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

Optimal. Leaf size=72 \[ \frac{a}{\sqrt{\frac{1-a x}{a x+1}}+1}-\frac{a}{\left (\sqrt{\frac{1-a x}{a x+1}}+1\right )^2}+a \left (-\tanh ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )\right ) \]

[Out]

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

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Rubi [A]  time = 0.382769, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6337, 77, 207} \[ \frac{a}{\sqrt{\frac{1-a x}{a x+1}}+1}-\frac{a}{\left (\sqrt{\frac{1-a x}{a x+1}}+1\right )^2}+a \left (-\tanh ^{-1}\left (\sqrt{\frac{1-a x}{a x+1}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcSech[a*x]*x^2),x]

[Out]

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

Rule 6337

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 207

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

Rubi steps

\begin{align*} \int \frac{e^{-\text{sech}^{-1}(a x)}}{x^2} \, dx &=\int \frac{1}{x^2 \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 )} \, dx\\ &=(4 a) \operatorname{Subst}\left (\int \frac{x}{(-1+x) (1+x)^3} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )\\ &=(4 a) \operatorname{Subst}\left (\int \left (\frac{1}{2 (1+x)^3}-\frac{1}{4 (1+x)^2}+\frac{1}{4 \left (-1+x^2\right )}\right ) \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )\\ &=-\frac{a}{\left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^2}+\frac{a}{1+\sqrt{\frac{1-a x}{1+a x}}}+a \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{\frac{1-a x}{1+a x}}\right )\\ &=-\frac{a}{\left (1+\sqrt{\frac{1-a x}{1+a x}}\right )^2}+\frac{a}{1+\sqrt{\frac{1-a x}{1+a x}}}-a \tanh ^{-1}\left (\sqrt{\frac{1-a x}{1+a x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0573269, size = 92, normalized size = 1.28 \[ \frac{1}{2} \left (\frac{\sqrt{\frac{1-a x}{a x+1}} (a x+1)}{a x^2}-\frac{1}{a x^2}+a \log (x)-a \log \left (a x \sqrt{\frac{1-a x}{a x+1}}+\sqrt{\frac{1-a x}{a x+1}}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcSech[a*x]*x^2),x]

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ({\frac{1}{ax}}+\sqrt{{\frac{1}{ax}}-1}\sqrt{1+{\frac{1}{ax}}} \right ) ^{-1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02791, size = 279, normalized size = 3.88 \begin{align*} -\frac{a^{2} x^{2} \log \left (a x \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} + 1\right ) - a^{2} x^{2} \log \left (a x \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} - 1\right ) - 2 \, a x \sqrt{\frac{a x + 1}{a x}} \sqrt{-\frac{a x - 1}{a x}} + 2}{4 \, a x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^2*x^2*log(a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 1) - a^2*x^2*log(a*x*sqrt((a*x + 1)/(a*x)
)*sqrt(-(a*x - 1)/(a*x)) - 1) - 2*a*x*sqrt((a*x + 1)/(a*x))*sqrt(-(a*x - 1)/(a*x)) + 2)/(a*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef