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

Optimal. Leaf size=66 \[ \frac{1}{3} x^3 \text{PolyLog}(2,a x)-\frac{x}{9 a^2}-\frac{\log (1-a x)}{9 a^3}-\frac{x^2}{18 a}+\frac{1}{9} x^3 \log (1-a x)-\frac{x^3}{27} \]

[Out]

-x/(9*a^2) - x^2/(18*a) - x^3/27 - Log[1 - a*x]/(9*a^3) + (x^3*Log[1 - a*x])/9 + (x^3*PolyLog[2, a*x])/3

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Rubi [A]  time = 0.0396548, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6591, 2395, 43} \[ \frac{1}{3} x^3 \text{PolyLog}(2,a x)-\frac{x}{9 a^2}-\frac{\log (1-a x)}{9 a^3}-\frac{x^2}{18 a}+\frac{1}{9} x^3 \log (1-a x)-\frac{x^3}{27} \]

Antiderivative was successfully verified.

[In]

Int[x^2*PolyLog[2, a*x],x]

[Out]

-x/(9*a^2) - x^2/(18*a) - x^3/27 - Log[1 - a*x]/(9*a^3) + (x^3*Log[1 - a*x])/9 + (x^3*PolyLog[2, a*x])/3

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^2 \text{Li}_2(a x) \, dx &=\frac{1}{3} x^3 \text{Li}_2(a x)+\frac{1}{3} \int x^2 \log (1-a x) \, dx\\ &=\frac{1}{9} x^3 \log (1-a x)+\frac{1}{3} x^3 \text{Li}_2(a x)+\frac{1}{9} a \int \frac{x^3}{1-a x} \, dx\\ &=\frac{1}{9} x^3 \log (1-a x)+\frac{1}{3} x^3 \text{Li}_2(a x)+\frac{1}{9} a \int \left (-\frac{1}{a^3}-\frac{x}{a^2}-\frac{x^2}{a}-\frac{1}{a^3 (-1+a x)}\right ) \, dx\\ &=-\frac{x}{9 a^2}-\frac{x^2}{18 a}-\frac{x^3}{27}-\frac{\log (1-a x)}{9 a^3}+\frac{1}{9} x^3 \log (1-a x)+\frac{1}{3} x^3 \text{Li}_2(a x)\\ \end{align*}

Mathematica [A]  time = 0.0292432, size = 57, normalized size = 0.86 \[ \frac{18 a^3 x^3 \text{PolyLog}(2,a x)-a x \left (2 a^2 x^2+3 a x+6\right )+6 \left (a^3 x^3-1\right ) \log (1-a x)}{54 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*PolyLog[2, a*x],x]

[Out]

(-(a*x*(6 + 3*a*x + 2*a^2*x^2)) + 6*(-1 + a^3*x^3)*Log[1 - a*x] + 18*a^3*x^3*PolyLog[2, a*x])/(54*a^3)

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Maple [A]  time = 0.047, size = 60, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}{\it polylog} \left ( 2,ax \right ) }{3}}+{\frac{{x}^{3}\ln \left ( -ax+1 \right ) }{9}}-{\frac{\ln \left ( -ax+1 \right ) }{9\,{a}^{3}}}-{\frac{{x}^{3}}{27}}-{\frac{{x}^{2}}{18\,a}}-{\frac{x}{9\,{a}^{2}}}+{\frac{11}{54\,{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*polylog(2,a*x)+1/9*x^3*ln(-a*x+1)-1/9*ln(-a*x+1)/a^3-1/27*x^3-1/18*x^2/a-1/9*x/a^2+11/54/a^3

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Maxima [A]  time = 0.973299, size = 76, normalized size = 1.15 \begin{align*} \frac{18 \, a^{3} x^{3}{\rm Li}_2\left (a x\right ) - 2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - 6 \, a x + 6 \,{\left (a^{3} x^{3} - 1\right )} \log \left (-a x + 1\right )}{54 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(18*a^3*x^3*dilog(a*x) - 2*a^3*x^3 - 3*a^2*x^2 - 6*a*x + 6*(a^3*x^3 - 1)*log(-a*x + 1))/a^3

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Fricas [A]  time = 2.59888, size = 132, normalized size = 2. \begin{align*} \frac{18 \, a^{3} x^{3}{\rm Li}_2\left (a x\right ) - 2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - 6 \, a x + 6 \,{\left (a^{3} x^{3} - 1\right )} \log \left (-a x + 1\right )}{54 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(18*a^3*x^3*dilog(a*x) - 2*a^3*x^3 - 3*a^2*x^2 - 6*a*x + 6*(a^3*x^3 - 1)*log(-a*x + 1))/a^3

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Sympy [A]  time = 4.42809, size = 49, normalized size = 0.74 \begin{align*} \begin{cases} - \frac{x^{3} \operatorname{Li}_{1}\left (a x\right )}{9} + \frac{x^{3} \operatorname{Li}_{2}\left (a x\right )}{3} - \frac{x^{3}}{27} - \frac{x^{2}}{18 a} - \frac{x}{9 a^{2}} + \frac{\operatorname{Li}_{1}\left (a x\right )}{9 a^{3}} & \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**2*polylog(2,a*x),x)

[Out]

Piecewise((-x**3*polylog(1, a*x)/9 + x**3*polylog(2, a*x)/3 - x**3/27 - x**2/(18*a) - x/(9*a**2) + polylog(1,
a*x)/(9*a**3), Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm Li}_2\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*dilog(a*x), x)