### 3.4 $$\int e^{\coth ^{-1}(a x)} \, dx$$

Optimal. Leaf size=36 $x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}$

[Out]

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

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Rubi [A]  time = 0.0358052, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 6, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.833, Rules used = {6168, 807, 266, 63, 208} $x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6168

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.)), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^2*(1 - x/a)^((n - 1)/2)*Sq
rt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0218327, size = 41, normalized size = 1.14 $x \sqrt{1-\frac{1}{a^2 x^2}}+\frac{\log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{a}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.12, size = 97, normalized size = 2.7 \begin{align*}{\frac{ax-1}{a} \left ( \sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }+a\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x-1)*((a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)+a*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2)))/((
a*x-1)/(a*x+1))^(1/2)/((a*x-1)*(a*x+1))^(1/2)/a/(a^2)^(1/2)

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Maxima [B]  time = 1.02547, size = 122, normalized size = 3.39 \begin{align*} -a{\left (\frac{2 \, \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.5687, size = 155, normalized size = 4.31 \begin{align*} \frac{{\left (a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}} + \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*x + 1)*sqrt((a*x - 1)/(a*x + 1)) + log(sqrt((a*x - 1)/(a*x + 1)) + 1) - log(sqrt((a*x - 1)/(a*x + 1)) - 1)
)/a

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15749, size = 77, normalized size = 2.14 \begin{align*} -\frac{\log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1} \right |}\right )}{{\left | a \right |} \mathrm{sgn}\left (a x + 1\right )} + \frac{\sqrt{a^{2} x^{2} - 1}}{a \mathrm{sgn}\left (a x + 1\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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