3.74 \(\int \frac{\log (2+e x)}{x} \, dx\)

Optimal. Leaf size=16 \[ \log (2) \log (x)-\text{PolyLog}\left (2,-\frac{e x}{2}\right ) \]

[Out]

Log[2]*Log[x] - PolyLog[2, -(e*x)/2]

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Rubi [A]  time = 0.0149935, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2392, 2391} \[ \log (2) \log (x)-\text{PolyLog}\left (2,-\frac{e x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[2 + e*x]/x,x]

[Out]

Log[2]*Log[x] - PolyLog[2, -(e*x)/2]

Rule 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2391

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

Rubi steps

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

Mathematica [A]  time = 0.0017052, size = 16, normalized size = 1. \[ \log (2) \log (x)-\text{PolyLog}\left (2,-\frac{e x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[2 + e*x]/x,x]

[Out]

Log[2]*Log[x] - PolyLog[2, -(e*x)/2]

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Maple [B]  time = 0.057, size = 33, normalized size = 2.1 \begin{align*} \left ( \ln \left ( ex+2 \right ) -\ln \left ({\frac{ex}{2}}+1 \right ) \right ) \ln \left ( -{\frac{ex}{2}} \right ) -{\it dilog} \left ({\frac{ex}{2}}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(e*x+2)/x,x)

[Out]

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

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Maxima [A]  time = 1.20486, size = 27, normalized size = 1.69 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left (e x + 2\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(e*x + 2)/x, x)

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Sympy [C]  time = 3.02837, size = 68, normalized size = 4.25 \begin{align*} \begin{cases} \log{\left (2 \right )} \log{\left (x \right )} - \operatorname{Li}_{2}\left (\frac{e x e^{i \pi }}{2}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (2 \right )} \log{\left (\frac{1}{x} \right )} - \operatorname{Li}_{2}\left (\frac{e x e^{i \pi }}{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 )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (2 \right )} - \operatorname{Li}_{2}\left (\frac{e x e^{i \pi }}{2}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(e*x+2)/x,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left (e x + 2\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(e*x + 2)/x, x)