3.35 \(\int e^{-\coth ^{-1}(a x)} x \, dx\)

Optimal. Leaf size=64 \[ \frac{1}{2} x^2 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{a}+\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0677775, antiderivative size = 64, 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, 835, 807, 266, 63, 208} \[ \frac{1}{2} x^2 \sqrt{1-\frac{1}{a^2 x^2}}-\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{a}+\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 - 1/(a^2*x^2)]*x)/a) + (Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 + ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]/(2*a^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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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

Mathematica [A]  time = 0.0354753, size = 49, normalized size = 0.77 \[ \frac{a x \sqrt{1-\frac{1}{a^2 x^2}} (a x-2)+\log \left (x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.128, size = 152, normalized size = 2.4 \begin{align*}{\frac{ax+1}{2\,{a}^{2}}\sqrt{{\frac{ax-1}{ax+1}}} \left ( \sqrt{{a}^{2}}\sqrt{{a}^{2}{x}^{2}-1}xa-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) a-2\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }+2\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.00163, size = 176, normalized size = 2.75 \begin{align*} -\frac{1}{2} \, a{\left (\frac{2 \,{\left (3 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} - \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{\frac{2 \,{\left (a x - 1\right )} a^{3}}{a x + 1} - \frac{{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{3}} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(2*(3*((a*x - 1)/(a*x + 1))^(3/2) - sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^3/(a*x + 1) - (a*x - 1)^2
*a^3/(a*x + 1)^2 - a^3) - log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^3 + log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^3)

________________________________________________________________________________________

Fricas [A]  time = 1.83951, size = 177, normalized size = 2.77 \begin{align*} \frac{{\left (a^{2} x^{2} - a x - 2\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 )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2*x^2 - a*x - 2)*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^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21142, size = 96, normalized size = 1.5 \begin{align*} \frac{1}{2} \, \sqrt{a^{2} x^{2} - 1}{\left (\frac{x \mathrm{sgn}\left (a x + 1\right )}{a} - \frac{2 \, \mathrm{sgn}\left (a x + 1\right )}{a^{2}}\right )} - \frac{\log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (a x + 1\right )}{2 \, a{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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