\(\int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx\) [32]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [F(-2)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 10, antiderivative size = 38 \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=-\sqrt {1+\frac {1}{a^2 x^2}}-\frac {1}{a x}+\text {arctanh}\left (\sqrt {1+\frac {1}{a^2 x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6471, 30, 272, 52, 65, 214} \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=\text {arctanh}\left (\sqrt {\frac {1}{a^2 x^2}+1}\right )-\sqrt {\frac {1}{a^2 x^2}+1}-\frac {1}{a x} \]

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 6471

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=-\sqrt {1+\frac {1}{a^2 x^2}}-\frac {1}{a x}+\log \left (\left (1+\sqrt {1+\frac {1}{a^2 x^2}}\right ) x\right ) \]

[In]

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

[Out]

-Sqrt[1 + 1/(a^2*x^2)] - 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. \(108\) vs. \(2(34)=68\).

Time = 0.05 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.87

method result size
default \(-\frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, \left (a^{2} \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}}-\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2} x^{2}-\ln \left (x +\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\right ) x \right )}{\sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}}-\frac {1}{a x}\) \(109\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=-\frac {a x \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right ) + a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + a x + 1}{a x} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=- \frac {a x}{\sqrt {a^{2} x^{2} + 1}} + \operatorname {asinh}{\left (a x \right )} - \frac {1}{a x} - \frac {1}{a x \sqrt {a^{2} x^{2} + 1}} \]

[In]

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

[Out]

-a*x/sqrt(a**2*x**2 + 1) + asinh(a*x) - 1/(a*x) - 1/(a*x*sqrt(a**2*x**2 + 1))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=-\sqrt {\frac {1}{a^{2} x^{2}} + 1} - \frac {1}{a x} + \frac {1}{2} \, \log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right ) \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [B] (verification not implemented)

Time = 5.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\text {csch}^{-1}(a x)}}{x} \, dx=\mathrm {atanh}\left (\sqrt {\frac {1}{a^2\,x^2}+1}\right )-\sqrt {\frac {1}{a^2\,x^2}+1}-\frac {1}{a\,x} \]

[In]

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

[Out]

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