\(\int \log (-2+\sqrt {\frac {1+x}{x}}) \, dx\) [298]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 14, antiderivative size = 69 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{2} \log \left (1-\sqrt {1+\frac {1}{x}}\right )-\frac {1}{3} \log \left (2-\sqrt {1+\frac {1}{x}}\right )-\frac {1}{6} \log \left (1+\sqrt {1+\frac {1}{x}}\right )+x \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \]

[Out]

1/2*ln(1-(1+1/x)^(1/2))-1/3*ln(2-(1+1/x)^(1/2))-1/6*ln(1+(1+1/x)^(1/2))+x*ln(-2+((1+x)/x)^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2628, 720, 31, 647} \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{2} \log \left (1-\sqrt {\frac {1}{x}+1}\right )-\frac {1}{3} \log \left (2-\sqrt {\frac {1}{x}+1}\right )-\frac {1}{6} \log \left (\sqrt {\frac {1}{x}+1}+1\right )+x \log \left (\sqrt {\frac {x+1}{x}}-2\right ) \]

[In]

Int[Log[-2 + Sqrt[(1 + x)/x]],x]

[Out]

Log[1 - Sqrt[1 + x^(-1)]]/2 - Log[2 - Sqrt[1 + x^(-1)]]/3 - Log[1 + Sqrt[1 + x^(-1)]]/6 + x*Log[-2 + Sqrt[(1 +
 x)/x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 720

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
 x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2628

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/u), x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps \begin{align*} \text {integral}& = x \log \left (-2+\sqrt {\frac {1+x}{x}}\right )-\int \frac {1}{-2+\left (-2+4 \sqrt {1+\frac {1}{x}}\right ) x} \, dx \\ & = x \log \left (-2+\sqrt {\frac {1+x}{x}}\right )-\text {Subst}\left (\int \frac {1}{(-2+x) \left (-1+x^2\right )} \, dx,x,\sqrt {1+\frac {1}{x}}\right ) \\ & = x \log \left (-2+\sqrt {\frac {1+x}{x}}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-2+x} \, dx,x,\sqrt {1+\frac {1}{x}}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {-2-x}{-1+x^2} \, dx,x,\sqrt {1+\frac {1}{x}}\right ) \\ & = -\frac {1}{3} \log \left (2-\sqrt {1+\frac {1}{x}}\right )+x \log \left (-2+\sqrt {\frac {1+x}{x}}\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\sqrt {1+\frac {1}{x}}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-1+x} \, dx,x,\sqrt {1+\frac {1}{x}}\right ) \\ & = \frac {1}{2} \log \left (1-\sqrt {1+\frac {1}{x}}\right )-\frac {1}{3} \log \left (2-\sqrt {1+\frac {1}{x}}\right )-\frac {1}{6} \log \left (1+\sqrt {1+\frac {1}{x}}\right )+x \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.93 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{6} \left (-6 \text {arctanh}\left (3-2 \sqrt {1+\frac {1}{x}}\right )+\log \left (2-\sqrt {1+\frac {1}{x}}\right )+6 x \log \left (-2+\sqrt {1+\frac {1}{x}}\right )-\log \left (1+\sqrt {1+\frac {1}{x}}\right )\right ) \]

[In]

Integrate[Log[-2 + Sqrt[(1 + x)/x]],x]

[Out]

(-6*ArcTanh[3 - 2*Sqrt[1 + x^(-1)]] + Log[2 - Sqrt[1 + x^(-1)]] + 6*x*Log[-2 + Sqrt[1 + x^(-1)]] - Log[1 + Sqr
t[1 + x^(-1)]])/6

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.57

method result size
default \(x \ln \left (-2+\sqrt {\frac {x +1}{x}}\right )-\frac {3 \sqrt {\frac {x +1}{x}}\, x \ln \left (-3 x +1\right )-\sqrt {9}\, \ln \left (\frac {4 \sqrt {9}\, \sqrt {x^{2}+x}+15 x +3}{9 x -3}\right ) \sqrt {x \left (x +1\right )}+6 \ln \left (\frac {1}{2}+x +\sqrt {x^{2}+x}\right ) \sqrt {x \left (x +1\right )}}{18 \sqrt {\frac {x +1}{x}}\, x}\) \(108\)
parts \(x \ln \left (-2+\sqrt {\frac {x +1}{x}}\right )-\frac {3 \sqrt {\frac {x +1}{x}}\, x \ln \left (-3 x +1\right )-\sqrt {9}\, \ln \left (\frac {4 \sqrt {9}\, \sqrt {x^{2}+x}+15 x +3}{9 x -3}\right ) \sqrt {x \left (x +1\right )}+6 \ln \left (\frac {1}{2}+x +\sqrt {x^{2}+x}\right ) \sqrt {x \left (x +1\right )}}{18 \sqrt {\frac {x +1}{x}}\, x}\) \(108\)

