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

Optimal. Leaf size=21 \[ x \log \left (\sqrt {\frac {1}{x}+1}\right )+\frac {1}{2} \log (x+1) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2453, 2448, 263, 31} \[ x \log \left (\sqrt {\frac {1}{x}+1}\right )+\frac {1}{2} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[Log[Sqrt[(1 + x)/x]],x]

[Out]

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

Rule 31

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

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2448

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

Rule 2453

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.), x_Symbol] :> Int[(a + b*Log[c*ExpandToSum[v, x]^p])^q, x] /;
FreeQ[{a, b, c, p, q}, x] && BinomialQ[v, x] &&  !BinomialMatchQ[v, x]

Rubi steps

\begin {align*} \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx &=\int \log \left (\sqrt {1+\frac {1}{x}}\right ) \, dx\\ &=x \log \left (\sqrt {1+\frac {1}{x}}\right )+\frac {1}{2} \int \frac {1}{\left (1+\frac {1}{x}\right ) x} \, dx\\ &=x \log \left (\sqrt {1+\frac {1}{x}}\right )+\frac {1}{2} \int \frac {1}{1+x} \, dx\\ &=x \log \left (\sqrt {1+\frac {1}{x}}\right )+\frac {1}{2} \log (1+x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 0.90 \[ \frac {1}{2} \left (\log (x)+(x+1) \log \left (\frac {x+1}{x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[Sqrt[(1 + x)/x]],x]

[Out]

(Log[x] + (1 + x)*Log[(1 + x)/x])/2

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fricas [A]  time = 0.40, size = 18, normalized size = 0.86 \[ \frac {1}{2} \, x \log \left (\frac {x + 1}{x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*log((x + 1)/x) + 1/2*log(x + 1)

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giac [B]  time = 0.16, size = 47, normalized size = 2.24 \[ \frac {\log \left (\frac {x + 1}{x}\right )}{2 \, {\left (\frac {x + 1}{x} - 1\right )}} + \frac {1}{2} \, \log \left (\frac {{\left | x + 1 \right |}}{{\left | x \right |}}\right ) - \frac {1}{2} \, \log \left ({\left | \frac {x + 1}{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log((x + 1)/x)/((x + 1)/x - 1) + 1/2*log(abs(x + 1)/abs(x)) - 1/2*log(abs((x + 1)/x - 1))

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maple [A]  time = 0.17, size = 22, normalized size = 1.05 \[ \frac {\left (\frac {1}{x}+1\right ) x \ln \left (\frac {1}{x}+1\right )}{2}-\frac {\ln \left (\frac {1}{x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 18, normalized size = 0.86 \[ \frac {1}{2} \, x \log \left (\frac {x + 1}{x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*log((x + 1)/x) + 1/2*log(x + 1)

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mupad [B]  time = 0.06, size = 18, normalized size = 0.86 \[ \frac {\ln \left (x+1\right )}{2}+\frac {x\,\ln \left (\frac {x+1}{x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((x + 1)/x)/2,x)

[Out]

log(x + 1)/2 + (x*log((x + 1)/x))/2

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sympy [A]  time = 0.11, size = 17, normalized size = 0.81 \[ \frac {x \log {\left (\frac {x + 1}{x} \right )}}{2} + \frac {\log {\left (2 x + 2 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log((x + 1)/x)/2 + log(2*x + 2)/2

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