### 3.5 $$\int \frac{e^{\coth ^{-1}(a x)}}{x} \, dx$$

Optimal. Leaf size=22 $\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )-\csc ^{-1}(a x)$

[Out]

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

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Rubi [A]  time = 0.0434277, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {6169, 844, 216, 266, 63, 208} $\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )-\csc ^{-1}(a x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6169

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 216

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

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

Mathematica [A]  time = 0.0131496, size = 36, normalized size = 1.64 $\log \left (x \left (\sqrt{\frac{a^2 x^2-1}{a^2 x^2}}+1\right )\right )-\sin ^{-1}\left (\frac{1}{a x}\right )$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.128, size = 132, normalized size = 6. \begin{align*} -{(ax-1) \left ( \sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}+\arctan \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}-1}}} \right ) \sqrt{{a}^{2}}-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 ) -\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \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,x)

[Out]

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

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Maxima [B]  time = 1.56288, size = 93, normalized size = 4.23 \begin{align*} a{\left (\frac{2 \, \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )}{a} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a}\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,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.53101, size = 150, normalized size = 6.82 \begin{align*} 2 \, \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) + \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 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,x, algorithm="fricas")

[Out]

2*arctan(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \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,x)

[Out]

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

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Giac [B]  time = 1.17373, size = 95, normalized size = 4.32 \begin{align*} a{\left (\frac{2 \, \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )}{a} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a} - \frac{\log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a}\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,x, algorithm="giac")

[Out]

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