3.49 \(\int \frac{e^{\sec ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=84 \[ \frac{1}{8} a^3 \sqrt{1-\frac{1}{a^2 x^2}} e^{\sec ^{-1}(a x)}-\frac{a^2 e^{\sec ^{-1}(a x)}}{8 x}-\frac{3}{40} a^3 e^{\sec ^{-1}(a x)} \cos \left (3 \sec ^{-1}(a x)\right )+\frac{1}{40} a^3 e^{\sec ^{-1}(a x)} \sin \left (3 \sec ^{-1}(a x)\right ) \]

[Out]

(a^3*E^ArcSec[a*x]*Sqrt[1 - 1/(a^2*x^2)])/8 - (a^2*E^ArcSec[a*x])/(8*x) - (3*a^3*E^ArcSec[a*x]*Cos[3*ArcSec[a*
x]])/40 + (a^3*E^ArcSec[a*x]*Sin[3*ArcSec[a*x]])/40

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Rubi [A]  time = 0.065074, antiderivative size = 84, 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, 4469, 4432} \[ \frac{1}{8} a^3 \sqrt{1-\frac{1}{a^2 x^2}} e^{\sec ^{-1}(a x)}-\frac{a^2 e^{\sec ^{-1}(a x)}}{8 x}-\frac{3}{40} a^3 e^{\sec ^{-1}(a x)} \cos \left (3 \sec ^{-1}(a x)\right )+\frac{1}{40} a^3 e^{\sec ^{-1}(a x)} \sin \left (3 \sec ^{-1}(a x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSec[a*x]/x^4,x]

[Out]

(a^3*E^ArcSec[a*x]*Sqrt[1 - 1/(a^2*x^2)])/8 - (a^2*E^ArcSec[a*x])/(8*x) - (3*a^3*E^ArcSec[a*x]*Cos[3*ArcSec[a*
x]])/40 + (a^3*E^ArcSec[a*x]*Sin[3*ArcSec[a*x]])/40

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^4} \, dx &=\frac{\operatorname{Subst}\left (\int a^4 e^x \cos ^2(x) \sin (x) \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=a^3 \operatorname{Subst}\left (\int e^x \cos ^2(x) \sin (x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=a^3 \operatorname{Subst}\left (\int \left (\frac{1}{4} e^x \sin (x)+\frac{1}{4} e^x \sin (3 x)\right ) \, dx,x,\sec ^{-1}(a x)\right )\\ &=\frac{1}{4} a^3 \operatorname{Subst}\left (\int e^x \sin (x) \, dx,x,\sec ^{-1}(a x)\right )+\frac{1}{4} a^3 \operatorname{Subst}\left (\int e^x \sin (3 x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=\frac{1}{8} a^3 e^{\sec ^{-1}(a x)} \sqrt{1-\frac{1}{a^2 x^2}}-\frac{a^2 e^{\sec ^{-1}(a x)}}{8 x}-\frac{3}{40} a^3 e^{\sec ^{-1}(a x)} \cos \left (3 \sec ^{-1}(a x)\right )+\frac{1}{40} a^3 e^{\sec ^{-1}(a x)} \sin \left (3 \sec ^{-1}(a x)\right )\\ \end{align*}

Mathematica [A]  time = 0.140966, size = 54, normalized size = 0.64 \[ \frac{1}{40} a^3 e^{\sec ^{-1}(a x)} \left (5 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{5}{a x}-3 \cos \left (3 \sec ^{-1}(a x)\right )+\sin \left (3 \sec ^{-1}(a x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcSec[a*x]/x^4,x]

[Out]

(a^3*E^ArcSec[a*x]*(5*Sqrt[1 - 1/(a^2*x^2)] - 5/(a*x) - 3*Cos[3*ArcSec[a*x]] + Sin[3*ArcSec[a*x]]))/40

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Maple [F]  time = 0.179, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{{\rm arcsec} \left (ax\right )}}}{{x}^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(arcsec(a*x))/x^4,x)

[Out]

int(exp(arcsec(a*x))/x^4,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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))/x^4,x, algorithm="maxima")

[Out]

integrate(e^(arcsec(a*x))/x^4, x)

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Fricas [A]  time = 2.71103, size = 100, normalized size = 1.19 \begin{align*} \frac{{\left (a^{2} x^{2} +{\left (a^{2} x^{2} + 1\right )} \sqrt{a^{2} x^{2} - 1} - 3\right )} e^{\left (\operatorname{arcsec}\left (a x\right )\right )}}{10 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))/x^4,x, algorithm="fricas")

[Out]

1/10*(a^2*x^2 + (a^2*x^2 + 1)*sqrt(a^2*x^2 - 1) - 3)*e^(arcsec(a*x))/x^3

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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^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(asec(a*x))/x**4,x)

[Out]

Integral(exp(asec(a*x))/x**4, 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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(arcsec(a*x))/x^4,x, algorithm="giac")

[Out]

integrate(e^(arcsec(a*x))/x^4, x)