### 3.18 $$\int \frac{\text{csch}^{-1}(\sqrt{x})}{x} \, dx$$

Optimal. Leaf size=46 $-\text{PolyLog}\left (2,e^{2 \text{csch}^{-1}\left (\sqrt{x}\right )}\right )+\text{csch}^{-1}\left (\sqrt{x}\right )^2-2 \text{csch}^{-1}\left (\sqrt{x}\right ) \log \left (1-e^{2 \text{csch}^{-1}\left (\sqrt{x}\right )}\right )$

[Out]

ArcCsch[Sqrt[x]]^2 - 2*ArcCsch[Sqrt[x]]*Log[1 - E^(2*ArcCsch[Sqrt[x]])] - PolyLog[2, E^(2*ArcCsch[Sqrt[x]])]

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Rubi [A]  time = 0.0965395, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {6282, 5659, 3716, 2190, 2279, 2391} $-\text{PolyLog}\left (2,e^{2 \text{csch}^{-1}\left (\sqrt{x}\right )}\right )+\text{csch}^{-1}\left (\sqrt{x}\right )^2-2 \text{csch}^{-1}\left (\sqrt{x}\right ) \log \left (1-e^{2 \text{csch}^{-1}\left (\sqrt{x}\right )}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[ArcCsch[Sqrt[x]]/x,x]

[Out]

ArcCsch[Sqrt[x]]^2 - 2*ArcCsch[Sqrt[x]]*Log[1 - E^(2*ArcCsch[Sqrt[x]])] - PolyLog[2, E^(2*ArcCsch[Sqrt[x]])]

Rule 6282

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcSinh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rule 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\text{csch}^{-1}\left (\sqrt{x}\right )}{x} \, dx &=2 \operatorname{Subst}\left (\int \frac{\text{csch}^{-1}(x)}{x} \, dx,x,\sqrt{x}\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{x} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=-\left (2 \operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )\right )\right )\\ &=\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2+4 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )\right )\\ &=\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )+2 \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )\right )\\ &=\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )+\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )\\ &=\sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )-\text{Li}_2\left (e^{2 \sinh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0353073, size = 45, normalized size = 0.98 $\text{PolyLog}\left (2,e^{-2 \text{csch}^{-1}\left (\sqrt{x}\right )}\right )-\text{csch}^{-1}\left (\sqrt{x}\right ) \left (\text{csch}^{-1}\left (\sqrt{x}\right )+2 \log \left (1-e^{-2 \text{csch}^{-1}\left (\sqrt{x}\right )}\right )\right )$

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCsch[Sqrt[x]]/x,x]

[Out]

-(ArcCsch[Sqrt[x]]*(ArcCsch[Sqrt[x]] + 2*Log[1 - E^(-2*ArcCsch[Sqrt[x]])])) + PolyLog[2, E^(-2*ArcCsch[Sqrt[x]
])]

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Maple [F]  time = 0.17, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}{\rm arccsch} \left (\sqrt{x}\right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(arccsch(x^(1/2))/x,x)

[Out]

int(arccsch(x^(1/2))/x,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsch}\left (\sqrt{x}\right )}{x}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsch(sqrt(x))/x, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcsch}\left (\sqrt{x}\right )}{x}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccsch(sqrt(x))/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acsch}{\left (\sqrt{x} \right )}}{x}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(x**(1/2))/x,x)

[Out]

Integral(acsch(sqrt(x))/x, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcsch}\left (\sqrt{x}\right )}{x}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsch(sqrt(x))/x, x)