\(\int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx\) [47]

Optimal result

Integrand size = 10, antiderivative size = 39 \[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\frac {1}{2} a e^{\sec ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}}-\frac {e^{\sec ^{-1}(a x)}}{2 x} \]

[Out]

-1/2*exp(arcsec(a*x))/x+1/2*a*exp(arcsec(a*x))*(1-1/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5374, 12, 4517} \[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\frac {1}{2} a \sqrt {1-\frac {1}{a^2 x^2}} e^{\sec ^{-1}(a x)}-\frac {e^{\sec ^{-1}(a x)}}{2 x} \]

[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 12

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

Rule 4517

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 5374

Int[(u_.)*(f_)^(ArcSec[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[(u /. x -> -a/b + Se
c[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*} \text {integral}& = \frac {\text {Subst}\left (\int a^2 e^x \sin (x) \, dx,x,\sec ^{-1}(a x)\right )}{a} \\ & = a \text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\frac {1}{2} a e^{\sec ^{-1}(a x)} \left (\sqrt {1-\frac {1}{a^2 x^2}}-\frac {1}{a x}\right ) \]

[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

Maple [F]

\[\int \frac {{\mathrm e}^{\operatorname {arcsec}\left (a x \right )}}{x^{2}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.59 \[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\frac {{\left (\sqrt {a^{2} x^{2} - 1} - 1\right )} e^{\left (\operatorname {arcsec}\left (a x\right )\right )}}{2 \, x} \]

[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

Sympy [F]

\[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\int \frac {e^{\operatorname {asec}{\left (a x \right )}}}{x^{2}}\, dx \]

[In]

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

[Out]

Integral(exp(asec(a*x))/x**2, x)

Maxima [F]

\[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\int { \frac {e^{\left (\operatorname {arcsec}\left (a x\right )\right )}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate(e^(arcsec(a*x))/x^2, x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\frac {1}{2} \, {\left (\sqrt {-\frac {1}{a^{2} x^{2}} + 1} e^{\left (\arccos \left (\frac {1}{a x}\right )\right )} - \frac {e^{\left (\arccos \left (\frac {1}{a x}\right )\right )}}{a x}\right )} a \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\sec ^{-1}(a x)}}{x^2} \, dx=\int \frac {{\mathrm {e}}^{\mathrm {acos}\left (\frac {1}{a\,x}\right )}}{x^2} \,d x \]

[In]

int(exp(acos(1/(a*x)))/x^2,x)

[Out]

int(exp(acos(1/(a*x)))/x^2, x)