3.24 \(\int \frac{\text{sech}^{-1}(\sqrt{x})}{x} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0952382, 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 = {6281, 5660, 3718, 2190, 2279, 2391} \[ -\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (\sqrt{x}\right )}\right )+\text{sech}^{-1}\left (\sqrt{x}\right )^2-2 \text{sech}^{-1}\left (\sqrt{x}\right ) \log \left (e^{2 \text{sech}^{-1}\left (\sqrt{x}\right )}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[Sqrt[x]]/x,x]

[Out]

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

Rule 6281

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

Rule 5660

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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{sech}^{-1}\left (\sqrt{x}\right )}{x} \, dx &=2 \operatorname{Subst}\left (\int \frac{\text{sech}^{-1}(x)}{x} \, dx,x,\sqrt{x}\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{\cosh ^{-1}(x)}{x} \, dx,x,\frac{1}{\sqrt{x}}\right )\right )\\ &=-\left (2 \operatorname{Subst}\left (\int x \tanh (x) \, dx,x,\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )\right )\right )\\ &=\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-4 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1+e^{2 x}} \, dx,x,\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )\right )\\ &=\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )+2 \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )\right )\\ &=\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )+\operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )\\ &=\cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )^2-2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )-\text{Li}_2\left (-e^{2 \cosh ^{-1}\left (\frac{1}{\sqrt{x}}\right )}\right )\\ \end{align*}

Mathematica [A]  time = 0.0359203, size = 45, normalized size = 0.98 \[ \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}\left (\sqrt{x}\right )}\right )-\text{sech}^{-1}\left (\sqrt{x}\right ) \left (\text{sech}^{-1}\left (\sqrt{x}\right )+2 \log \left (e^{-2 \text{sech}^{-1}\left (\sqrt{x}\right )}+1\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSech[Sqrt[x]]/x,x]

[Out]

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

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Maple [A]  time = 0.16, size = 65, normalized size = 1.4 \begin{align*} \left ({\rm arcsech} \left (\sqrt{x}\right ) \right ) ^{2}-2\,{\rm arcsech} \left (\sqrt{x}\right )\ln \left ( 1+ \left ({\frac{1}{\sqrt{x}}}+\sqrt{-1+{\frac{1}{\sqrt{x}}}}\sqrt{1+{\frac{1}{\sqrt{x}}}} \right ) ^{2} \right ) -{\it polylog} \left ( 2,- \left ({\frac{1}{\sqrt{x}}}+\sqrt{-1+{\frac{1}{\sqrt{x}}}}\sqrt{1+{\frac{1}{\sqrt{x}}}} \right ) ^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arcsech(x^(1/2))^2-2*arcsech(x^(1/2))*ln(1+(1/x^(1/2)+(-1+1/x^(1/2))^(1/2)*(1+1/x^(1/2))^(1/2))^2)-polylog(2,-
(1/x^(1/2)+(-1+1/x^(1/2))^(1/2)*(1+1/x^(1/2))^(1/2))^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{4} \, \log \left (x\right )^{2} + \log \left (x\right ) \log \left (\sqrt{\sqrt{x} + 1} \sqrt{-\sqrt{x} + 1} + 1\right ) - \log \left (\sqrt{x} + 1\right ) \log \left (\sqrt{x}\right ) - \log \left (\sqrt{x}\right ) \log \left (-\sqrt{x} + 1\right ) -{\rm Li}_2\left (-\sqrt{x}\right ) -{\rm Li}_2\left (\sqrt{x}\right ) + \int \frac{\log \left (x\right )}{2 \,{\left ({\left (x - 1\right )} e^{\left (\frac{1}{2} \, \log \left (\sqrt{x} + 1\right ) + \frac{1}{2} \, \log \left (-\sqrt{x} + 1\right )\right )} + x - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(x)^2 + log(x)*log(sqrt(sqrt(x) + 1)*sqrt(-sqrt(x) + 1) + 1) - log(sqrt(x) + 1)*log(sqrt(x)) - log(sqr
t(x))*log(-sqrt(x) + 1) - dilog(-sqrt(x)) - dilog(sqrt(x)) + integrate(1/2*log(x)/((x - 1)*e^(1/2*log(sqrt(x)
+ 1) + 1/2*log(-sqrt(x) + 1)) + x - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asech(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(sqrt(x))/x, x)