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

Optimal. Leaf size=42 \[ -\frac {1}{2} i \text {Li}_2\left (-i \sqrt {x}\right )+\frac {1}{2} i \text {Li}_2\left (i \sqrt {x}\right )+\frac {1}{4} \pi \log (x) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 42, normalized size of antiderivative = 1.00, 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.20, size = 84, normalized size = 2.00 \[ -\log (x) \tan ^{-1}\left (\sqrt {x}-\sqrt {x+1}\right )+\frac {1}{4} i \left (-2 \text {Li}_2\left (-i \sqrt {x}\right )+2 \text {Li}_2\left (i \sqrt {x}\right )+\left (\log \left (1-i \sqrt {x}\right )-\log \left (1+i \sqrt {x}\right )\right ) \log (x)\right ) \]

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]])

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (\sqrt {x + 1} - \sqrt {x}\right )}{x}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\arctan \left (-\sqrt {x + 1} + \sqrt {x}\right )}{x}\,{d x} \]

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)

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maple [B]  time = 1.40, size = 194, normalized size = 4.62 \[ 2 \arctan \left (\sqrt {x}-\sqrt {x +1}\right ) \ln \left (1-\frac {\left (1+i \left (\sqrt {x}-\sqrt {x +1}\right )\right )^{4}}{\left (\left (\sqrt {x}-\sqrt {x +1}\right )^{2}+1\right )^{2}}\right )-2 \arctan \left (\sqrt {x}-\sqrt {x +1}\right ) \ln \left (1+\frac {\left (1+i \left (\sqrt {x}-\sqrt {x +1}\right )\right )^{4}}{\left (\left (\sqrt {x}-\sqrt {x +1}\right )^{2}+1\right )^{2}}\right )+\frac {i \dilog \left (1+\frac {\left (1+i \left (\sqrt {x}-\sqrt {x +1}\right )\right )^{4}}{\left (\left (\sqrt {x}-\sqrt {x +1}\right )^{2}+1\right )^{2}}\right )}{2}-\frac {i \dilog \left (1-\frac {\left (1+i \left (\sqrt {x}-\sqrt {x +1}\right )\right )^{4}}{\left (\left (\sqrt {x}-\sqrt {x +1}\right )^{2}+1\right )^{2}}\right )}{2} \]

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)))^4/((x^(1/2)-(x+1)^(1/2))^2+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*dilog(1+(1+I*(x^(1/2)-
(x+1)^(1/2)))^4/((x^(1/2)-(x+1)^(1/2))^2+1)^2)-1/2*I*dilog(1-(1+I*(x^(1/2)-(x+1)^(1/2)))^4/((x^(1/2)-(x+1)^(1/
2))^2+1)^2)

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maxima [A]  time = 0.43, size = 43, normalized size = 1.02 \[ \frac {1}{4} \, \pi \log \left (x + 1\right ) + \arctan \left (\sqrt {x + 1} - \sqrt {x}\right ) \log \relax (x) + \frac {1}{2} i \, {\rm Li}_2\left (i \, \sqrt {x} + 1\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-i \, \sqrt {x} + 1\right ) \]

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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