\(\int \frac {1+x}{\log (x) (x+\log (x))} \, dx\) [293]

Optimal result

Integrand size = 14, antiderivative size = 13 \[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=\log (\log (x))-\log (x+\log (x))+\operatorname {LogIntegral}(x) \]

[Out]

Li(x)+ln(ln(x))-ln(x+ln(x))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6874, 2395, 2335, 2339, 29, 6816} \[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=\operatorname {LogIntegral}(x)+\log (\log (x))-\log (x+\log (x)) \]

[In]

Int[(1 + x)/(Log[x]*(x + Log[x])),x]

[Out]

Log[Log[x]] - Log[x + Log[x]] + LogIntegral[x]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2339

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

Rule 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1+x}{x \log (x)}+\frac {-1-x}{x (x+\log (x))}\right ) \, dx \\ & = \int \frac {1+x}{x \log (x)} \, dx+\int \frac {-1-x}{x (x+\log (x))} \, dx \\ & = -\log (x+\log (x))+\int \left (\frac {1}{\log (x)}+\frac {1}{x \log (x)}\right ) \, dx \\ & = -\log (x+\log (x))+\int \frac {1}{\log (x)} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -\log (x+\log (x))+\text {li}(x)+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \log (\log (x))-\log (x+\log (x))+\text {li}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=\log (\log (x))-\log (x+\log (x))+\operatorname {LogIntegral}(x) \]

[In]

Integrate[(1 + x)/(Log[x]*(x + Log[x])),x]

[Out]

Log[Log[x]] - Log[x + Log[x]] + LogIntegral[x]

Maple [A] (verified)

Time = 1.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.54

method	result	size
default	\(-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )\right )-\ln \left (x +\ln \left (x \right )\right )\)	\(20\)
risch	\(-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+\ln \left (\ln \left (x \right )\right )-\ln \left (x +\ln \left (x \right )\right )\)	\(20\)

[In]

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

[Out]

-Ei(1,-ln(x))+ln(ln(x))-ln(x+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=-\log \left (x + \log \left (x\right )\right ) + \log \left (\log \left (x\right )\right ) + \operatorname {log\_integral}\left (x\right ) \]

[In]

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

[Out]

-log(x + log(x)) + log(log(x)) + log_integral(x)

Sympy [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=- \log {\left (x + \log {\left (x \right )} \right )} + \log {\left (\log {\left (x \right )} \right )} + \operatorname {Ei}{\left (\log {\left (x \right )} \right )} \]

[In]

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

[Out]

-log(x + log(x)) + log(log(x)) + Ei(log(x))

Maxima [F]

\[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=\int { \frac {x + 1}{{\left (x + \log \left (x\right )\right )} \log \left (x\right )} \,d x } \]

[In]

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

[Out]

integrate((x + 1)/(x*log(x)), x) - log(x + log(x))

Giac [F]

\[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=\int { \frac {x + 1}{{\left (x + \log \left (x\right )\right )} \log \left (x\right )} \,d x } \]

[In]

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

[Out]

integrate((x + 1)/((x + log(x))*log(x)), x)

Mupad [B] (verification not implemented)

Time = 1.55 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{\log (x) (x+\log (x))} \, dx=\ln \left (\ln \left (x\right )\right )-\ln \left (x+\ln \left (x\right )\right )+\mathrm {logint}\left (x\right ) \]

[In]

int((x + 1)/(log(x)*(x + log(x))),x)

[Out]

log(log(x)) - log(x + log(x)) + logint(x)