3.295 \(\int \frac {\sinh ^{-1}(\sqrt {x})}{x} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5890, 3716, 2190, 2279, 2391} \[ \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\sqrt {x}\right )}\right )-\sinh ^{-1}\left (\sqrt {x}\right )^2+2 \sinh ^{-1}\left (\sqrt {x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\sqrt {x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[Sqrt[x]]/x,x]

[Out]

-ArcSinh[Sqrt[x]]^2 + 2*ArcSinh[Sqrt[x]]*Log[1 - E^(2*ArcSinh[Sqrt[x]])] + PolyLog[2, E^(2*ArcSinh[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 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 5890

Int[ArcSinh[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Coth[x], x], x, ArcSinh[a*x^p]],
x] /; FreeQ[{a, p}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\sqrt {x}\right )}\right )-\sinh ^{-1}\left (\sqrt {x}\right )^2+2 \sinh ^{-1}\left (\sqrt {x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\sqrt {x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[Sqrt[x]]/x,x]

[Out]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(sqrt(x))/x, x)

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maple [A]  time = 0.07, size = 78, normalized size = 1.70 \[ -\arcsinh \left (\sqrt {x}\right )^{2}+2 \arcsinh \left (\sqrt {x}\right ) \ln \left (1+\sqrt {x}+\sqrt {1+x}\right )+2 \polylog \left (2, -\sqrt {x}-\sqrt {1+x}\right )+2 \arcsinh \left (\sqrt {x}\right ) \ln \left (1-\sqrt {x}-\sqrt {1+x}\right )+2 \polylog \left (2, \sqrt {x}+\sqrt {1+x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(x^(1/2))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(sqrt(x))/x, x)

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