\(\int e^{-\coth ^{-1}(a x)} \, dx\) [36]

Optimal result

Integrand size = 8, antiderivative size = 37 \[ \int e^{-\coth ^{-1}(a x)} \, dx=\sqrt {1-\frac {1}{a^2 x^2}} x-\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6303, 821, 272, 65, 214} \[ \int e^{-\coth ^{-1}(a x)} \, dx=x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \]

[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 65

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 214

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

Rule 272

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 821

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 6303

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]

Rubi steps \begin{align*} \text {integral}& = -\text {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 {\text {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 {\text {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 \text {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 {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int e^{-\coth ^{-1}(a x)} \, dx=\sqrt {1-\frac {1}{a^2 x^2}} x-\frac {\log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{a} \]

[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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(90\) vs. \(2(33)=66\).

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.46

method	result	size
risch	\(\frac {\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{a}-\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}\, \left (a x -1\right )}\)	\(91\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )-\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\right )}{\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a \sqrt {a^{2}}}\)	\(99\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.73 \[ \int e^{-\coth ^{-1}(a x)} \, dx=\frac {{\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}} - \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a} \]

[In]

integrate(((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

Sympy [F]

\[ \int e^{-\coth ^{-1}(a x)} \, dx=\int \sqrt {\frac {a x - 1}{a x + 1}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (33) = 66\).

Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.43 \[ \int e^{-\coth ^{-1}(a x)} \, dx=-a {\left (\frac {2 \, \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \]

[In]

integrate(((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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.41 \[ \int e^{-\coth ^{-1}(a x)} \, dx=\frac {\log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{{\left | a \right |}} + \frac {\sqrt {a^{2} x^{2} - 1} \mathrm {sgn}\left (a x + 1\right )}{a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.63 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57 \[ \int e^{-\coth ^{-1}(a x)} \, dx=\frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}-\frac {2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \]

[In]

int(((a*x - 1)/(a*x + 1))^(1/2),x)

[Out]

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