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

Optimal. Leaf size=56 \[ \frac{1}{2} x^2 \text{PolyLog}(2,a x)-\frac{\log (1-a x)}{4 a^2}+\frac{1}{4} x^2 \log (1-a x)-\frac{x}{4 a}-\frac{x^2}{8} \]

[Out]

-x/(4*a) - x^2/8 - Log[1 - a*x]/(4*a^2) + (x^2*Log[1 - a*x])/4 + (x^2*PolyLog[2, a*x])/2

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Rubi [A]  time = 0.0282725, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6591, 2395, 43} \[ \frac{1}{2} x^2 \text{PolyLog}(2,a x)-\frac{\log (1-a x)}{4 a^2}+\frac{1}{4} x^2 \log (1-a x)-\frac{x}{4 a}-\frac{x^2}{8} \]

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[2, a*x],x]

[Out]

-x/(4*a) - x^2/8 - Log[1 - a*x]/(4*a^2) + (x^2*Log[1 - a*x])/4 + (x^2*PolyLog[2, a*x])/2

Rule 6591

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

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x \text{Li}_2(a x) \, dx &=\frac{1}{2} x^2 \text{Li}_2(a x)+\frac{1}{2} \int x \log (1-a x) \, dx\\ &=\frac{1}{4} x^2 \log (1-a x)+\frac{1}{2} x^2 \text{Li}_2(a x)+\frac{1}{4} a \int \frac{x^2}{1-a x} \, dx\\ &=\frac{1}{4} x^2 \log (1-a x)+\frac{1}{2} x^2 \text{Li}_2(a x)+\frac{1}{4} a \int \left (-\frac{1}{a^2}-\frac{x}{a}-\frac{1}{a^2 (-1+a x)}\right ) \, dx\\ &=-\frac{x}{4 a}-\frac{x^2}{8}-\frac{\log (1-a x)}{4 a^2}+\frac{1}{4} x^2 \log (1-a x)+\frac{1}{2} x^2 \text{Li}_2(a x)\\ \end{align*}

Mathematica [A]  time = 0.0238254, size = 48, normalized size = 0.86 \[ \frac{4 a^2 x^2 \text{PolyLog}(2,a x)+2 \left (a^2 x^2-1\right ) \log (1-a x)-a x (a x+2)}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[2, a*x],x]

[Out]

(-(a*x*(2 + a*x)) + 2*(-1 + a^2*x^2)*Log[1 - a*x] + 4*a^2*x^2*PolyLog[2, a*x])/(8*a^2)

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Maple [A]  time = 0.046, size = 52, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}{\it polylog} \left ( 2,ax \right ) }{2}}+{\frac{{x}^{2}\ln \left ( -ax+1 \right ) }{4}}-{\frac{\ln \left ( -ax+1 \right ) }{4\,{a}^{2}}}-{\frac{{x}^{2}}{8}}-{\frac{x}{4\,a}}+{\frac{3}{8\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*polylog(2,a*x)+1/4*x^2*ln(-a*x+1)-1/4*ln(-a*x+1)/a^2-1/8*x^2-1/4*x/a+3/8/a^2

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Maxima [A]  time = 0.965084, size = 65, normalized size = 1.16 \begin{align*} \frac{4 \, a^{2} x^{2}{\rm Li}_2\left (a x\right ) - a^{2} x^{2} - 2 \, a x + 2 \,{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*a^2*x^2*dilog(a*x) - a^2*x^2 - 2*a*x + 2*(a^2*x^2 - 1)*log(-a*x + 1))/a^2

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Fricas [A]  time = 2.58187, size = 111, normalized size = 1.98 \begin{align*} \frac{4 \, a^{2} x^{2}{\rm Li}_2\left (a x\right ) - a^{2} x^{2} - 2 \, a x + 2 \,{\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*a^2*x^2*dilog(a*x) - a^2*x^2 - 2*a*x + 2*(a^2*x^2 - 1)*log(-a*x + 1))/a^2

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Sympy [A]  time = 2.43658, size = 41, normalized size = 0.73 \begin{align*} \begin{cases} - \frac{x^{2} \operatorname{Li}_{1}\left (a x\right )}{4} + \frac{x^{2} \operatorname{Li}_{2}\left (a x\right )}{2} - \frac{x^{2}}{8} - \frac{x}{4 a} + \frac{\operatorname{Li}_{1}\left (a x\right )}{4 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**2*polylog(1, a*x)/4 + x**2*polylog(2, a*x)/2 - x**2/8 - x/(4*a) + polylog(1, a*x)/(4*a**2), Ne(
a, 0)), (0, True))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*dilog(a*x), x)