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

Optimal. Leaf size=31 \[ \log \left (\frac {e x}{2}\right ) (a+b \log (e x-2))+b \text {Li}_2\left (1-\frac {e x}{2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2394, 2315} \[ b \text {PolyLog}\left (2,1-\frac {e x}{2}\right )+\log \left (\frac {e x}{2}\right ) (a+b \log (e x-2)) \]

Antiderivative was successfully verified.

[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 2315

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

Rule 2394

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*} \int \frac {a+b \log (-2+e x)}{x} \, dx &=\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*}

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Mathematica [A]  time = 0.00, size = 34, normalized size = 1.10 \[ a \log (x)+b \text {Li}_2\left (\frac {1}{2} (2-e x)\right )+b \log \left (\frac {e x}{2}\right ) \log (e x-2) \]

Antiderivative was successfully verified.

[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]

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (e x - 2\right ) + a}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(e*x - 2) + a)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \log \left (e x - 2\right ) + a}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(e*x - 2) + a)/x, x)

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maple [A]  time = 0.04, size = 28, normalized size = 0.90 \[ b \ln \left (\frac {e x}{2}\right ) \ln \left (e x -2\right )+a \ln \left (e x \right )+b \dilog \left (\frac {e x}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 27, normalized size = 0.87 \[ {\left (\log \left (e x - 2\right ) \log \left (\frac {1}{2} \, e x\right ) + {\rm Li}_2\left (-\frac {1}{2} \, e x + 1\right )\right )} b + a \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.16, size = 25, normalized size = 0.81 \[ b\,{\mathrm {Li}}_{\mathrm {2}}\left (\frac {e\,x}{2}\right )+a\,\ln \relax (x)+b\,\ln \left (e\,x-2\right )\,\ln \left (\frac {e\,x}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 4.89, size = 95, normalized size = 3.06 \[ a \log {\relax (x )} + b \left (\begin {cases} \log {\relax (2 )} \log {\relax (x )} + 3 i \pi \log {\relax (x )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \left |{x}\right | < 1 \\- \log {\relax (2 )} \log {\left (\frac {1}{x} \right )} - 3 i \pi \log {\left (\frac {1}{x} \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\relax (2 )} - 3 i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\relax (2 )} + 3 i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} - \operatorname {Li}_{2}\left (\frac {e x}{2}\right ) & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) + b*Piecewise((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*mei
jerg(((), (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))

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