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

Optimal. Leaf size=45 \[ 2 i e^{\sec ^{-1}(a x)} \, _2F_1\left (-\frac {i}{2},1;1-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )-i e^{\sec ^{-1}(a x)} \]

[Out]

-I*exp(arcsec(a*x))+2*I*exp(arcsec(a*x))*hypergeom([1, -1/2*I],[1-1/2*I],-(1/a/x+I*(1-1/a^2/x^2)^(1/2))^2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5266, 12, 4442, 2194, 2251} \[ 2 i e^{\sec ^{-1}(a x)} \, _2F_1\left (-\frac {i}{2},1;1-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )-i e^{\sec ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-I)*E^ArcSec[a*x] + (2*I)*E^ArcSec[a*x]*Hypergeometric2F1[-I/2, 1, 1 - I/2, -E^((2*I)*ArcSec[a*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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rule 4442

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Dist[I^n, Int[ExpandIntegran
d[(F^(c*(a + b*x))*(1 - E^(2*I*(d + e*x)))^n)/(1 + E^(2*I*(d + e*x)))^n, x], x], x] /; FreeQ[{F, a, b, c, d, e
}, x] && IntegerQ[n]

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} \, dx &=\frac {\operatorname {Subst}\left (\int a e^x \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=\operatorname {Subst}\left (\int e^x \tan (x) \, dx,x,\sec ^{-1}(a x)\right )\\ &=i \operatorname {Subst}\left (\int \left (-e^x+\frac {2 e^x}{1+e^{2 i x}}\right ) \, dx,x,\sec ^{-1}(a x)\right )\\ &=-\left (i \operatorname {Subst}\left (\int e^x \, dx,x,\sec ^{-1}(a x)\right )\right )+2 i \operatorname {Subst}\left (\int \frac {e^x}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a x)\right )\\ &=-i e^{\sec ^{-1}(a x)}+2 i e^{\sec ^{-1}(a x)} \, _2F_1\left (-\frac {i}{2},1;1-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 79, normalized size = 1.76 \[ -i \left (\left (\frac {1}{5}-\frac {2 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \, _2F_1\left (1,1-\frac {i}{2};2-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )-e^{\sec ^{-1}(a x)} \, _2F_1\left (-\frac {i}{2},1;1-\frac {i}{2};-e^{2 i \sec ^{-1}(a x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-I)*(-(E^ArcSec[a*x]*Hypergeometric2F1[-1/2*I, 1, 1 - I/2, -E^((2*I)*ArcSec[a*x])]) + (1/5 - (2*I)/5)*E^((1 +
2*I)*ArcSec[a*x])*Hypergeometric2F1[1, 1 - I/2, 2 - I/2, -E^((2*I)*ArcSec[a*x])])

________________________________________________________________________________________

fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{\left (\operatorname {arcsec}\left (a x\right )\right )}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(e^(arcsec(a*x))/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arcsec(a*x))/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arcsec(a*x))/x, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {e}}^{\mathrm {acos}\left (\frac {1}{a\,x}\right )}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(acos(1/(a*x)))/x, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(asec(a*x))/x, x)

________________________________________________________________________________________