\(\int e^{\sec ^{-1}(a x)} x \, dx\) [44]

Optimal result

Integrand size = 8, antiderivative size = 91 \[ \int e^{\sec ^{-1}(a x)} x \, dx=-\frac {\left (\frac {8}{5}+\frac {4 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}+\frac {\left (\frac {16}{5}+\frac {8 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},3,2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^2} \]

[Out]

(-8/5-4/5*I)*exp((1+2*I)*arcsec(a*x))*hypergeom([2, 1-1/2*I],[2-1/2*I],-(1/a/x+I*(1-1/a^2/x^2)^(1/2))^2)/a^2+(
16/5+8/5*I)*exp((1+2*I)*arcsec(a*x))*hypergeom([3, 1-1/2*I],[2-1/2*I],-(1/a/x+I*(1-1/a^2/x^2)^(1/2))^2)/a^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5374, 12, 4559, 2283} \[ \int e^{\sec ^{-1}(a x)} x \, dx=\frac {\left (\frac {16}{5}+\frac {8 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},3,2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}-\frac {\left (\frac {8}{5}+\frac {4 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^2} \]

[In]

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

[Out]

((-8/5 - (4*I)/5)*E^((1 + 2*I)*ArcSec[a*x])*Hypergeometric2F1[1 - I/2, 2, 2 - I/2, -E^((2*I)*ArcSec[a*x])])/a^
2 + ((16/5 + (8*I)/5)*E^((1 + 2*I)*ArcSec[a*x])*Hypergeometric2F1[1 - I/2, 3, 2 - I/2, -E^((2*I)*ArcSec[a*x])]
)/a^2

Rule 12

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

Rule 2283

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

Rule 4559

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol]
:> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
IGtQ[m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]

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 \frac {e^x \sec ^2(x) \tan (x)}{a} \, dx,x,\sec ^{-1}(a x)\right )}{a} \\ & = \frac {\text {Subst}\left (\int e^x \sec ^2(x) \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a^2} \\ & = \frac {\text {Subst}\left (\int \left (\frac {8 i e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^3}-\frac {4 i e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^2}\right ) \, dx,x,\sec ^{-1}(a x)\right )}{a^2} \\ & = -\frac {(4 i) \text {Subst}\left (\int \frac {e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^2} \, dx,x,\sec ^{-1}(a x)\right )}{a^2}+\frac {(8 i) \text {Subst}\left (\int \frac {e^{(1+2 i) x}}{\left (1+e^{2 i x}\right )^3} \, dx,x,\sec ^{-1}(a x)\right )}{a^2} \\ & = -\frac {\left (\frac {8}{5}+\frac {4 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},2,2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^2}+\frac {\left (\frac {16}{5}+\frac {8 i}{5}\right ) e^{(1+2 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1-\frac {i}{2},3,2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.18 \[ \int e^{\sec ^{-1}(a x)} x \, dx=\frac {\left (\frac {1}{5}+\frac {i}{10}\right ) e^{\sec ^{-1}(a x)} \left ((-2+i) a x \left (\sqrt {1-\frac {1}{a^2 x^2}}-a x\right )+(1+2 i) \operatorname {Hypergeometric2F1}\left (-\frac {i}{2},1,1-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )-e^{2 i \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (1,1-\frac {i}{2},2-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )\right )}{a^2} \]

[In]

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

[Out]

((1/5 + I/10)*E^ArcSec[a*x]*((-2 + I)*a*x*(Sqrt[1 - 1/(a^2*x^2)] - a*x) + (1 + 2*I)*Hypergeometric2F1[-1/2*I,
1, 1 - I/2, -E^((2*I)*ArcSec[a*x])] - E^((2*I)*ArcSec[a*x])*Hypergeometric2F1[1, 1 - I/2, 2 - I/2, -E^((2*I)*A
rcSec[a*x])]))/a^2

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(x*e^(arcsec(a*x)), x)

Sympy [F]

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

[In]

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

[Out]

Integral(x*exp(asec(a*x)), x)

Maxima [F]

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

[In]

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

[Out]

integrate(x*e^(arcsec(a*x)), x)

Giac [F]

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

[In]

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

[Out]

integrate(x*e^(arcsec(a*x)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(x*exp(acos(1/(a*x))), x)