3.1.78 \(\int \frac {\log (-2+e x)}{x} \, dx\) [78]

3.1.78.1 Optimal result
3.1.78.2 Mathematica [A] (verified)
3.1.78.3 Rubi [A] (verified)
3.1.78.4 Maple [A] (verified)
3.1.78.5 Fricas [F]
3.1.78.6 Sympy [C] (verification not implemented)
3.1.78.7 Maxima [A] (verification not implemented)
3.1.78.8 Giac [F]
3.1.78.9 Mupad [B] (verification not implemented)

3.1.78.1 Optimal result

Integrand size = 10, antiderivative size = 25 \[ \int \frac {\log (-2+e x)}{x} \, dx=\log \left (\frac {e x}{2}\right ) \log (-2+e x)+\operatorname {PolyLog}\left (2,1-\frac {e x}{2}\right ) \]

output
ln(1/2*e*x)*ln(e*x-2)+polylog(2,1-1/2*e*x)
 
3.1.78.2 Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\log (-2+e x)}{x} \, dx=\log \left (\frac {e x}{2}\right ) \log (-2+e x)+\operatorname {PolyLog}\left (2,\frac {1}{2} (2-e x)\right ) \]

input
Integrate[Log[-2 + e*x]/x,x]
 
output
Log[(e*x)/2]*Log[-2 + e*x] + PolyLog[2, (2 - e*x)/2]
 
3.1.78.3 Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2841, 25, 2752}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\log (e x-2)}{x} \, dx\)

\(\Big \downarrow \) 2841

\(\displaystyle \log \left (\frac {e x}{2}\right ) \log (e x-2)-e \int -\frac {\log \left (\frac {e x}{2}\right )}{2-e x}dx\)

\(\Big \downarrow \) 25

\(\displaystyle e \int \frac {\log \left (\frac {e x}{2}\right )}{2-e x}dx+\log \left (\frac {e x}{2}\right ) \log (e x-2)\)

\(\Big \downarrow \) 2752

\(\displaystyle \operatorname {PolyLog}\left (2,1-\frac {e x}{2}\right )+\log \left (\frac {e x}{2}\right ) \log (e x-2)\)

input
Int[Log[-2 + e*x]/x,x]
 
output
Log[(e*x)/2]*Log[-2 + e*x] + PolyLog[2, 1 - (e*x)/2]
 

3.1.78.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2752
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2841
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] - Simp[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]
 
3.1.78.4 Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method result size
derivativedivides \(\operatorname {dilog}\left (\frac {e x}{2}\right )+\ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )\) \(19\)
default \(\operatorname {dilog}\left (\frac {e x}{2}\right )+\ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )\) \(19\)
risch \(\operatorname {dilog}\left (\frac {e x}{2}\right )+\ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )\) \(19\)
parts \(\ln \left (e x -2\right ) \ln \left (x \right )-e \left (\frac {\left (\ln \left (x \right )-\ln \left (\frac {e x}{2}\right )\right ) \ln \left (1-\frac {e x}{2}\right )}{e}-\frac {\operatorname {dilog}\left (\frac {e x}{2}\right )}{e}\right )\) \(46\)

input
int(ln(e*x-2)/x,x,method=_RETURNVERBOSE)
 
output
dilog(1/2*e*x)+ln(1/2*e*x)*ln(e*x-2)
 
3.1.78.5 Fricas [F]

\[ \int \frac {\log (-2+e x)}{x} \, dx=\int { \frac {\log \left (e x - 2\right )}{x} \,d x } \]

input
integrate(log(e*x-2)/x,x, algorithm="fricas")
 
output
integral(log(e*x - 2)/x, x)
 
3.1.78.6 Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.97 (sec) , antiderivative size = 102, normalized size of antiderivative = 4.08 \[ \int \frac {\log (-2+e x)}{x} \, dx=\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (2 \right )} \log {\left (x \right )} + 3 i \pi \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (2 \right )} \log {\left (\frac {1}{x} \right )} - 3 i \pi \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (2 \right )} - 3 i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (2 \right )} + 3 i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {otherwise} \end {cases} \]

input
integrate(ln(e*x-2)/x,x)
 
output
Piecewise((-polylog(2, e*x/2), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(2)*log 
(x) + 3*I*pi*log(x) - polylog(2, e*x/2), Abs(x) < 1), (-log(2)*log(1/x) - 
3*I*pi*log(1/x) - polylog(2, e*x/2), 1/Abs(x) < 1), (-meijerg(((), (1, 1)) 
, ((0, 0), ()), x)*log(2) - 3*I*pi*meijerg(((), (1, 1)), ((0, 0), ()), x) 
+ meijerg(((1, 1), ()), ((), (0, 0)), x)*log(2) + 3*I*pi*meijerg(((1, 1), 
()), ((), (0, 0)), x) - polylog(2, e*x/2), True))
 
3.1.78.7 Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {\log (-2+e x)}{x} \, dx=\log \left (e x - 2\right ) \log \left (\frac {1}{2} \, e x\right ) + {\rm Li}_2\left (-\frac {1}{2} \, e x + 1\right ) \]

input
integrate(log(e*x-2)/x,x, algorithm="maxima")
 
output
log(e*x - 2)*log(1/2*e*x) + dilog(-1/2*e*x + 1)
 
3.1.78.8 Giac [F]

\[ \int \frac {\log (-2+e x)}{x} \, dx=\int { \frac {\log \left (e x - 2\right )}{x} \,d x } \]

input
integrate(log(e*x-2)/x,x, algorithm="giac")
 
output
integrate(log(e*x - 2)/x, x)
 
3.1.78.9 Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {\log (-2+e x)}{x} \, dx={\mathrm {Li}}_{\mathrm {2}}\left (\frac {e\,x}{2}\right )+\ln \left (e\,x-2\right )\,\ln \left (\frac {e\,x}{2}\right ) \]

input
int(log(e*x - 2)/x,x)
 
output
dilog((e*x)/2) + log(e*x - 2)*log((e*x)/2)