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

Optimal. Leaf size=46 \[ -\text {Li}_2\left (-e^{2 \text {sech}^{-1}\left (\sqrt {x}\right )}\right )+\text {sech}^{-1}\left (\sqrt {x}\right )^2-2 \text {sech}^{-1}\left (\sqrt {x}\right ) \log \left (e^{2 \text {sech}^{-1}\left (\sqrt {x}\right )}+1\right ) \]

[Out]

arcsech(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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Rubi [A]  time = 0.10, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.04, size = 45, normalized size = 0.98 \[ \text {Li}_2\left (-e^{-2 \text {sech}^{-1}\left (\sqrt {x}\right )}\right )-\text {sech}^{-1}\left (\sqrt {x}\right ) \left (\text {sech}^{-1}\left (\sqrt {x}\right )+2 \log \left (e^{-2 \text {sech}^{-1}\left (\sqrt {x}\right )}+1\right )\right ) \]

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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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x}\,{d x} \]

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)

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maple [A]  time = 0.10, size = 65, normalized size = 1.41 \[ \mathrm {arcsech}\left (\sqrt {x}\right )^{2}-2 \,\mathrm {arcsech}\left (\sqrt {x}\right ) \ln \left (1+\left (\frac {1}{\sqrt {x}}+\sqrt {-1+\frac {1}{\sqrt {x}}}\, \sqrt {1+\frac {1}{\sqrt {x}}}\right )^{2}\right )-\polylog \left (2, -\left (\frac {1}{\sqrt {x}}+\sqrt {-1+\frac {1}{\sqrt {x}}}\, \sqrt {1+\frac {1}{\sqrt {x}}}\right )^{2}\right ) \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, \log \relax (x)^{2} + \log \relax (x) \log \left (\sqrt {\sqrt {x} + 1} \sqrt {-\sqrt {x} + 1} + 1\right ) - \log \left (\sqrt {x} + 1\right ) \log \left (\sqrt {x}\right ) - \log \left (\sqrt {x}\right ) \log \left (-\sqrt {x} + 1\right ) - {\rm Li}_2\left (-\sqrt {x}\right ) - {\rm Li}_2\left (\sqrt {x}\right ) + \int \frac {\log \relax (x)}{2 \, {\left ({\left (x - 1\right )} e^{\left (\frac {1}{2} \, \log \left (\sqrt {x} + 1\right ) + \frac {1}{2} \, \log \left (-\sqrt {x} + 1\right )\right )} + x - 1\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(1/x^(1/2))/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asech}{\left (\sqrt {x} \right )}}{x}\, dx \]

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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