3.11 \(\int x^3 \text {Li}_3(a x) \, dx\)

Optimal. Leaf size=88 \[ \frac {\log (1-a x)}{64 a^4}+\frac {x}{64 a^3}+\frac {x^2}{128 a^2}-\frac {1}{16} x^4 \text {Li}_2(a x)+\frac {1}{4} x^4 \text {Li}_3(a x)-\frac {1}{64} x^4 \log (1-a x)+\frac {x^3}{192 a}+\frac {x^4}{256} \]

[Out]

1/64*x/a^3+1/128*x^2/a^2+1/192*x^3/a+1/256*x^4+1/64*ln(-a*x+1)/a^4-1/64*x^4*ln(-a*x+1)-1/16*x^4*polylog(2,a*x)
+1/4*x^4*polylog(3,a*x)

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Rubi [A]  time = 0.06, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.01, size = 86, normalized size = 0.98 \[ \frac {-48 a^4 x^4 \text {Li}_2(a x)+192 a^4 x^4 \text {Li}_3(a x)+3 a^4 x^4-12 a^4 x^4 \log (1-a x)+4 a^3 x^3+6 a^2 x^2+12 a x+12 \log (1-a x)}{768 a^4} \]

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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fricas [C]  time = 0.54, size = 77, normalized size = 0.88 \[ -\frac {48 \, a^{4} x^{4} {\rm Li}_2\left (a x\right ) - 192 \, a^{4} x^{4} {\rm polylog}\left (3, a x\right ) - 3 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 12 \, a x + 12 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )}{768 \, a^{4}} \]

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*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Li}_{3}(a x)\,{d x} \]

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)

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maple [A]  time = 0.15, size = 78, normalized size = 0.89 \[ -\frac {-\frac {x a \left (15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{3840}-\frac {\left (-5 a^{4} x^{4}+5\right ) \ln \left (-a x +1\right )}{320}+\frac {x^{4} a^{4} \polylog \left (2, a x \right )}{16}-\frac {x^{4} a^{4} \polylog \left (3, a x \right )}{4}}{a^{4}} \]

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 = 0.33, size = 77, normalized size = 0.88 \[ -\frac {48 \, a^{4} x^{4} {\rm Li}_2\left (a x\right ) - 192 \, a^{4} x^{4} {\rm Li}_{3}(a x) - 3 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 12 \, a x + 12 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )}{768 \, a^{4}} \]

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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mupad [B]  time = 0.82, size = 71, normalized size = 0.81 \[ \frac {\ln \left (a\,x-1\right )}{64\,a^4}-\frac {x^4\,\ln \left (1-a\,x\right )}{64}+\frac {x}{64\,a^3}+\frac {x^4}{256}-\frac {x^4\,\mathrm {polylog}\left (2,a\,x\right )}{16}+\frac {x^4\,\mathrm {polylog}\left (3,a\,x\right )}{4}+\frac {x^3}{192\,a}+\frac {x^2}{128\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(3, a*x),x)

[Out]

log(a*x - 1)/(64*a^4) - (x^4*log(1 - a*x))/64 + x/(64*a^3) + x^4/256 - (x^4*polylog(2, a*x))/16 + (x^4*polylog
(3, a*x))/4 + x^3/(192*a) + x^2/(128*a^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {Li}_{3}\left (a x\right )\, dx \]

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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