3.30.99 \(\int \frac {2+\log (x)}{x+x \log (x)} \, dx\)

Optimal. Leaf size=9 \[ 4+\log (x+x \log (x)) \]

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Rubi [A]  time = 0.02, antiderivative size = 8, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \begin {gather*} \log (x)+\log (\log (x)+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + Log[x])/(x + x*Log[x]),x]

[Out]

Log[x] + Log[1 + Log[x]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {2+x}{1+x} \, dx,x,\log (x)\right )\\ &=\operatorname {Subst}\left (\int \left (1+\frac {1}{1+x}\right ) \, dx,x,\log (x)\right )\\ &=\log (x)+\log (1+\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 8, normalized size = 0.89 \begin {gather*} \log (x)+\log (1+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + Log[x])/(x + x*Log[x]),x]

[Out]

Log[x] + Log[1 + Log[x]]

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fricas [A]  time = 0.76, size = 8, normalized size = 0.89 \begin {gather*} \log \relax (x) + \log \left (\log \relax (x) + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + log(log(x) + 1)

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giac [B]  time = 0.24, size = 25, normalized size = 2.78 \begin {gather*} \frac {1}{2} \, \log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\relax (x) - 1\right )}^{2} + {\left (\log \left ({\left | x \right |}\right ) + 1\right )}^{2}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 8, normalized size = 0.89

method	result	size
derivativedivides	\(\ln \left (x \ln \relax (x )+x \right )\)	\(8\)
default	\(\ln \left (x \ln \relax (x )+x \right )\)	\(8\)
norman	\(\ln \relax (x )+\ln \left (\ln \relax (x )+1\right )\)	\(9\)
risch	\(\ln \relax (x )+\ln \left (\ln \relax (x )+1\right )\)	\(9\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x*ln(x)+x)

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maxima [A]  time = 0.65, size = 7, normalized size = 0.78 \begin {gather*} \log \left (x \log \relax (x) + x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x*log(x) + x)

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mupad [B]  time = 1.77, size = 8, normalized size = 0.89 \begin {gather*} \ln \left (\ln \relax (x)+1\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) + 2)/(x + x*log(x)),x)

[Out]

log(log(x) + 1) + log(x)

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sympy [A]  time = 0.09, size = 8, normalized size = 0.89 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\relax (x )} + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + log(log(x) + 1)

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