3.72 \(\int \frac{a+b \log (c (\frac{1}{c}+e x))}{x} \, dx\)

Optimal. Leaf size=15 \[ a \log (x)-b \text{PolyLog}(2,-c e x) \]

[Out]

a*Log[x] - b*PolyLog[2, -(c*e*x)]

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Rubi [A]  time = 0.0193654, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2392, 2391} \[ a \log (x)-b \text{PolyLog}(2,-c e x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(c^(-1) + e*x)])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(c*e*x)]

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

Mathematica [A]  time = 0.0027481, size = 15, normalized size = 1. \[ a \log (x)-b \text{PolyLog}(2,-c e x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(c^(-1) + e*x)])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(c*e*x)]

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Maple [A]  time = 0.06, size = 19, normalized size = 1.3 \begin{align*} a\ln \left ( cex \right ) -b{\it dilog} \left ( cex+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(1/c+e*x)))/x,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\log \left (c e x + 1\right )}{x}\,{d x} + a \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(log(c*e*x + 1)/x, x) + a*log(x)

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Fricas [A]  time = 1.93335, size = 39, normalized size = 2.6 \begin{align*} -b{\rm Li}_2\left (-c e x\right ) + a \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*dilog(-c*e*x) + a*log(x)

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Sympy [C]  time = 5.34438, size = 17, normalized size = 1.13 \begin{align*} a \log{\left (x \right )} - b \operatorname{Li}_{2}\left (c e x e^{i \pi }\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - b*polylog(2, c*e*x*exp_polar(I*pi))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + \frac{1}{c}\right )} c\right ) + a}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log((e*x + 1/c)*c) + a)/x, x)