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

Optimal. Leaf size=91 \[ \frac{(2+2 i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},2;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a}-\frac{(1+i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.0922147, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5266, 4471, 2251} \[ \frac{(2+2 i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},2;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a}-\frac{(1+i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4471

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 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])

Rubi steps

\begin{align*} \int e^{\sec ^{-1}(a x)} \, dx &=\frac{\operatorname{Subst}\left (\int e^x \sec (x) \tan (x) \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{4 i e^{(1+i) x}}{\left (1+e^{2 i x}\right )^2}-\frac{2 i e^{(1+i) x}}{1+e^{2 i x}}\right ) \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=-\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{(1+i) x}}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(a x)\right )}{a}+\frac{(4 i) \operatorname{Subst}\left (\int \frac{e^{(1+i) x}}{\left (1+e^{2 i x}\right )^2} \, dx,x,\sec ^{-1}(a x)\right )}{a}\\ &=-\frac{(1+i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a}+\frac{(2+2 i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},2;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0604628, size = 54, normalized size = 0.59 \[ x e^{\sec ^{-1}(a x)}-\frac{(1-i) e^{(1+i) \sec ^{-1}(a x)} \, _2F_1\left (\frac{1}{2}-\frac{i}{2},1;\frac{3}{2}-\frac{i}{2};-e^{2 i \sec ^{-1}(a x)}\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.175, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{\rm arcsec} \left (ax\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (\operatorname{arcsec}\left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arcsec(a*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (e^{\left (\operatorname{arcsec}\left (a x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(e^(arcsec(a*x)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\operatorname{asec}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(asec(a*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (\operatorname{arcsec}\left (a x\right )\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(arcsec(a*x)), x)