3.266 \(\int \frac {\log (1+\sqrt {x}-x)}{x} \, dx\)

Optimal. Leaf size=122 \[ 2 \text {Li}_2\left (1-\frac {2 \sqrt {x}}{1+\sqrt {5}}\right )-2 \text {Li}_2\left (\frac {2 \sqrt {x}}{1-\sqrt {5}}\right )-2 \log \left (\frac {1}{2} \left (1+\sqrt {5}\right )\right ) \log \left (-2 \sqrt {x}+\sqrt {5}+1\right )-2 \log \left (1-\frac {2 \sqrt {x}}{1-\sqrt {5}}\right ) \log \left (\sqrt {x}\right )+2 \log \left (-x+\sqrt {x}+1\right ) \log \left (\sqrt {x}\right ) \]

[Out]

-2*ln(1/2+1/2*5^(1/2))*ln(1+5^(1/2)-2*x^(1/2))+ln(x)*ln(1-x+x^(1/2))-ln(x)*ln(1-2*x^(1/2)/(-5^(1/2)+1))-2*poly
log(2,2*x^(1/2)/(-5^(1/2)+1))+2*polylog(2,1-2*x^(1/2)/(5^(1/2)+1))

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Rubi [A]  time = 0.15, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {2530, 2524, 2357, 2317, 2391, 2316, 2315} \[ 2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {x}}{1+\sqrt {5}}\right )-2 \text {PolyLog}\left (2,\frac {2 \sqrt {x}}{1-\sqrt {5}}\right )-2 \log \left (\frac {1}{2} \left (1+\sqrt {5}\right )\right ) \log \left (-2 \sqrt {x}+\sqrt {5}+1\right )-2 \log \left (1-\frac {2 \sqrt {x}}{1-\sqrt {5}}\right ) \log \left (\sqrt {x}\right )+2 \log \left (-x+\sqrt {x}+1\right ) \log \left (\sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[1 + Sqrt[x] - x]/x,x]

[Out]

-2*Log[(1 + Sqrt[5])/2]*Log[1 + Sqrt[5] - 2*Sqrt[x]] - 2*Log[1 - (2*Sqrt[x])/(1 - Sqrt[5])]*Log[Sqrt[x]] + 2*L
og[1 + Sqrt[x] - x]*Log[Sqrt[x]] + 2*PolyLog[2, 1 - (2*Sqrt[x])/(1 + Sqrt[5])] - 2*PolyLog[2, (2*Sqrt[x])/(1 -
Sqrt[5])]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 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 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2530

Int[((a_.) + Log[u_]*(b_.))*(RFx_), x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[RFx*(a + b*Log[u]
), x]}, Dist[lst[[2]]*lst[[4]], Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x] /;  !FalseQ[lst]] /; Fre
eQ[{a, b}, x] && RationalFunctionQ[RFx, x]

