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

Optimal. Leaf size=21 \[ \log (x) (a+b \log (2))-b \text{PolyLog}\left (2,-\frac{e x}{2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0214168, antiderivative size = 21, normalized size of antiderivative = 1., 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 = {2392, 2391} \[ \log (x) (a+b \log (2))-b \text{PolyLog}\left (2,-\frac{e x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[2 + e*x])/x,x]

[Out]

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

Mathematica [A]  time = 0.0019993, size = 22, normalized size = 1.05 \[ -b \text{PolyLog}\left (2,-\frac{e x}{2}\right )+a \log (x)+b \log (2) \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[2 + e*x])/x,x]

[Out]

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

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Maple [B]  time = 0.066, size = 46, normalized size = 2.2 \begin{align*} a\ln \left ( ex \right ) +\ln \left ( ex+2 \right ) \ln \left ( -{\frac{ex}{2}} \right ) b-\ln \left ({\frac{ex}{2}}+1 \right ) \ln \left ( -{\frac{ex}{2}} \right ) b-{\it dilog} \left ({\frac{ex}{2}}+1 \right ) b \end{align*}

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

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Maxima [A]  time = 1.77969, size = 36, normalized size = 1.71 \begin{align*}{\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 ) \end{align*}

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

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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Sympy [A]  time = 3.93505, size = 75, normalized size = 3.57 \begin{align*} a \log{\left (x \right )} + b \left (\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}\right ) \end{align*}

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) - polylog(2, e*x*exp_polar(I*pi)/2), Abs(x) < 1), (-log(2)*log(1/x) - po
lylog(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{b \log \left (e x + 2\right ) + a}{x}\,{d x} \end{align*}

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)