3.43 \(\int e^{\sec ^{-1}(a x)} x^2 \, dx\)

Optimal. Leaf size=99 \[ \frac{\left (\frac{24}{5}+\frac{8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{3}{2}-\frac{i}{2},4;\frac{5}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}-\frac{\left (\frac{12}{5}+\frac{4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{3}{2}-\frac{i}{2},3;\frac{5}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \]

[Out]

((-12/5 - (4*I)/5)*E^((1 + 3*I)*ArcSec[a*x])*Hypergeometric2F1[3/2 - I/2, 3, 5/2 - I/2, -E^((2*I)*ArcSec[a*x])
])/a^3 + ((24/5 + (8*I)/5)*E^((1 + 3*I)*ArcSec[a*x])*Hypergeometric2F1[3/2 - I/2, 4, 5/2 - I/2, -E^((2*I)*ArcS
ec[a*x])])/a^3

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Rubi [A]  time = 0.117391, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5266, 12, 4471, 2251} \[ \frac{\left (\frac{24}{5}+\frac{8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{3}{2}-\frac{i}{2},4;\frac{5}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}-\frac{\left (\frac{12}{5}+\frac{4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{3}{2}-\frac{i}{2},3;\frac{5}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSec[a*x]*x^2,x]

[Out]

((-12/5 - (4*I)/5)*E^((1 + 3*I)*ArcSec[a*x])*Hypergeometric2F1[3/2 - I/2, 3, 5/2 - I/2, -E^((2*I)*ArcSec[a*x])
])/a^3 + ((24/5 + (8*I)/5)*E^((1 + 3*I)*ArcSec[a*x])*Hypergeometric2F1[3/2 - I/2, 4, 5/2 - I/2, -E^((2*I)*ArcS
ec[a*x])])/a^3

Rule 5266

Int[(u_.)*(f_)^(ArcSec[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[(u /. x -> -(a/b) +
Sec[x]/b)*f^(c*x^n)*Sec[x]*Tan[x], x], x, ArcSec[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[n, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4471

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int e^{\sec ^{-1}(a x)} x^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^x \sec ^3(x) \tan (x)}{a^2} \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int e^x \sec ^3(x) \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a^3}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{16 i e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^4}-\frac{8 i e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^3}\right ) \, dx,x,\sec ^{-1}(a x)\right )}{a^3}\\ &=-\frac{(8 i) \operatorname{Subst}\left (\int \frac{e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^3} \, dx,x,\sec ^{-1}(a x)\right )}{a^3}+\frac{(16 i) \operatorname{Subst}\left (\int \frac{e^{(1+3 i) x}}{\left (1+e^{2 i x}\right )^4} \, dx,x,\sec ^{-1}(a x)\right )}{a^3}\\ &=-\frac{\left (\frac{12}{5}+\frac{4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{3}{2}-\frac{i}{2},3;\frac{5}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}+\frac{\left (\frac{24}{5}+\frac{8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{3}{2}-\frac{i}{2},4;\frac{5}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}\\ \end{align*}

Mathematica [A]  time = 0.301252, size = 95, normalized size = 0.96 \[ \frac{e^{\sec ^{-1}(a x)} \left (a^4 x^4 \left (\cos \left (2 \sec ^{-1}(a x)\right )-\sin \left (2 \sec ^{-1}(a x)\right )+5\right )-(4+4 i) \left (a x \sqrt{1-\frac{1}{a^2 x^2}}-i\right ) \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )\right )}{12 a^4 x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSec[a*x]*x^2,x]

[Out]

(E^ArcSec[a*x]*((-4 - 4*I)*(-I + a*Sqrt[1 - 1/(a^2*x^2)]*x)*Hypergeometric2F1[1/2 - I/2, 1, 3/2 - I/2, -E^((2*
I)*ArcSec[a*x])] + a^4*x^4*(5 + Cos[2*ArcSec[a*x]] - Sin[2*ArcSec[a*x]])))/(12*a^4*x)

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Maple [F]  time = 0.181, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{\rm arcsec} \left (ax\right )}}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsec(a*x))*x^2,x)

[Out]

int(exp(arcsec(a*x))*x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\left (\operatorname{arcsec}\left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))*x^2,x, algorithm="maxima")

[Out]

integrate(x^2*e^(arcsec(a*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} e^{\left (\operatorname{arcsec}\left (a x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))*x^2,x, algorithm="fricas")

[Out]

integral(x^2*e^(arcsec(a*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asec(a*x))*x**2,x)

[Out]

Integral(x**2*exp(asec(a*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} e^{\left (\operatorname{arcsec}\left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))*x^2,x, algorithm="giac")

[Out]

integrate(x^2*e^(arcsec(a*x)), x)