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

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

[Out]

-1/8*a^2*exp(arcsec(a*x))/x-3/40*a^3*exp(arcsec(a*x))*cos(3*arcsec(a*x))+1/40*a^3*exp(arcsec(a*x))*sin(3*arcse
c(a*x))+1/8*a^3*exp(arcsec(a*x))*(1-1/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.14, 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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fricas [A]  time = 1.84, size = 40, normalized size = 0.48 \[ \frac {{\left (a^{2} x^{2} + {\left (a^{2} x^{2} + 1\right )} \sqrt {a^{2} x^{2} - 1} - 3\right )} e^{\left (\operatorname {arcsec}\left (a x\right )\right )}}{10 \, x^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (\operatorname {arcsec}\left (a x\right )\right )}}{x^{4}}\,{d x} \]

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)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{\mathrm {arcsec}\left (a x \right )}}{x^{4}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\left (\operatorname {arcsec}\left (a x\right )\right )}}{x^{4}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{\mathrm {acos}\left (\frac {1}{a\,x}\right )}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(acos(1/(a*x)))/x^4,x)

[Out]

int(exp(acos(1/(a*x)))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{\operatorname {asec}{\left (a x \right )}}}{x^{4}}\, dx \]

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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