\(\int \frac {a+b \log (-2+e x)}{x} \, dx\) [84]

Optimal result

Integrand size = 14, antiderivative size = 31 \[ \int \frac {a+b \log (-2+e x)}{x} \, dx=\log \left (\frac {e x}{2}\right ) (a+b \log (-2+e x))+b \operatorname {PolyLog}\left (2,1-\frac {e x}{2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, 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} \[ \int \frac {a+b \log (-2+e x)}{x} \, dx=\log \left (\frac {e x}{2}\right ) (a+b \log (e x-2))+b \operatorname {PolyLog}\left (2,1-\frac {e x}{2}\right ) \]

[In]

Int[(a + b*Log[-2 + e*x])/x,x]

[Out]

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

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 \left (\frac {e x}{2}\right ) (a+b \log (-2+e x))-(b e) \int \frac {\log \left (\frac {e x}{2}\right )}{-2+e x} \, dx \\ & = \log \left (\frac {e x}{2}\right ) (a+b \log (-2+e x))+b \text {Li}_2\left (1-\frac {e x}{2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \log (-2+e x)}{x} \, dx=a \log (x)+b \log \left (\frac {e x}{2}\right ) \log (-2+e x)+b \operatorname {PolyLog}\left (2,\frac {1}{2} (2-e x)\right ) \]

[In]

Integrate[(a + b*Log[-2 + e*x])/x,x]

[Out]

a*Log[x] + b*Log[(e*x)/2]*Log[-2 + e*x] + b*PolyLog[2, (2 - e*x)/2]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
risch	\(\ln \left (x \right ) a +\ln \left (e x -2\right ) \ln \left (\frac {e x}{2}\right ) b +\operatorname {dilog}\left (\frac {e x}{2}\right ) b\)	\(26\)
parts	\(\ln \left (x \right ) a +b \left (\operatorname {dilog}\left (\frac {e x}{2}\right )+\ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )\right )\)	\(26\)
derivativedivides	\(a \ln \left (e x \right )+b \left (\operatorname {dilog}\left (\frac {e x}{2}\right )+\ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )\right )\)	\(28\)
default	\(a \ln \left (e x \right )+b \left (\operatorname {dilog}\left (\frac {e x}{2}\right )+\ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )\right )\)	\(28\)

[In]

int((a+b*ln(e*x-2))/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*a+ln(e*x-2)*ln(1/2*e*x)*b+dilog(1/2*e*x)*b

Fricas [F]

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

[In]

integrate((a+b*log(e*x-2))/x,x, algorithm="fricas")

[Out]

integral((b*log(e*x - 2) + a)/x, x)

Sympy [A] (verification not implemented)

Time = 2.86 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.52 \[ \int \frac {a+b \log (-2+e x)}{x} \, dx=a \log {\left (x \right )} + b \left (\begin {cases} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (2 \right )} \log {\left (x \right )} + 3 i \pi \log {\left (x \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (2 \right )} \log {\left (\frac {1}{x} \right )} - 3 i \pi \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (2 \right )} - 3 i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (2 \right )} + 3 i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((a+b*ln(e*x-2))/x,x)

[Out]

a*log(x) + b*Piecewise((-polylog(2, e*x/2), (Abs(x) < 1) & (1/Abs(x) < 1)), (log(2)*log(x) + 3*I*pi*log(x) - p
olylog(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), (-meije
rg(((), (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))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {a+b \log (-2+e x)}{x} \, dx={\left (\log \left (e x - 2\right ) \log \left (\frac {1}{2} \, e x\right ) + {\rm Li}_2\left (-\frac {1}{2} \, e x + 1\right )\right )} b + a \log \left (x\right ) \]

[In]

integrate((a+b*log(e*x-2))/x,x, algorithm="maxima")

[Out]

(log(e*x - 2)*log(1/2*e*x) + dilog(-1/2*e*x + 1))*b + a*log(x)

Giac [F]

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

[In]

integrate((a+b*log(e*x-2))/x,x, algorithm="giac")

[Out]

integrate((b*log(e*x - 2) + a)/x, x)

Mupad [B] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {a+b \log (-2+e x)}{x} \, dx=b\,{\mathrm {Li}}_{\mathrm {2}}\left (\frac {e\,x}{2}\right )+a\,\ln \left (x\right )+b\,\ln \left (e\,x-2\right )\,\ln \left (\frac {e\,x}{2}\right ) \]

[In]

int((a + b*log(e*x - 2))/x,x)

[Out]

b*dilog((e*x)/2) + a*log(x) + b*log(e*x - 2)*log((e*x)/2)