\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx\) [54]

Optimal result

Integrand size = 12, antiderivative size = 46 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=-\frac {4 a \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}-\csc ^{-1}(a x)+\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right ) \]

[Out]

-arccsc(a*x)+arctanh((1-1/a^2/x^2)^(1/2))-4*a*(1-1/a^2/x^2)^(1/2)/(a+1/x)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6304, 6874, 222, 272, 65, 214, 665} \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-\frac {4 a \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}-\csc ^{-1}(a x) \]

[In]

Int[1/(E^(3*ArcCoth[a*x])*x),x]

[Out]

(-4*a*Sqrt[1 - 1/(a^2*x^2)])/(a + x^(-1)) - ArcCsc[a*x] + ArcTanh[Sqrt[1 - 1/(a^2*x^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 222

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 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 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^2}{x \left (1+\frac {x}{a}\right ) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\text {Subst}\left (\int \left (\frac {1}{a \sqrt {1-\frac {x^2}{a^2}}}+\frac {1}{x \sqrt {1-\frac {x^2}{a^2}}}-\frac {4}{(a+x) \sqrt {1-\frac {x^2}{a^2}}}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = 4 \text {Subst}\left (\int \frac {1}{(a+x) \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a}-\text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {4 a \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}-\csc ^{-1}(a x)-\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {4 a \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}-\csc ^{-1}(a x)+a^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 {4 a \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}-\csc ^{-1}(a x)+\text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.20 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=-\frac {4 a \sqrt {1-\frac {1}{a^2 x^2}} x}{1+a x}-\arcsin \left (\frac {1}{a x}\right )+\log \left (\left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right ) \]

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*x),x]

[Out]

(-4*a*Sqrt[1 - 1/(a^2*x^2)]*x)/(1 + a*x) - ArcSin[1/(a*x)] + Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(42)=84\).

Time = 0.12 (sec) , antiderivative size = 366, normalized size of antiderivative = 7.96

method	result	size
default	\(-\frac {\left (\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{2} x^{2}+a^{2} x^{2} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-\ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{2} x^{2}+2 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x +2 a x \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-2 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -2 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+2 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}+\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}-a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right )+\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{\sqrt {a^{2}}\, \left (a x -1\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}}\)	\(366\)

[In]

int(((a*x-1)/(a*x+1))^(3/2)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=-4 \, \sqrt {\frac {a x - 1}{a x + 1}} + 2 \, \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x,x, algorithm="fricas")

[Out]

-4*sqrt((a*x - 1)/(a*x + 1)) + 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)

Sympy [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=\int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x}\, dx \]

[In]

integrate(((a*x-1)/(a*x+1))**(3/2)/x,x)

[Out]

Integral(((a*x - 1)/(a*x + 1))**(3/2)/x, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (42) = 84\).

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.93 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=a {\left (\frac {2 \, \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a} - \frac {4 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a}\right )} \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/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 - 4*sqrt((a*x - 1)/(a*x + 1))/a)

Giac [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=\int { \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x} \,d x } \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x,x, algorithm="giac")

[Out]

undef

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{x} \, dx=2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )+2\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )-4\,\sqrt {\frac {a\,x-1}{a\,x+1}} \]

[In]

int(((a*x - 1)/(a*x + 1))^(3/2)/x,x)

[Out]

2*atan(((a*x - 1)/(a*x + 1))^(1/2)) + 2*atanh(((a*x - 1)/(a*x + 1))^(1/2)) - 4*((a*x - 1)/(a*x + 1))^(1/2)