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

Optimal. Leaf size=39 \[ \frac{1}{2} a \sqrt{1-\frac{1}{a^2 x^2}} e^{\sec ^{-1}(a x)}-\frac{e^{\sec ^{-1}(a x)}}{2 x} \]

[Out]

(a*E^ArcSec[a*x]*Sqrt[1 - 1/(a^2*x^2)])/2 - E^ArcSec[a*x]/(2*x)

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Rubi [A]  time = 0.0298828, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5266, 12, 4432} \[ \frac{1}{2} a \sqrt{1-\frac{1}{a^2 x^2}} e^{\sec ^{-1}(a x)}-\frac{e^{\sec ^{-1}(a x)}}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*E^ArcSec[a*x]*Sqrt[1 - 1/(a^2*x^2)])/2 - E^ArcSec[a*x]/(2*x)

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 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
 /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int \frac{e^{\sec ^{-1}(a x)}}{x^2} \, dx &=\frac{\operatorname{Subst}\left (\int a^2 e^x \sin (x) \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=a \operatorname{Subst}\left (\int e^x \sin (x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=\frac{1}{2} a e^{\sec ^{-1}(a x)} \sqrt{1-\frac{1}{a^2 x^2}}-\frac{e^{\sec ^{-1}(a x)}}{2 x}\\ \end{align*}

Mathematica [A]  time = 0.0410404, size = 34, normalized size = 0.87 \[ \frac{1}{2} a \left (\sqrt{1-\frac{1}{a^2 x^2}}-\frac{1}{a x}\right ) e^{\sec ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*E^ArcSec[a*x]*(Sqrt[1 - 1/(a^2*x^2)] - 1/(a*x)))/2

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Maple [F]  time = 0.176, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\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 \frac{e^{\left (\operatorname{arcsec}\left (a x\right )\right )}}{x^{2}}\,{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(e^(arcsec(a*x))/x^2, x)

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Fricas [A]  time = 2.43419, size = 63, normalized size = 1.62 \begin{align*} \frac{{\left (\sqrt{a^{2} x^{2} - 1} - 1\right )} e^{\left (\operatorname{arcsec}\left (a x\right )\right )}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(a^2*x^2 - 1) - 1)*e^(arcsec(a*x))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(asec(a*x))/x**2, x)

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Giac [A]  time = 1.15866, size = 69, normalized size = 1.77 \begin{align*} \frac{1}{2} \, a \sqrt{-\frac{1}{a^{2} x^{2}} + 1} e^{\left (\frac{1}{2} \, \pi - \arcsin \left (\frac{1}{a x}\right )\right )} - \frac{e^{\left (\frac{1}{2} \, \pi - \arcsin \left (\frac{1}{a x}\right )\right )}}{2 \, 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]

1/2*a*sqrt(-1/(a^2*x^2) + 1)*e^(1/2*pi - arcsin(1/(a*x))) - 1/2*e^(1/2*pi - arcsin(1/(a*x)))/x