Rubi steps

\begin {align*} \int \frac {\log \left (1+\sqrt {x}-x\right )}{x} \, dx &=2 \operatorname {Subst}\left (\int \frac {\log \left (1+x-x^2\right )}{x} \, dx,x,\sqrt {x}\right )\\ &=2 \log \left (1+\sqrt {x}-x\right ) \log \left (\sqrt {x}\right )-2 \operatorname {Subst}\left (\int \frac {(1-2 x) \log (x)}{1+x-x^2} \, dx,x,\sqrt {x}\right )\\ &=2 \log \left (1+\sqrt {x}-x\right ) \log \left (\sqrt {x}\right )-2 \operatorname {Subst}\left (\int \left (-\frac {2 \log (x)}{1-\sqrt {5}-2 x}-\frac {2 \log (x)}{1+\sqrt {5}-2 x}\right ) \, dx,x,\sqrt {x}\right )\\ &=2 \log \left (1+\sqrt {x}-x\right ) \log \left (\sqrt {x}\right )+4 \operatorname {Subst}\left (\int \frac {\log (x)}{1-\sqrt {5}-2 x} \, dx,x,\sqrt {x}\right )+4 \operatorname {Subst}\left (\int \frac {\log (x)}{1+\sqrt {5}-2 x} \, dx,x,\sqrt {x}\right )\\ &=-2 \log \left (\frac {1}{2} \left (1+\sqrt {5}\right )\right ) \log \left (1+\sqrt {5}-2 \sqrt {x}\right )-2 \log \left (1-\frac {2 \sqrt {x}}{1-\sqrt {5}}\right ) \log \left (\sqrt {x}\right )+2 \log \left (1+\sqrt {x}-x\right ) \log \left (\sqrt {x}\right )+2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {2 x}{1-\sqrt {5}}\right )}{x} \, dx,x,\sqrt {x}\right )+4 \operatorname {Subst}\left (\int \frac {\log \left (\frac {2 x}{1+\sqrt {5}}\right )}{1+\sqrt {5}-2 x} \, dx,x,\sqrt {x}\right )\\ &=-2 \log \left (\frac {1}{2} \left (1+\sqrt {5}\right )\right ) \log \left (1+\sqrt {5}-2 \sqrt {x}\right )-2 \log \left (1-\frac {2 \sqrt {x}}{1-\sqrt {5}}\right ) \log \left (\sqrt {x}\right )+2 \log \left (1+\sqrt {x}-x\right ) \log \left (\sqrt {x}\right )+2 \text {Li}_2\left (1-\frac {2 \sqrt {x}}{1+\sqrt {5}}\right )-2 \text {Li}_2\left (\frac {2 \sqrt {x}}{1-\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 121, normalized size = 0.99 \[ 2 \text {Li}_2\left (\frac {-2 \sqrt {x}+\sqrt {5}+1}{1+\sqrt {5}}\right )-2 \text {Li}_2\left (-\frac {2 \sqrt {x}}{-1+\sqrt {5}}\right )-2 \log \left (\frac {1}{2} \left (1+\sqrt {5}\right )\right ) \log \left (-2 \sqrt {x}+\sqrt {5}+1\right )+\left (\log \left (\sqrt {5}-1\right )-\log \left (2 \sqrt {x}+\sqrt {5}-1\right )\right ) \log (x)+\log \left (-x+\sqrt {x}+1\right ) \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[1 + Sqrt[x] - x]/x,x]

[Out]

-2*Log[(1 + Sqrt[5])/2]*Log[1 + Sqrt[5] - 2*Sqrt[x]] + (Log[-1 + Sqrt[5]] - Log[-1 + Sqrt[5] + 2*Sqrt[x]])*Log
[x] + Log[1 + Sqrt[x] - x]*Log[x] + 2*PolyLog[2, (1 + Sqrt[5] - 2*Sqrt[x])/(1 + Sqrt[5])] - 2*PolyLog[2, (-2*S
qrt[x])/(-1 + Sqrt[5])]

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (-x + \sqrt {x} + 1\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(-x + sqrt(x) + 1)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(-x + sqrt(x) + 1)/x, x)

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maple [A]  time = 0.07, size = 102, normalized size = 0.84 \[ -\ln \relax (x ) \ln \left (\frac {2 \sqrt {x}-1+\sqrt {5}}{\sqrt {5}-1}\right )-\ln \relax (x ) \ln \left (\frac {-2 \sqrt {x}+1+\sqrt {5}}{\sqrt {5}+1}\right )+\ln \relax (x ) \ln \left (-x +\sqrt {x}+1\right )-2 \dilog \left (\frac {2 \sqrt {x}-1+\sqrt {5}}{\sqrt {5}-1}\right )-2 \dilog \left (\frac {-2 \sqrt {x}+1+\sqrt {5}}{\sqrt {5}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*ln(1-x+x^(1/2))-ln(x)*ln((1+5^(1/2)-2*x^(1/2))/(5^(1/2)+1))-ln(x)*ln((-1+5^(1/2)+2*x^(1/2))/(5^(1/2)-1))
-2*dilog((1+5^(1/2)-2*x^(1/2))/(5^(1/2)+1))-2*dilog((-1+5^(1/2)+2*x^(1/2))/(5^(1/2)-1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(1-x+x^(1/2))/x,x, algorithm="maxima")

[Out]

integrate(log(-x + sqrt(x) + 1)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (\sqrt {x}-x+1\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x^(1/2) - x + 1)/x,x)

[Out]

int(log(x^(1/2) - x + 1)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(sqrt(x) - x + 1)/x, x)

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