3.2 \(\int x^3 \text{PolyLog}(2,a x) \, dx\)

Optimal. Leaf size=76 \[ \frac{1}{4} x^4 \text{PolyLog}(2,a x)-\frac{x^2}{32 a^2}-\frac{x}{16 a^3}-\frac{\log (1-a x)}{16 a^4}-\frac{x^3}{48 a}+\frac{1}{16} x^4 \log (1-a x)-\frac{x^4}{64} \]

[Out]

-x/(16*a^3) - x^2/(32*a^2) - x^3/(48*a) - x^4/64 - Log[1 - a*x]/(16*a^4) + (x^4*Log[1 - a*x])/16 + (x^4*PolyLo
g[2, a*x])/4

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Rubi [A]  time = 0.0447573, antiderivative size = 76, 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} \[ \frac{1}{4} x^4 \text{PolyLog}(2,a x)-\frac{x^2}{32 a^2}-\frac{x}{16 a^3}-\frac{\log (1-a x)}{16 a^4}-\frac{x^3}{48 a}+\frac{1}{16} x^4 \log (1-a x)-\frac{x^4}{64} \]

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[2, a*x],x]

[Out]

-x/(16*a^3) - x^2/(32*a^2) - x^3/(48*a) - x^4/64 - Log[1 - a*x]/(16*a^4) + (x^4*Log[1 - a*x])/16 + (x^4*PolyLo
g[2, a*x])/4

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^3 \text{Li}_2(a x) \, dx &=\frac{1}{4} x^4 \text{Li}_2(a x)+\frac{1}{4} \int x^3 \log (1-a x) \, dx\\ &=\frac{1}{16} x^4 \log (1-a x)+\frac{1}{4} x^4 \text{Li}_2(a x)+\frac{1}{16} a \int \frac{x^4}{1-a x} \, dx\\ &=\frac{1}{16} x^4 \log (1-a x)+\frac{1}{4} x^4 \text{Li}_2(a x)+\frac{1}{16} a \int \left (-\frac{1}{a^4}-\frac{x}{a^3}-\frac{x^2}{a^2}-\frac{x^3}{a}-\frac{1}{a^4 (-1+a x)}\right ) \, dx\\ &=-\frac{x}{16 a^3}-\frac{x^2}{32 a^2}-\frac{x^3}{48 a}-\frac{x^4}{64}-\frac{\log (1-a x)}{16 a^4}+\frac{1}{16} x^4 \log (1-a x)+\frac{1}{4} x^4 \text{Li}_2(a x)\\ \end{align*}

Mathematica [A]  time = 0.0327425, size = 65, normalized size = 0.86 \[ \frac{48 a^4 x^4 \text{PolyLog}(2,a x)-a x \left (3 a^3 x^3+4 a^2 x^2+6 a x+12\right )+12 \left (a^4 x^4-1\right ) \log (1-a x)}{192 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*PolyLog[2, a*x],x]

[Out]

(-(a*x*(12 + 6*a*x + 4*a^2*x^2 + 3*a^3*x^3)) + 12*(-1 + a^4*x^4)*Log[1 - a*x] + 48*a^4*x^4*PolyLog[2, a*x])/(1
92*a^4)

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Maple [A]  time = 0.046, size = 68, normalized size = 0.9 \begin{align*}{\frac{{x}^{4}{\it polylog} \left ( 2,ax \right ) }{4}}+{\frac{{x}^{4}\ln \left ( -ax+1 \right ) }{16}}-{\frac{\ln \left ( -ax+1 \right ) }{16\,{a}^{4}}}-{\frac{{x}^{4}}{64}}-{\frac{{x}^{3}}{48\,a}}-{\frac{{x}^{2}}{32\,{a}^{2}}}-{\frac{x}{16\,{a}^{3}}}+{\frac{25}{192\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(2,a*x),x)

[Out]

1/4*x^4*polylog(2,a*x)+1/16*x^4*ln(-a*x+1)-1/16*ln(-a*x+1)/a^4-1/64*x^4-1/48*x^3/a-1/32*x^2/a^2-1/16*x/a^3+25/
192/a^4

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Maxima [A]  time = 0.98341, size = 86, normalized size = 1.13 \begin{align*} \frac{48 \, a^{4} x^{4}{\rm Li}_2\left (a x\right ) - 3 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 12 \, a x + 12 \,{\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )}{192 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(2,a*x),x, algorithm="maxima")

[Out]

1/192*(48*a^4*x^4*dilog(a*x) - 3*a^4*x^4 - 4*a^3*x^3 - 6*a^2*x^2 - 12*a*x + 12*(a^4*x^4 - 1)*log(-a*x + 1))/a^
4

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Fricas [A]  time = 2.66203, size = 153, normalized size = 2.01 \begin{align*} \frac{48 \, a^{4} x^{4}{\rm Li}_2\left (a x\right ) - 3 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 12 \, a x + 12 \,{\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )}{192 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(2,a*x),x, algorithm="fricas")

[Out]

1/192*(48*a^4*x^4*dilog(a*x) - 3*a^4*x^4 - 4*a^3*x^3 - 6*a^2*x^2 - 12*a*x + 12*(a^4*x^4 - 1)*log(-a*x + 1))/a^
4

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Sympy [A]  time = 8.74118, size = 58, normalized size = 0.76 \begin{align*} \begin{cases} - \frac{x^{4} \operatorname{Li}_{1}\left (a x\right )}{16} + \frac{x^{4} \operatorname{Li}_{2}\left (a x\right )}{4} - \frac{x^{4}}{64} - \frac{x^{3}}{48 a} - \frac{x^{2}}{32 a^{2}} - \frac{x}{16 a^{3}} + \frac{\operatorname{Li}_{1}\left (a x\right )}{16 a^{4}} & \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**3*polylog(2,a*x),x)

[Out]

Piecewise((-x**4*polylog(1, a*x)/16 + x**4*polylog(2, a*x)/4 - x**4/64 - x**3/(48*a) - x**2/(32*a**2) - x/(16*
a**3) + polylog(1, a*x)/(16*a**4), Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm Li}_2\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(2,a*x),x, algorithm="giac")

[Out]

integrate(x^3*dilog(a*x), x)