\(\int \frac {a+b \log (-1+e x)}{x} \, dx\) [83]

Optimal result

Integrand size = 14, antiderivative size = 26 \[ \int \frac {a+b \log (-1+e x)}{x} \, dx=\log (e x) (a+b \log (-1+e x))+b \operatorname {PolyLog}(2,1-e x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2441, 2352} \[ \int \frac {a+b \log (-1+e x)}{x} \, dx=\log (e x) (a+b \log (e x-1))+b \operatorname {PolyLog}(2,1-e x) \]

[In]

Int[(a + b*Log[-1 + e*x])/x,x]

[Out]

Log[e*x]*(a + b*Log[-1 + e*x]) + b*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) (a+b \log (-1+e x))-(b e) \int \frac {\log (e x)}{-1+e x} \, dx \\ & = \log (e x) (a+b \log (-1+e x))+b \text {Li}_2(1-e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \log (-1+e x)}{x} \, dx=a \log (x)+b \log (e x) \log (-1+e x)+b \operatorname {PolyLog}(2,1-e x) \]

[In]

Integrate[(a + b*Log[-1 + e*x])/x,x]

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92

method	result	size
risch	\(\ln \left (x \right ) a +\ln \left (e x -1\right ) \ln \left (e x \right ) b +\operatorname {dilog}\left (e x \right ) b\)	\(24\)
parts	\(\ln \left (x \right ) a +b \left (\operatorname {dilog}\left (e x \right )+\ln \left (e x \right ) \ln \left (e x -1\right )\right )\)	\(24\)
derivativedivides	\(a \ln \left (e x \right )+b \left (\operatorname {dilog}\left (e x \right )+\ln \left (e x \right ) \ln \left (e x -1\right )\right )\)	\(26\)
default	\(a \ln \left (e x \right )+b \left (\operatorname {dilog}\left (e x \right )+\ln \left (e x \right ) \ln \left (e x -1\right )\right )\)	\(26\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*log(e*x - 1) + a)/x, x)

Sympy [A] (verification not implemented)

Time = 2.72 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.54 \[ \int \frac {a+b \log (-1+e x)}{x} \, dx=a \log {\left (x \right )} + b \left (\begin {cases} - \operatorname {Li}_{2}\left (e x\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\i \pi \log {\left (x \right )} - \operatorname {Li}_{2}\left (e x\right ) & \text {for}\: \left |{x}\right | < 1 \\- i \pi \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (e x\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} - \operatorname {Li}_{2}\left (e x\right ) & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((a+b*ln(e*x-1))/x,x)

[Out]

a*log(x) + b*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*p
i*meijerg(((1, 1), ()), ((), (0, 0)), x) - polylog(2, e*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log (-1+e x)}{x} \, dx={\left (\log \left (e x - 1\right ) \log \left (e x\right ) + {\rm Li}_2\left (-e x + 1\right )\right )} b + a \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((b*log(e*x - 1) + a)/x, x)

Mupad [B] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \log (-1+e x)}{x} \, dx=b\,{\mathrm {Li}}_{\mathrm {2}}\left (e\,x\right )+a\,\ln \left (x\right )+b\,\ln \left (e\,x-1\right )\,\ln \left (e\,x\right ) \]

[In]

int((a + b*log(e*x - 1))/x,x)

[Out]

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