\(\int \frac {\log (-1+e x)}{x} \, dx\) [77]

Optimal result

Integrand size = 10, antiderivative size = 20 \[ \int \frac {\log (-1+e x)}{x} \, dx=\log (e x) \log (-1+e x)+\operatorname {PolyLog}(2,1-e x) \]

[Out]

ln(e*x)*ln(e*x-1)+polylog(2,-e*x+1)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2441, 2352} \[ \int \frac {\log (-1+e x)}{x} \, dx=\operatorname {PolyLog}(2,1-e x)+\log (e x) \log (e x-1) \]

[In]

Int[Log[-1 + e*x]/x,x]

[Out]

Log[e*x]*Log[-1 + e*x] + PolyLog[2, 1 - e*x]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = \log (e x) \log (-1+e x)-e \int \frac {\log (e x)}{-1+e x} \, dx \\ & = \log (e x) \log (-1+e x)+\text {Li}_2(1-e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\log (-1+e x)}{x} \, dx=\log (e x) \log (-1+e x)+\operatorname {PolyLog}(2,1-e x) \]

[In]

Integrate[Log[-1 + e*x]/x,x]

[Out]

Log[e*x]*Log[-1 + e*x] + PolyLog[2, 1 - e*x]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\operatorname {dilog}\left (e x \right )+\ln \left (e x \right ) \ln \left (e x -1\right )\)	\(17\)
default	\(\operatorname {dilog}\left (e x \right )+\ln \left (e x \right ) \ln \left (e x -1\right )\)	\(17\)
risch	\(\operatorname {dilog}\left (e x \right )+\ln \left (e x \right ) \ln \left (e x -1\right )\)	\(17\)
parts	\(\ln \left (e x -1\right ) \ln \left (x \right )-e \left (\frac {\left (\ln \left (x \right )-\ln \left (e x \right )\right ) \ln \left (-e x +1\right )}{e}-\frac {\operatorname {dilog}\left (e x \right )}{e}\right )\)	\(44\)

[In]

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

[Out]

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

Fricas [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="fricas")

[Out]

integral(log(e*x - 1)/x, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.94 (sec) , antiderivative size = 60, normalized size of antiderivative = 3.00 \[ \int \frac {\log (-1+e x)}{x} \, dx=\begin {cases} - \operatorname {Li}_{2}\left (e x\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\i \pi \log {\left (x \right )} - \operatorname {Li}_{2}\left (e x\right ) & \text {for}\: \left |{x}\right | < 1 \\- i \pi \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (e x\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} - \operatorname {Li}_{2}\left (e x\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-polylog(2, e*x), (Abs(x) < 1) & (1/Abs(x) < 1)), (I*pi*log(x) - polylog(2, e*x), Abs(x) < 1), (-I*
pi*log(1/x) - polylog(2, e*x), 1/Abs(x) < 1), (-I*pi*meijerg(((), (1, 1)), ((0, 0), ()), x) + I*pi*meijerg(((1
, 1), ()), ((), (0, 0)), x) - polylog(2, e*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {\log (-1+e x)}{x} \, dx=\log \left (e x - 1\right ) \log \left (e x\right ) + {\rm Li}_2\left (-e x + 1\right ) \]

[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 = 16, normalized size of antiderivative = 0.80 \[ \int \frac {\log (-1+e x)}{x} \, dx={\mathrm {Li}}_{\mathrm {2}}\left (e\,x\right )+\ln \left (e\,x-1\right )\,\ln \left (e\,x\right ) \]

[In]

int(log(e*x - 1)/x,x)

[Out]

dilog(e*x) + log(e*x - 1)*log(e*x)