3.5 \(\int \text{PolyLog}(2,a x) \, dx\)

Optimal. Leaf size=29 \[ x \text{PolyLog}(2,a x)-\frac{(1-a x) \log (1-a x)}{a}-x \]

[Out]

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

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Rubi [A]  time = 0.0085078, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 5, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6586, 2389, 2295} \[ x \text{PolyLog}(2,a x)-\frac{(1-a x) \log (1-a x)}{a}-x \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x],x]

[Out]

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

Rule 6586

Int[PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[x*PolyLog[n, a*(b*x^p)^q], x] - Dist[p*q, I
nt[PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ[{a, b, p, q}, x] && GtQ[n, 0]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin{align*} \int \text{Li}_2(a x) \, dx &=x \text{Li}_2(a x)+\int \log (1-a x) \, dx\\ &=x \text{Li}_2(a x)-\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-a x)}{a}\\ &=-x-\frac{(1-a x) \log (1-a x)}{a}+x \text{Li}_2(a x)\\ \end{align*}

Mathematica [A]  time = 0.011601, size = 26, normalized size = 0.9 \[ x \text{PolyLog}(2,a x)+\left (x-\frac{1}{a}\right ) \log (1-a x)-x \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x],x]

[Out]

-x + (-a^(-1) + x)*Log[1 - a*x] + x*PolyLog[2, a*x]

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Maple [A]  time = 0.045, size = 36, normalized size = 1.2 \begin{align*} x{\it polylog} \left ( 2,ax \right ) +\ln \left ( -ax+1 \right ) x-x-{\frac{\ln \left ( -ax+1 \right ) }{a}}+{a}^{-1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x),x)

[Out]

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

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Maxima [A]  time = 0.980908, size = 39, normalized size = 1.34 \begin{align*} \frac{a x{\rm Li}_2\left (a x\right ) - a x +{\left (a x - 1\right )} \log \left (-a x + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x),x, algorithm="maxima")

[Out]

(a*x*dilog(a*x) - a*x + (a*x - 1)*log(-a*x + 1))/a

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Fricas [A]  time = 2.63468, size = 70, normalized size = 2.41 \begin{align*} \frac{a x{\rm Li}_2\left (a x\right ) - a x +{\left (a x - 1\right )} \log \left (-a x + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x),x, algorithm="fricas")

[Out]

(a*x*dilog(a*x) - a*x + (a*x - 1)*log(-a*x + 1))/a

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Sympy [A]  time = 1.20944, size = 22, normalized size = 0.76 \begin{align*} \begin{cases} - x \operatorname{Li}_{1}\left (a x\right ) + x \operatorname{Li}_{2}\left (a x\right ) - x + \frac{\operatorname{Li}_{1}\left (a x\right )}{a} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x),x)

[Out]

Piecewise((-x*polylog(1, a*x) + x*polylog(2, a*x) - x + polylog(1, a*x)/a, Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\rm Li}_2\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x),x, algorithm="giac")

[Out]

integrate(dilog(a*x), x)