3.69 \(\int e^{2 \text{sech}^{-1}(a x)} \, dx\)

Optimal. Leaf size=57 \[ -\frac{4}{a \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )}+\frac{4 \tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{a}-x \]

[Out]

-x - 4/(a*(1 - Sqrt[(1 - a*x)/(1 + a*x)])) + (4*ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]])/a

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Rubi [A]  time = 0.173586, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6332, 1647, 12, 801, 203} \[ -\frac{4}{a \left (1-\sqrt{\frac{1-a x}{a x+1}}\right )}+\frac{4 \tan ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )}{a}-x \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcSech[a*x]),x]

[Out]

-x - 4/(a*(1 - Sqrt[(1 - a*x)/(1 + a*x)])) + (4*ArcTan[Sqrt[(1 - a*x)/(1 + a*x)]])/a

Rule 6332

Int[E^(ArcSech[u_]*(n_.)), x_Symbol] :> Int[(1/u + Sqrt[(1 - u)/(1 + u)] + (1*Sqrt[(1 - u)/(1 + u)])/u)^n, x]
/; IntegerQ[n]

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 12

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int e^{2 \text{sech}^{-1}(a x)} \, dx &=\int \left (\frac{1}{a x}+\sqrt{\frac{1-a x}{1+a x}}+\frac{\sqrt{\frac{1-a x}{1+a x}}}{a x}\right )^2 \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{x (1+x)^2}{(-1+x)^2 \left (1+x^2\right )^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a}\\ &=-x+\frac{2 \operatorname{Subst}\left (\int -\frac{4 x}{(-1+x)^2 \left (1+x^2\right )} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a}\\ &=-x-\frac{8 \operatorname{Subst}\left (\int \frac{x}{(-1+x)^2 \left (1+x^2\right )} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a}\\ &=-x-\frac{8 \operatorname{Subst}\left (\int \left (\frac{1}{2 (-1+x)^2}-\frac{1}{2 \left (1+x^2\right )}\right ) \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a}\\ &=-x-\frac{4}{a \left (1-\sqrt{\frac{1-a x}{1+a x}}\right )}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )}{a}\\ &=-x-\frac{4}{a \left (1-\sqrt{\frac{1-a x}{1+a x}}\right )}+\frac{4 \tan ^{-1}\left (\sqrt{\frac{1-a x}{1+a x}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0948087, size = 75, normalized size = 1.32 \[ -\frac{a^2 x^2+2 \sqrt{\frac{1-a x}{a x+1}} (a x+1)+2 a x \tan ^{-1}\left (\frac{a x}{\sqrt{\frac{1-a x}{a x+1}} (a x+1)}\right )+2}{a^2 x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcSech[a*x]),x]

[Out]

-((2 + a^2*x^2 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x) + 2*a*x*ArcTan[(a*x)/(Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*
x))])/(a^2*x))

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Maple [C]  time = 0.209, size = 98, normalized size = 1.7 \begin{align*} -x-2\,{\frac{1}{{a}^{2}x}}-2\,{\frac{{\it csgn} \left ( a \right ) }{a\sqrt{-{a}^{2}{x}^{2}+1}}\sqrt{-{\frac{ax-1}{ax}}}\sqrt{{\frac{ax+1}{ax}}} \left ( \arctan \left ({\frac{{\it csgn} \left ( a \right ) ax}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) xa+{\it csgn} \left ( a \right ) \sqrt{-{a}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2,x)

[Out]

-x-2/a^2/x-2/a*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*(arctan(csgn(a)*a*x/(-a^2*x^2+1)^(1/2))*x*a+csgn(a)*(-
a^2*x^2+1)^(1/2))*csgn(a)/(-a^2*x^2+1)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -x + \frac{2 \,{\left (-\frac{a^{2} \arcsin \left (\frac{a^{2} x}{\sqrt{a^{2}}}\right )}{\sqrt{a^{2}}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{x}\right )}}{a^{2}} + \frac{-\frac{1}{x}}{a^{2}} - \frac{1}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2,x, algorithm="maxima")

[Out]

-x + 2*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)/x^2, x)/a^2 + integrate(x^(-2), x)/a^2 - 1/(a^2*x)

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Fricas [A]  time = 2.3013, size = 186, normalized size = 3.26 \begin{align*} -\frac{a^{2} x^{2} + 2 \, a x \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} - 2 \, a x \arctan \left (\sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}}\right ) + 2}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int - a^{2}\, dx + \int \frac{2}{x^{2}}\, dx + \int \frac{2 a \sqrt{-1 + \frac{1}{a x}} \sqrt{1 + \frac{1}{a x}}}{x}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)**(1/2)*(1+1/a/x)**(1/2))**2,x)

[Out]

(Integral(-a**2, x) + Integral(2/x**2, x) + Integral(2*a*sqrt(-1 + 1/(a*x))*sqrt(1 + 1/(a*x))/x, x))/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\sqrt{\frac{1}{a x} + 1} \sqrt{\frac{1}{a x} - 1} + \frac{1}{a x}\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1/a/x-1)^(1/2)*(1+1/a/x)^(1/2))^2,x, algorithm="giac")

[Out]

integrate((sqrt(1/(a*x) + 1)*sqrt(1/(a*x) - 1) + 1/(a*x))^2, x)