\(\int e^{\coth ^{-1}(a x)} x \, dx\) [3]

Optimal result

Integrand size = 8, antiderivative size = 63 \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{a}+\frac {1}{2} \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6304, 849, 821, 272, 65, 214} \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a^2}+\frac {1}{2} x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{a} \]

[In]

Int[E^ArcCoth[a*x]*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 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 849

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 6304

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x (2+a x)+\log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{2 a^2} \]

[In]

Integrate[E^ArcCoth[a*x]*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 [A] (verified)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.59

method	result	size
risch	\(\frac {\left (a x +2\right ) \left (a x -1\right )}{2 a^{2} \sqrt {\frac {a x -1}{a x +1}}}+\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{2 a \sqrt {a^{2}}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right )}\)	\(100\)
default	\(\frac {\left (a x -1\right ) \left (\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x -\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +2 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}+2 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )\right )}{2 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} \sqrt {a^{2}}}\)	\(152\)

[In]

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

[Out]

1/2*(a*x+2)*(a*x-1)/a^2/((a*x-1)/(a*x+1))^(1/2)+1/2/a*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 = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.16 \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {{\left (a^{2} x^{2} + 3 \, a x + 2\right )} \sqrt {\frac {a x - 1}{a x + 1}} + \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{2 \, a^{2}} \]

[In]

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

[Out]

1/2*((a^2*x^2 + 3*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]

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

[In]

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

[Out]

Integral(x/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. 128 vs. \(2 (53) = 106\).

Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.03 \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {1}{2} \, a {\left (\frac {2 \, {\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 3 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{3}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{3}}\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.22 \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {1}{2} \, \sqrt {a^{2} x^{2} - 1} {\left (\frac {x}{a \mathrm {sgn}\left (a x + 1\right )} + \frac {2}{a^{2} \mathrm {sgn}\left (a x + 1\right )}\right )} - \frac {\log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{2 \, a {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int e^{\coth ^{-1}(a x)} x \, dx=\frac {3\,\sqrt {\frac {a\,x-1}{a\,x+1}}-{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}}+\frac {\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a^2} \]

[In]

int(x/((a*x - 1)/(a*x + 1))^(1/2),x)

[Out]

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