3.48 \(\int \frac{e^{\sec ^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=41 \[ \frac{1}{10} a^2 e^{\sec ^{-1}(a x)} \sin \left (2 \sec ^{-1}(a x)\right )-\frac{1}{5} a^2 e^{\sec ^{-1}(a x)} \cos \left (2 \sec ^{-1}(a x)\right ) \]

[Out]

-(a^2*E^ArcSec[a*x]*Cos[2*ArcSec[a*x]])/5 + (a^2*E^ArcSec[a*x]*Sin[2*ArcSec[a*x]])/10

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Rubi [A]  time = 0.0416936, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5266, 12, 4469, 4432} \[ \frac{1}{10} a^2 e^{\sec ^{-1}(a x)} \sin \left (2 \sec ^{-1}(a x)\right )-\frac{1}{5} a^2 e^{\sec ^{-1}(a x)} \cos \left (2 \sec ^{-1}(a x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSec[a*x]/x^3,x]

[Out]

-(a^2*E^ArcSec[a*x]*Cos[2*ArcSec[a*x]])/5 + (a^2*E^ArcSec[a*x]*Sin[2*ArcSec[a*x]])/10

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 4469

Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :
> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g
}, x] && IGtQ[m, 0] && IGtQ[n, 0]

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^3} \, dx &=\frac{\operatorname{Subst}\left (\int a^3 e^x \cos (x) \sin (x) \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=a^2 \operatorname{Subst}\left (\int e^x \cos (x) \sin (x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=a^2 \operatorname{Subst}\left (\int \frac{1}{2} e^x \sin (2 x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=\frac{1}{2} a^2 \operatorname{Subst}\left (\int e^x \sin (2 x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=-\frac{1}{5} a^2 e^{\sec ^{-1}(a x)} \cos \left (2 \sec ^{-1}(a x)\right )+\frac{1}{10} a^2 e^{\sec ^{-1}(a x)} \sin \left (2 \sec ^{-1}(a x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0501803, size = 30, normalized size = 0.73 \[ \frac{1}{10} a^2 e^{\sec ^{-1}(a x)} \left (\sin \left (2 \sec ^{-1}(a x)\right )-2 \cos \left (2 \sec ^{-1}(a x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSec[a*x]/x^3,x]

[Out]

(a^2*E^ArcSec[a*x]*(-2*Cos[2*ArcSec[a*x]] + Sin[2*ArcSec[a*x]]))/10

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Maple [F]  time = 0.177, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{{\rm arcsec} \left (ax\right )}}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsec(a*x))/x^3,x)

[Out]

int(exp(arcsec(a*x))/x^3,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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))/x^3,x, algorithm="maxima")

[Out]

integrate(e^(arcsec(a*x))/x^3, x)

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Fricas [A]  time = 2.48443, size = 80, normalized size = 1.95 \begin{align*} \frac{{\left (a^{2} x^{2} + \sqrt{a^{2} x^{2} - 1} - 2\right )} e^{\left (\operatorname{arcsec}\left (a x\right )\right )}}{5 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))/x^3,x, algorithm="fricas")

[Out]

1/5*(a^2*x^2 + sqrt(a^2*x^2 - 1) - 2)*e^(arcsec(a*x))/x^2

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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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asec(a*x))/x**3,x)

[Out]

Integral(exp(asec(a*x))/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{\left (\operatorname{arcsec}\left (a x\right )\right )}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))/x^3,x, algorithm="giac")

[Out]

integrate(e^(arcsec(a*x))/x^3, x)