3.157 \(\int \frac{\tan ^{-1}(\sqrt{x})}{x} \, dx\)

Optimal. Leaf size=31 \[ i \text{PolyLog}\left (2,-i \sqrt{x}\right )-i \text{PolyLog}\left (2,i \sqrt{x}\right ) \]

[Out]

I*PolyLog[2, (-I)*Sqrt[x]] - I*PolyLog[2, I*Sqrt[x]]

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Rubi [A]  time = 0.0353228, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {5031, 4848, 2391} \[ i \text{PolyLog}\left (2,-i \sqrt{x}\right )-i \text{PolyLog}\left (2,i \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[Sqrt[x]]/x,x]

[Out]

I*PolyLog[2, (-I)*Sqrt[x]] - I*PolyLog[2, I*Sqrt[x]]

Rule 5031

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTan[c*x])^p
/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

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

Mathematica [A]  time = 0.0044137, size = 31, normalized size = 1. \[ i \text{PolyLog}\left (2,-i \sqrt{x}\right )-i \text{PolyLog}\left (2,i \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Sqrt[x]]/x,x]

[Out]

I*PolyLog[2, (-I)*Sqrt[x]] - I*PolyLog[2, I*Sqrt[x]]

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Maple [B]  time = 0.036, size = 61, normalized size = 2. \begin{align*} \ln \left ( x \right ) \arctan \left ( \sqrt{x} \right ) +{\frac{i}{2}}\ln \left ( x \right ) \ln \left ( 1+i\sqrt{x} \right ) -{\frac{i}{2}}\ln \left ( x \right ) \ln \left ( 1-i\sqrt{x} \right ) +i{\it dilog} \left ( 1+i\sqrt{x} \right ) -i{\it dilog} \left ( 1-i\sqrt{x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*arctan(x^(1/2))+1/2*I*ln(x)*ln(1+I*x^(1/2))-1/2*I*ln(x)*ln(1-I*x^(1/2))+I*dilog(1+I*x^(1/2))-I*dilog(1-I
*x^(1/2))

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Maxima [B]  time = 1.53405, size = 47, normalized size = 1.52 \begin{align*} -\frac{1}{2} \, \pi \log \left (x + 1\right ) + \arctan \left (\sqrt{x}\right ) \log \left (x\right ) - i \,{\rm Li}_2\left (i \, \sqrt{x} + 1\right ) + i \,{\rm Li}_2\left (-i \, \sqrt{x} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*pi*log(x + 1) + arctan(sqrt(x))*log(x) - I*dilog(I*sqrt(x) + 1) + I*dilog(-I*sqrt(x) + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(sqrt(x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(sqrt(x))/x, x)