\(\int \log (\sqrt {\frac {1+x}{x}}) \, dx\) [296]

Optimal result

Integrand size = 12, antiderivative size = 21 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=x \log \left (\sqrt {1+\frac {1}{x}}\right )+\frac {1}{2} \log (1+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2503, 2498, 269, 31} \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=x \log \left (\sqrt {\frac {1}{x}+1}\right )+\frac {1}{2} \log (x+1) \]

[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 269

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 2498

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 2503

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{2} \left (x \log \left (1+\frac {1}{x}\right )+\log (1+x)\right ) \]

[In]

Integrate[Log[Sqrt[(1 + x)/x]],x]

[Out]

(x*Log[1 + x^(-1)] + Log[1 + x])/2

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {x \ln \left (\frac {x +1}{x}\right )}{2}+\frac {\ln \left (x +1\right )}{2}\)	\(19\)
parts	\(\frac {x \ln \left (\frac {x +1}{x}\right )}{2}+\frac {\ln \left (x +1\right )}{2}\)	\(19\)
derivativedivides	\(-\frac {\ln \left (\frac {1}{x}\right )}{2}+\frac {\ln \left (1+\frac {1}{x}\right ) \left (1+\frac {1}{x}\right ) x}{2}\)	\(22\)
default	\(-\frac {\ln \left (\frac {1}{x}\right )}{2}+\frac {\ln \left (1+\frac {1}{x}\right ) \left (1+\frac {1}{x}\right ) x}{2}\)	\(22\)
parallelrisch	\(\frac {x \ln \left (\frac {x +1}{x}\right )}{2}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (\frac {x +1}{x}\right )}{2}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{2} \, x \log \left (\frac {x + 1}{x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

1/2*x*log((x + 1)/x) + 1/2*log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {x \log {\left (\frac {x + 1}{x} \right )}}{2} + \frac {\log {\left (2 x + 2 \right )}}{2} \]

[In]

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

[Out]

x*log((x + 1)/x)/2 + log(2*x + 2)/2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {1}{2} \, x \log \left (\frac {x + 1}{x}\right ) + \frac {1}{2} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

1/2*x*log((x + 1)/x) + 1/2*log(x + 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (16) = 32\).

Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.24 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {\log \left (\frac {x + 1}{x}\right )}{2 \, {\left (\frac {x + 1}{x} - 1\right )}} + \frac {1}{2} \, \log \left (\frac {{\left | x + 1 \right |}}{{\left | x \right |}}\right ) - \frac {1}{2} \, \log \left ({\left | \frac {x + 1}{x} - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \log \left (\sqrt {\frac {1+x}{x}}\right ) \, dx=\frac {\ln \left (x+1\right )}{2}+\frac {x\,\ln \left (\frac {x+1}{x}\right )}{2} \]

[In]

int(log((x + 1)/x)/2,x)

[Out]

log(x + 1)/2 + (x*log((x + 1)/x))/2