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

Optimal. Leaf size=42 \[ -\frac{1}{2} i \text{PolyLog}\left (2,-i \sqrt{x}\right )+\frac{1}{2} i \text{PolyLog}\left (2,i \sqrt{x}\right )+\frac{1}{4} \pi \log (x) \]

[Out]

(Pi*Log[x])/4 - (I/2)*PolyLog[2, (-I)*Sqrt[x]] + (I/2)*PolyLog[2, I*Sqrt[x]]

________________________________________________________________________________________

Rubi [A]  time = 0.0422181, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {5159, 29, 5031, 4848, 2391} \[ -\frac{1}{2} i \text{PolyLog}\left (2,-i \sqrt{x}\right )+\frac{1}{2} i \text{PolyLog}\left (2,i \sqrt{x}\right )+\frac{1}{4} \pi \log (x) \]

Antiderivative was successfully verified.

[In]

Int[-(ArcTan[Sqrt[x] - Sqrt[1 + x]]/x),x]

[Out]

(Pi*Log[x])/4 - (I/2)*PolyLog[2, (-I)*Sqrt[x]] + (I/2)*PolyLog[2, I*Sqrt[x]]

Rule 5159

Int[ArcTan[(v_) + (s_.)*Sqrt[w_]]*(u_.), x_Symbol] :> Dist[(Pi*s)/4, Int[u, x], x] + Dist[1/2, Int[u*ArcTan[v]
, x], x] /; EqQ[s^2, 1] && EqQ[w, v^2 + 1]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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}-\sqrt{1+x}\right )}{x} \, dx &=-\left (\frac{1}{2} \int \frac{\tan ^{-1}\left (\sqrt{x}\right )}{x} \, dx\right )+\frac{1}{4} \pi \int \frac{1}{x} \, dx\\ &=\frac{1}{4} \pi \log (x)-\operatorname{Subst}\left (\int \frac{\tan ^{-1}(x)}{x} \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{4} \pi \log (x)-\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,\sqrt{x}\right )+\frac{1}{2} i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,\sqrt{x}\right )\\ &=\frac{1}{4} \pi \log (x)-\frac{1}{2} i \text{Li}_2\left (-i \sqrt{x}\right )+\frac{1}{2} i \text{Li}_2\left (i \sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.184779, size = 84, normalized size = 2. \[ -\log (x) \tan ^{-1}\left (\sqrt{x}-\sqrt{x+1}\right )+\frac{1}{4} i \left (-2 \text{PolyLog}\left (2,-i \sqrt{x}\right )+2 \text{PolyLog}\left (2,i \sqrt{x}\right )+\left (\log \left (1-i \sqrt{x}\right )-\log \left (1+i \sqrt{x}\right )\right ) \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[-(ArcTan[Sqrt[x] - Sqrt[1 + x]]/x),x]

[Out]

-(ArcTan[Sqrt[x] - Sqrt[1 + x]]*Log[x]) + (I/4)*((Log[1 - I*Sqrt[x]] - Log[1 + I*Sqrt[x]])*Log[x] - 2*PolyLog[
2, (-I)*Sqrt[x]] + 2*PolyLog[2, I*Sqrt[x]])

________________________________________________________________________________________

Maple [B]  time = 1.115, size = 374, normalized size = 8.9 \begin{align*} 2\,\arctan \left ( \sqrt{x}-\sqrt{x+1} \right ) \ln \left ( 1-{\frac{1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) }{\sqrt{ \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1}}} \right ) -2\,i{\it polylog} \left ( 2,{ \left ( 1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) \right ){\frac{1}{\sqrt{ \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1}}}} \right ) +2\,\arctan \left ( \sqrt{x}-\sqrt{x+1} \right ) \ln \left ( 1+{\frac{1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) }{\sqrt{ \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1}}} \right ) -2\,i{\it polylog} \left ( 2,-{ \left ( 1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) \right ){\frac{1}{\sqrt{ \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1}}}} \right ) -2\,\arctan \left ( \sqrt{x}-\sqrt{x+1} \right ) \ln \left ( 1+{\frac{ \left ( 1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) \right ) ^{4}}{ \left ( \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1 \right ) ^{2}}} \right ) +{\frac{i}{2}}{\it polylog} \left ( 2,-{ \left ( 1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) \right ) ^{4} \left ( \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1 \right ) ^{-2}} \right ) +2\,\arctan \left ( \sqrt{x}-\sqrt{x+1} \right ) \ln \left ({\frac{ \left ( 1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) \right ) ^{2}}{ \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1}}+1 \right ) -i{\it polylog} \left ( 2,-{ \left ( 1+i \left ( \sqrt{x}-\sqrt{x+1} \right ) \right ) ^{2} \left ( \left ( \sqrt{x}-\sqrt{x+1} \right ) ^{2}+1 \right ) ^{-1}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(x^(1/2)-(x+1)^(1/2))*ln(1-(1+I*(x^(1/2)-(x+1)^(1/2)))/((x^(1/2)-(x+1)^(1/2))^2+1)^(1/2))-2*I*polylog(
2,(1+I*(x^(1/2)-(x+1)^(1/2)))/((x^(1/2)-(x+1)^(1/2))^2+1)^(1/2))+2*arctan(x^(1/2)-(x+1)^(1/2))*ln(1+(1+I*(x^(1
/2)-(x+1)^(1/2)))/((x^(1/2)-(x+1)^(1/2))^2+1)^(1/2))-2*I*polylog(2,-(1+I*(x^(1/2)-(x+1)^(1/2)))/((x^(1/2)-(x+1
)^(1/2))^2+1)^(1/2))-2*arctan(x^(1/2)-(x+1)^(1/2))*ln(1+(1+I*(x^(1/2)-(x+1)^(1/2)))^4/((x^(1/2)-(x+1)^(1/2))^2
+1)^2)+1/2*I*polylog(2,-(1+I*(x^(1/2)-(x+1)^(1/2)))^4/((x^(1/2)-(x+1)^(1/2))^2+1)^2)+2*arctan(x^(1/2)-(x+1)^(1
/2))*ln((1+I*(x^(1/2)-(x+1)^(1/2)))^2/((x^(1/2)-(x+1)^(1/2))^2+1)+1)-I*polylog(2,-(1+I*(x^(1/2)-(x+1)^(1/2)))^
2/((x^(1/2)-(x+1)^(1/2))^2+1))

________________________________________________________________________________________

Maxima [A]  time = 1.54996, size = 58, normalized size = 1.38 \begin{align*} \frac{1}{4} \, \pi \log \left (x + 1\right ) + \arctan \left (\sqrt{x + 1} - \sqrt{x}\right ) \log \left (x\right ) + \frac{1}{2} i \,{\rm Li}_2\left (i \, \sqrt{x} + 1\right ) - \frac{1}{2} 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)-(1+x)^(1/2))/x,x, algorithm="maxima")

[Out]

1/4*pi*log(x + 1) + arctan(sqrt(x + 1) - sqrt(x))*log(x) + 1/2*I*dilog(I*sqrt(x) + 1) - 1/2*I*dilog(-I*sqrt(x)
+ 1)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (\sqrt{x + 1} - \sqrt{x}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arctan(sqrt(x + 1) - sqrt(x))/x, x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\arctan \left (-\sqrt{x + 1} + \sqrt{x}\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-arctan(-sqrt(x + 1) + sqrt(x))/x, x)