3.55 \(\int \frac {\log (x)}{(1+\log (x))^2} \, dx\)

Optimal. Leaf size=8 \[ \frac {x}{\log (x)+1} \]

[Out]

x/(1+ln(x))

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Rubi [A]  time = 0.04, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2360, 2297, 2299, 2178} \[ \frac {x}{\log (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[Log[x]/(1 + Log[x])^2,x]

[Out]

x/(1 + Log[x])

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2299

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 8, normalized size = 1.00 \[ \frac {x}{\log (x)+1} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]/(1 + Log[x])^2,x]

[Out]

x/(1 + Log[x])

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fricas [A]  time = 0.39, size = 8, normalized size = 1.00 \[ \frac {x}{\log \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(log(x) + 1)

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giac [A]  time = 0.96, size = 8, normalized size = 1.00 \[ \frac {x}{\log \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(log(x) + 1)

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maple [A]  time = 0.14, size = 9, normalized size = 1.12 \[ \frac {x}{\ln \relax (x )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(ln(x)+1)

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maxima [A]  time = 0.54, size = 8, normalized size = 1.00 \[ \frac {x}{\log \relax (x) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(log(x) + 1)

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mupad [B]  time = 0.26, size = 8, normalized size = 1.00 \[ \frac {x}{\ln \relax (x)+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)/(log(x) + 1)^2,x)

[Out]

x/(log(x) + 1)

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sympy [A]  time = 0.09, size = 5, normalized size = 0.62 \[ \frac {x}{\log {\relax (x )} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(log(x) + 1)

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