\(\int \frac {\log (1+e x)}{x} \, dx\) [75]

Optimal result

Integrand size = 10, antiderivative size = 8 \[ \int \frac {\log (1+e x)}{x} \, dx=-\operatorname {PolyLog}(2,-e x) \]

[Out]

-polylog(2,-e*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2438} \[ \int \frac {\log (1+e x)}{x} \, dx=-\operatorname {PolyLog}(2,-e x) \]

[In]

Int[Log[1 + e*x]/x,x]

[Out]

-PolyLog[2, -(e*x)]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps \begin{align*} \text {integral}& = -\text {Li}_2(-e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int \frac {\log (1+e x)}{x} \, dx=-\operatorname {PolyLog}(2,-e x) \]

[In]

Integrate[Log[1 + e*x]/x,x]

[Out]

-PolyLog[2, -(e*x)]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.12

method	result	size
derivativedivides	\(-\operatorname {dilog}\left (e x +1\right )\)	\(9\)
default	\(-\operatorname {dilog}\left (e x +1\right )\)	\(9\)
meijerg	\(-\operatorname {Li}_{2}\left (-e x \right )\)	\(9\)
risch	\(-\operatorname {dilog}\left (e x +1\right )\)	\(9\)
parts	\(\ln \left (e x +1\right ) \ln \left (x \right )-e \left (\frac {\operatorname {dilog}\left (e x +1\right )}{e}+\frac {\ln \left (x \right ) \ln \left (e x +1\right )}{e}\right )\)	\(37\)

[In]

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

[Out]

-dilog(e*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88 \[ \int \frac {\log (1+e x)}{x} \, dx=-{\rm Li}_2\left (-e x\right ) \]

[In]

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

[Out]

-dilog(-e*x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.34 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.25 \[ \int \frac {\log (1+e x)}{x} \, dx=- \operatorname {Li}_{2}\left (e x e^{i \pi }\right ) \]

[In]

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

[Out]

-polylog(2, e*x*exp_polar(I*pi))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 19 vs. \(2 (7) = 14\).

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 2.38 \[ \int \frac {\log (1+e x)}{x} \, dx=\log \left (e x + 1\right ) \log \left (-e x\right ) + {\rm Li}_2\left (e x + 1\right ) \]

[In]

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

[Out]

log(e*x + 1)*log(-e*x) + dilog(e*x + 1)

Giac [F]

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

[In]

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

[Out]

integrate(log(e*x + 1)/x, x)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.25 \[ \int \frac {\log (1+e x)}{x} \, dx={\mathrm {Li}}_{\mathrm {2}}\left (-e\,x\right )+\ln \left (e\,x+1\right )\,\ln \left (-e\,x\right ) \]

[In]

int(log(e*x + 1)/x,x)

[Out]

dilog(-e*x) + log(e*x + 1)*log(-e*x)