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

Optimal. Leaf size=38 \[ -\sqrt{\frac{1}{a^2 x^2}+1}+\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )-\frac{1}{a x} \]

[Out]

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

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Rubi [A]  time = 0.034632, antiderivative size = 38, 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 = {6336, 30, 266, 50, 63, 208} \[ -\sqrt{\frac{1}{a^2 x^2}+1}+\tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )-\frac{1}{a x} \]

Antiderivative was successfully verified.

[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 6336

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]

Rule 30

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

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 50

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 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 \frac{e^{\text{csch}^{-1}(a x)}}{x} \, dx &=\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} \operatorname{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} \operatorname{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 \operatorname{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}+\tanh ^{-1}\left (\sqrt{1+\frac{1}{a^2 x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0275416, size = 42, normalized size = 1.11 \[ -\sqrt{\frac{1}{a^2 x^2}+1}+\log \left (x \left (\sqrt{\frac{1}{a^2 x^2}+1}+1\right )\right )-\frac{1}{a x} \]

Warning: Unable to verify antiderivative.

[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]

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Maple [B]  time = 0.18, size = 107, normalized size = 2.8 \begin{align*}{\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}}}}{x}^{2}{a}^{2}+\ln \left ( x+\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}} \right ) x \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}}-{\frac{1}{ax}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*x^2*a^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

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Maxima [A]  time = 1.00055, size = 73, normalized size = 1.92 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 3.01638, size = 142, normalized size = 3.74 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 4.26471, size = 41, normalized size = 1.08 \begin{align*} - \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError