3.38.43 \(\int \frac {1-x+\log (\log (x))}{x} \, dx\)

Optimal. Leaf size=10 \[ -x+\log (x) \log (\log (x)) \]

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {14, 43, 2521} \begin {gather*} \log (x) \log (\log (x))-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x + Log[Log[x]])/x,x]

[Out]

-x + Log[x]*Log[Log[x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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])

Rule 2521

Int[((a_.) + Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)]*(b_.))/(x_), x_Symbol] :> Simp[(Log[d*x^n]*(a + b*Log[c*Lo
g[d*x^n]^p]))/n, x] - Simp[b*p*Log[x], x] /; FreeQ[{a, b, c, d, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 10, normalized size = 1.00 \begin {gather*} -x+\log (x) \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x + Log[Log[x]])/x,x]

[Out]

-x + Log[x]*Log[Log[x]]

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fricas [A]  time = 0.78, size = 10, normalized size = 1.00 \begin {gather*} \log \relax (x) \log \left (\log \relax (x)\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)*log(log(x)) - x

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giac [A]  time = 0.12, size = 10, normalized size = 1.00 \begin {gather*} \log \relax (x) \log \left (\log \relax (x)\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)*log(log(x)) - x

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maple [A]  time = 0.02, size = 11, normalized size = 1.10

method	result	size
default	\(\ln \relax (x ) \ln \left (\ln \relax (x )\right )-x\)	\(11\)
norman	\(\ln \relax (x ) \ln \left (\ln \relax (x )\right )-x\)	\(11\)
risch	\(\ln \relax (x ) \ln \left (\ln \relax (x )\right )-x\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*ln(ln(x))-x

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maxima [A]  time = 0.35, size = 10, normalized size = 1.00 \begin {gather*} \log \relax (x) \log \left (\log \relax (x)\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)*log(log(x)) - x

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mupad [B]  time = 2.13, size = 10, normalized size = 1.00 \begin {gather*} \ln \left (\ln \relax (x)\right )\,\ln \relax (x)-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x)) - x + 1)/x,x)

[Out]

log(log(x))*log(x) - x

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sympy [A]  time = 0.25, size = 8, normalized size = 0.80 \begin {gather*} - x + \log {\relax (x )} \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + log(x)*log(log(x))

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