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

Optimal. Leaf size=50 \[ -\frac {1}{2 \left (\sqrt {\frac {1}{x}+1}+1\right )}+x \log \left (\sqrt {\frac {x+1}{x}}+1\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt {\frac {x+1}{x}}\right ) \]

[Out]

1/2*arctanh(((1+x)/x)^(1/2))+x*ln(1+((1+x)/x)^(1/2))-1/2/(1+(1+1/x)^(1/2))

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Rubi [A]  time = 0.06, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2548, 12, 44, 207} \[ -\frac {1}{2 \left (\sqrt {\frac {1}{x}+1}+1\right )}+x \log \left (\sqrt {\frac {x+1}{x}}+1\right )+\frac {1}{2} \tanh ^{-1}\left (\sqrt {\frac {x+1}{x}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[1 + Sqrt[(1 + x)/x]],x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2548

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

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Mathematica [A]  time = 0.04, size = 53, normalized size = 1.06 \[ \frac {1}{4} \left (-2 \sqrt {\frac {1}{x}+1} x+2 x+4 x \log \left (\sqrt {\frac {1}{x}+1}+1\right )+\log \left (\left (2 \sqrt {\frac {1}{x}+1}+2\right ) x+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[1 + Sqrt[(1 + x)/x]],x]

[Out]

(2*x - 2*Sqrt[1 + x^(-1)]*x + 4*x*Log[1 + Sqrt[1 + x^(-1)]] + Log[1 + (2 + 2*Sqrt[1 + x^(-1)])*x])/4

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fricas [A]  time = 0.45, size = 49, normalized size = 0.98 \[ \frac {1}{4} \, {\left (4 \, x + 1\right )} \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) - \frac {1}{2} \, x \sqrt {\frac {x + 1}{x}} + \frac {1}{2} \, x - \frac {1}{4} \, \log \left (\sqrt {\frac {x + 1}{x}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x + 1)*log(sqrt((x + 1)/x) + 1) - 1/2*x*sqrt((x + 1)/x) + 1/2*x - 1/4*log(sqrt((x + 1)/x) - 1)

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giac [A]  time = 0.23, size = 53, normalized size = 1.06 \[ x \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) + \frac {1}{2} \, x - \frac {\log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right )}{4 \, \mathrm {sgn}\relax (x)} - \frac {\sqrt {x^{2} + x}}{2 \, \mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(sqrt((x + 1)/x) + 1) + 1/2*x - 1/4*log(abs(-2*x + 2*sqrt(x^2 + x) - 1))/sgn(x) - 1/2*sqrt(x^2 + x)/sgn(x
)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \ln \left (1+\sqrt {\frac {x +1}{x}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 68, normalized size = 1.36 \[ \frac {\log \left (\sqrt {\frac {x + 1}{x}} + 1\right )}{\frac {x + 1}{x} - 1} - \frac {1}{2 \, {\left (\sqrt {\frac {x + 1}{x}} + 1\right )}} + \frac {1}{4} \, \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) - \frac {1}{4} \, \log \left (\sqrt {\frac {x + 1}{x}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(sqrt((x + 1)/x) + 1)/((x + 1)/x - 1) - 1/2/(sqrt((x + 1)/x) + 1) + 1/4*log(sqrt((x + 1)/x) + 1) - 1/4*log(
sqrt((x + 1)/x) - 1)

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mupad [B]  time = 0.45, size = 38, normalized size = 0.76 \[ \frac {x}{2}+\frac {\mathrm {atanh}\left (\sqrt {\frac {1}{x}+1}\right )}{2}+x\,\ln \left (\sqrt {\frac {x+1}{x}}+1\right )-\frac {x\,\sqrt {\frac {1}{x}+1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(((x + 1)/x)^(1/2) + 1),x)

[Out]

x/2 + atanh((1/x + 1)^(1/2))/2 + x*log(((x + 1)/x)^(1/2) + 1) - (x*(1/x + 1)^(1/2))/2

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sympy [A]  time = 45.79, size = 53, normalized size = 1.06 \[ x \log {\left (\sqrt {\frac {x + 1}{x}} + 1 \right )} - \frac {\log {\left (\sqrt {1 + \frac {1}{x}} - 1 \right )}}{4} + \frac {\log {\left (\sqrt {1 + \frac {1}{x}} + 1 \right )}}{4} - \frac {1}{2 \left (\sqrt {1 + \frac {1}{x}} + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(sqrt((x + 1)/x) + 1) - log(sqrt(1 + 1/x) - 1)/4 + log(sqrt(1 + 1/x) + 1)/4 - 1/(2*(sqrt(1 + 1/x) + 1))

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