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

Optimal. Leaf size=88 \[ -\frac{1}{16} x^4 \text{PolyLog}(2,a x)+\frac{1}{4} x^4 \text{PolyLog}(3,a x)+\frac{x^2}{128 a^2}+\frac{x}{64 a^3}+\frac{\log (1-a x)}{64 a^4}+\frac{x^3}{192 a}-\frac{1}{64} x^4 \log (1-a x)+\frac{x^4}{256} \]

[Out]

x/(64*a^3) + x^2/(128*a^2) + x^3/(192*a) + x^4/256 + Log[1 - a*x]/(64*a^4) - (x^4*Log[1 - a*x])/64 - (x^4*Poly
Log[2, a*x])/16 + (x^4*PolyLog[3, a*x])/4

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Rubi [A]  time = 0.0573359, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6591, 2395, 43} \[ -\frac{1}{16} x^4 \text{PolyLog}(2,a x)+\frac{1}{4} x^4 \text{PolyLog}(3,a x)+\frac{x^2}{128 a^2}+\frac{x}{64 a^3}+\frac{\log (1-a x)}{64 a^4}+\frac{x^3}{192 a}-\frac{1}{64} x^4 \log (1-a x)+\frac{x^4}{256} \]

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[3, a*x],x]

[Out]

x/(64*a^3) + x^2/(128*a^2) + x^3/(192*a) + x^4/256 + Log[1 - a*x]/(64*a^4) - (x^4*Log[1 - a*x])/64 - (x^4*Poly
Log[2, a*x])/16 + (x^4*PolyLog[3, 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}_3(a x) \, dx &=\frac{1}{4} x^4 \text{Li}_3(a x)-\frac{1}{4} \int x^3 \text{Li}_2(a x) \, dx\\ &=-\frac{1}{16} x^4 \text{Li}_2(a x)+\frac{1}{4} x^4 \text{Li}_3(a x)-\frac{1}{16} \int x^3 \log (1-a x) \, dx\\ &=-\frac{1}{64} x^4 \log (1-a x)-\frac{1}{16} x^4 \text{Li}_2(a x)+\frac{1}{4} x^4 \text{Li}_3(a x)-\frac{1}{64} a \int \frac{x^4}{1-a x} \, dx\\ &=-\frac{1}{64} x^4 \log (1-a x)-\frac{1}{16} x^4 \text{Li}_2(a x)+\frac{1}{4} x^4 \text{Li}_3(a x)-\frac{1}{64} 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}{64 a^3}+\frac{x^2}{128 a^2}+\frac{x^3}{192 a}+\frac{x^4}{256}+\frac{\log (1-a x)}{64 a^4}-\frac{1}{64} x^4 \log (1-a x)-\frac{1}{16} x^4 \text{Li}_2(a x)+\frac{1}{4} x^4 \text{Li}_3(a x)\\ \end{align*}

Mathematica [A]  time = 0.0118425, size = 86, normalized size = 0.98 \[ \frac{-48 a^4 x^4 \text{PolyLog}(2,a x)+192 a^4 x^4 \text{PolyLog}(3,a x)+3 a^4 x^4+4 a^3 x^3+6 a^2 x^2-12 a^4 x^4 \log (1-a x)+12 a x+12 \log (1-a x)}{768 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*PolyLog[3, a*x],x]

[Out]

(12*a*x + 6*a^2*x^2 + 4*a^3*x^3 + 3*a^4*x^4 + 12*Log[1 - a*x] - 12*a^4*x^4*Log[1 - a*x] - 48*a^4*x^4*PolyLog[2
, a*x] + 192*a^4*x^4*PolyLog[3, a*x])/(768*a^4)

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Maple [A]  time = 0.161, size = 78, normalized size = 0.9 \begin{align*} -{\frac{1}{{a}^{4}} \left ( -{\frac{xa \left ( 15\,{x}^{3}{a}^{3}+20\,{a}^{2}{x}^{2}+30\,ax+60 \right ) }{3840}}-{\frac{ \left ( -5\,{x}^{4}{a}^{4}+5 \right ) \ln \left ( -ax+1 \right ) }{320}}+{\frac{{x}^{4}{a}^{4}{\it polylog} \left ( 2,ax \right ) }{16}}-{\frac{{x}^{4}{a}^{4}{\it polylog} \left ( 3,ax \right ) }{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^4*(-1/3840*x*a*(15*a^3*x^3+20*a^2*x^2+30*a*x+60)-1/320*(-5*a^4*x^4+5)*ln(-a*x+1)+1/16*x^4*a^4*polylog(2,a
*x)-1/4*x^4*a^4*polylog(3,a*x))

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Maxima [A]  time = 0.998854, size = 104, normalized size = 1.18 \begin{align*} -\frac{48 \, a^{4} x^{4}{\rm Li}_2\left (a x\right ) - 192 \, a^{4} x^{4}{\rm Li}_{3}(a x) - 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 )}{768 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(48*a^4*x^4*dilog(a*x) - 192*a^4*x^4*polylog(3, 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 [C]  time = 2.54882, size = 244, normalized size = 2.77 \begin{align*} -\frac{48 \, a^{4} x^{4}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x + 1\right )}{a}, -\frac{\log \left (-a x + 1\right )}{x}\right ) - 192 \, a^{4} x^{4}{\rm polylog}\left (3, 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 )}{768 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(48*a^4*x^4*\%iint(a, x, -log(-a*x + 1)/a, -log(-a*x + 1)/x) - 192*a^4*x^4*polylog(3, 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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{Li}_{3}\left (a x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*polylog(3,a*x),x)

[Out]

Integral(x**3*polylog(3, a*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*polylog(3, a*x), x)