[In]

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

[Out]

x*ln(-2+((x+1)/x)^(1/2))-1/18/((x+1)/x)^(1/2)/x*(3*((x+1)/x)^(1/2)*x*ln(-3*x+1)-9^(1/2)*ln(1/3*(4*9^(1/2)*(x^2
+x)^(1/2)+15*x+3)/(3*x-1))*(x*(x+1))^(1/2)+6*ln(1/2+x+(x^2+x)^(1/2))*(x*(x+1))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.70 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{3} \, {\left (3 \, x - 1\right )} \log \left (\sqrt {\frac {x + 1}{x}} - 2\right ) - \frac {1}{6} \, \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {x + 1}{x}} - 1\right ) \]

[In]

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

[Out]

1/3*(3*x - 1)*log(sqrt((x + 1)/x) - 2) - 1/6*log(sqrt((x + 1)/x) + 1) + 1/2*log(sqrt((x + 1)/x) - 1)

Sympy [A] (verification not implemented)

Time = 46.32 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.77 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=x \log {\left (\sqrt {\frac {x + 1}{x}} - 2 \right )} - \frac {\log {\left (\sqrt {1 + \frac {1}{x}} - 2 \right )}}{3} + \frac {\log {\left (\sqrt {1 + \frac {1}{x}} - 1 \right )}}{2} - \frac {\log {\left (\sqrt {1 + \frac {1}{x}} + 1 \right )}}{6} \]

[In]

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

[Out]

x*log(sqrt((x + 1)/x) - 2) - log(sqrt(1 + 1/x) - 2)/3 + log(sqrt(1 + 1/x) - 1)/2 - log(sqrt(1 + 1/x) + 1)/6

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {\log \left (\sqrt {\frac {x + 1}{x}} - 2\right )}{\frac {x + 1}{x} - 1} - \frac {1}{6} \, \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {x + 1}{x}} - 1\right ) - \frac {1}{3} \, \log \left (\sqrt {\frac {x + 1}{x}} - 2\right ) \]

[In]

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

[Out]

log(sqrt((x + 1)/x) - 2)/((x + 1)/x - 1) - 1/6*log(sqrt((x + 1)/x) + 1) + 1/2*log(sqrt((x + 1)/x) - 1) - 1/3*l
og(sqrt((x + 1)/x) - 2)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=x \log \left (\sqrt {\frac {x + 1}{x}} - 2\right ) + \frac {\log \left ({\left | -x + \sqrt {x^{2} + x} + 1 \right |}\right )}{6 \, \mathrm {sgn}\left (x\right )} + \frac {\log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right )}{3 \, \mathrm {sgn}\left (x\right )} - \frac {\log \left ({\left | -3 \, x + 3 \, \sqrt {x^{2} + x} - 1 \right |}\right )}{6 \, \mathrm {sgn}\left (x\right )} - \frac {1}{6} \, \log \left ({\left | 3 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

x*log(sqrt((x + 1)/x) - 2) + 1/6*log(abs(-x + sqrt(x^2 + x) + 1))/sgn(x) + 1/3*log(abs(-2*x + 2*sqrt(x^2 + x)
- 1))/sgn(x) - 1/6*log(abs(-3*x + 3*sqrt(x^2 + x) - 1))/sgn(x) - 1/6*log(abs(3*x - 1))

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \log \left (-2+\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {\ln \left (5-5\,\sqrt {\frac {x+1}{x}}\right )}{2}-\frac {\ln \left (\frac {\sqrt {\frac {x+1}{x}}}{9}+\frac {1}{9}\right )}{6}-\frac {\ln \left (\frac {10}{9}-\frac {5\,\sqrt {\frac {x+1}{x}}}{9}\right )}{3}+x\,\ln \left (\sqrt {\frac {x+1}{x}}-2\right ) \]

[In]

int(log(((x + 1)/x)^(1/2) - 2),x)

[Out]

log(5 - 5*((x + 1)/x)^(1/2))/2 - log(((x + 1)/x)^(1/2)/9 + 1/9)/6 - log(10/9 - (5*((x + 1)/x)^(1/2))/9)/3 + x*
log(((x + 1)/x)^(1/2) - 2)