}{f n (n+2) \sqrt{\sin ^2(e+f x)}}+\frac{a^4 \left (4 n^2+21 n+30\right ) \sin (e+f x) \sec ^{n+1}(e+f x)}{f (n+1) (n+2) (n+3)}+\frac{\sin (e+f x) \left (a^2 \sec (e+f x)+a^2\right )^2 \sec ^{n+1}(e+f x)}{f (n+3)}+\frac{2 (n+4) \sin (e+f x) \left (a^4 \sec (e+f x)+a^4\right ) \sec ^{n+1}(e+f x)}{f (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^n*(a + a*Sec[e + f*x])^4,x]

[Out]

(a^4*(30 + 21*n + 4*n^2)*Sec[e + f*x]^(1 + n)*Sin[e + f*x])/(f*(1 + n)*(2 + n)*(3 + n)) + (Sec[e + f*x]^(1 + n
)*(a^2 + a^2*Sec[e + f*x])^2*Sin[e + f*x])/(f*(3 + n)) + (2*(4 + n)*Sec[e + f*x]^(1 + n)*(a^4 + a^4*Sec[e + f*
x])*Sin[e + f*x])/(f*(2 + n)*(3 + n)) - (a^4*(3 + 24*n + 8*n^2)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, C
os[e + f*x]^2]*Sec[e + f*x]^(-1 + n)*Sin[e + f*x])/(f*(1 - n)*(1 + n)*(3 + n)*Sqrt[Sin[e + f*x]^2]) + (4*a^4*(
3 + 2*n)*Hypergeometric2F1[1/2, -n/2, (2 - n)/2, Cos[e + f*x]^2]*Sec[e + f*x]^n*Sin[e + f*x])/(f*n*(2 + n)*Sqr
t[Sin[e + f*x]^2])

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 4018

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*(m + n
)), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d*n
) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

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))^4 \, dx &=\frac{\sec ^{1+n}(e+f x) \left (a^2+a^2 \sec (e+f x)\right )^2 \sin (e+f x)}{f (3+n)}+\frac{a \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 (a (3+2 n)+a (8+2 n) \sec (e+f x)) \, dx}{3+n}\\ &=\frac{\sec ^{1+n}(e+f x) \left (a^2+a^2 \sec (e+f x)\right )^2 \sin (e+f x)}{f (3+n)}+\frac{2 (4+n) \sec ^{1+n}(e+f x) \left (a^4+a^4 \sec (e+f x)\right ) \sin (e+f x)}{f (2+n) (3+n)}+\frac{a \int \sec ^n(e+f x) (a+a \sec (e+f x)) \left (a^2 \left (6+15 n+4 n^2\right )+a^2 \left (30+21 n+4 n^2\right ) \sec (e+f x)\right ) \, dx}{6+5 n+n^2}\\ &=\frac{a^4 \left (30+21 n+4 n^2\right ) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n) \left (6+5 n+n^2\right )}+\frac{\sec ^{1+n}(e+f x) \left (a^2+a^2 \sec (e+f x)\right )^2 \sin (e+f x)}{f (3+n)}+\frac{2 (4+n) \sec ^{1+n}(e+f x) \left (a^4+a^4 \sec (e+f x)\right ) \sin (e+f x)}{f (2+n) (3+n)}+\frac{a \int \sec ^n(e+f x) \left (a^3 (2+n) \left (3+24 n+8 n^2\right )+4 a^3 (1+n) (3+n) (3+2 n) \sec (e+f x)\right ) \, dx}{6+11 n+6 n^2+n^3}\\ &=\frac{a^4 \left (30+21 n+4 n^2\right ) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n) \left (6+5 n+n^2\right )}+\frac{\sec ^{1+n}(e+f x) \left (a^2+a^2 \sec (e+f x)\right )^2 \sin (e+f x)}{f (3+n)}+\frac{2 (4+n) \sec ^{1+n}(e+f x) \left (a^4+a^4 \sec (e+f x)\right ) \sin (e+f x)}{f (2+n) (3+n)}+\frac{\left (4 a^4 (3+2 n)\right ) \int \sec ^{1+n}(e+f x) \, dx}{2+n}+\frac{\left (a^4 \left (3+24 n+8 n^2\right )\right ) \int \sec ^n(e+f x) \, dx}{3+4 n+n^2}\\ &=\frac{a^4 \left (30+21 n+4 n^2\right ) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n) \left (6+5 n+n^2\right )}+\frac{\sec ^{1+n}(e+f x) \left (a^2+a^2 \sec (e+f x)\right )^2 \sin (e+f x)}{f (3+n)}+\frac{2 (4+n) \sec ^{1+n}(e+f x) \left (a^4+a^4 \sec (e+f x)\right ) \sin (e+f x)}{f (2+n) (3+n)}+\frac{\left (4 a^4 (3+2 n) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-1-n}(e+f x) \, dx}{2+n}+\frac{\left (a^4 \left (3+24 n+8 n^2\right ) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-n}(e+f x) \, dx}{3+4 n+n^2}\\ &=\frac{a^4 \left (30+21 n+4 n^2\right ) \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n) \left (6+5 n+n^2\right )}+\frac{\sec ^{1+n}(e+f x) \left (a^2+a^2 \sec (e+f x)\right )^2 \sin (e+f x)}{f (3+n)}+\frac{2 (4+n) \sec ^{1+n}(e+f x) \left (a^4+a^4 \sec (e+f x)\right ) \sin (e+f x)}{f (2+n) (3+n)}-\frac{a^4 \left (3+24 n+8 n^2\right ) \, _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) \left (3+4 n+n^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{4 a^4 (3+2 n) \, _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 (2+n) \sqrt{\sin ^2(e+f x)}}\\ \end{align*}

Mathematica [F]  time = 0.590919, size = 0, normalized size = 0. \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^4 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^n*(a + a*Sec[e + f*x])^4,x]

[Out]

Integrate[Sec[e + f*x]^n*(a + a*Sec[e + f*x])^4, x]

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Maple [F]  time = 0.962, 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 ) ^{4}\, 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))^4,x)

[Out]

int(sec(f*x+e)^n*(a+a*sec(f*x+e))^4,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 )}^{4} \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))^4,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^4*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^{4} \sec \left (f x + e\right )^{4} + 4 \, a^{4} \sec \left (f x + e\right )^{3} + 6 \, a^{4} \sec \left (f x + e\right )^{2} + 4 \, a^{4} \sec \left (f x + e\right ) + a^{4}\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))^4,x, algorithm="fricas")

[Out]

integral((a^4*sec(f*x + e)^4 + 4*a^4*sec(f*x + e)^3 + 6*a^4*sec(f*x + e)^2 + 4*a^4*sec(f*x + e) + a^4)*sec(f*x
+ e)^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int 4 \sec{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx + \int 6 \sec ^{2}{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx + \int 4 \sec ^{3}{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx + \int \sec ^{4}{\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))**4,x)

[Out]

a**4*(Integral(4*sec(e + f*x)*sec(e + f*x)**n, x) + Integral(6*sec(e + f*x)**2*sec(e + f*x)**n, x) + Integral(
4*sec(e + f*x)**3*sec(e + f*x)**n, x) + Integral(sec(e + f*x)**4*sec(e + f*x)**n, x) + Integral(sec(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 )}^{4} \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))^4,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e) + a)^4*sec(f*x + e)^n, x)