3.13 \(\int x \text {Li}_3(a x) \, dx\)

Optimal. Leaf size=68 \[ \frac {\log (1-a x)}{8 a^2}-\frac {1}{4} x^2 \text {Li}_2(a x)+\frac {1}{2} x^2 \text {Li}_3(a x)-\frac {1}{8} x^2 \log (1-a x)+\frac {x}{8 a}+\frac {x^2}{16} \]

[Out]

1/8*x/a+1/16*x^2+1/8*ln(-a*x+1)/a^2-1/8*x^2*ln(-a*x+1)-1/4*x^2*polylog(2,a*x)+1/2*x^2*polylog(3,a*x)

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6591, 2395, 43} \[ -\frac {1}{4} x^2 \text {PolyLog}(2,a x)+\frac {1}{2} x^2 \text {PolyLog}(3,a x)+\frac {\log (1-a x)}{8 a^2}-\frac {1}{8} x^2 \log (1-a x)+\frac {x}{8 a}+\frac {x^2}{16} \]

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[3, a*x],x]

[Out]

x/(8*a) + x^2/16 + Log[1 - a*x]/(8*a^2) - (x^2*Log[1 - a*x])/8 - (x^2*PolyLog[2, a*x])/4 + (x^2*PolyLog[3, a*x
])/2

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 \text {Li}_3(a x) \, dx &=\frac {1}{2} x^2 \text {Li}_3(a x)-\frac {1}{2} \int x \text {Li}_2(a x) \, dx\\ &=-\frac {1}{4} x^2 \text {Li}_2(a x)+\frac {1}{2} x^2 \text {Li}_3(a x)-\frac {1}{4} \int x \log (1-a x) \, dx\\ &=-\frac {1}{8} x^2 \log (1-a x)-\frac {1}{4} x^2 \text {Li}_2(a x)+\frac {1}{2} x^2 \text {Li}_3(a x)-\frac {1}{8} a \int \frac {x^2}{1-a x} \, dx\\ &=-\frac {1}{8} x^2 \log (1-a x)-\frac {1}{4} x^2 \text {Li}_2(a x)+\frac {1}{2} x^2 \text {Li}_3(a x)-\frac {1}{8} a \int \left (-\frac {1}{a^2}-\frac {x}{a}-\frac {1}{a^2 (-1+a x)}\right ) \, dx\\ &=\frac {x}{8 a}+\frac {x^2}{16}+\frac {\log (1-a x)}{8 a^2}-\frac {1}{8} x^2 \log (1-a x)-\frac {1}{4} x^2 \text {Li}_2(a x)+\frac {1}{2} x^2 \text {Li}_3(a x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 69, normalized size = 1.01 \[ \frac {-4 a^2 x^2 \text {Li}_2(a x)+8 a^2 x^2 \text {Li}_3(a x)+a^2 x^2-2 a^2 x^2 \log (1-a x)+2 a x+2 \log (1-a x)}{16 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[3, a*x],x]

[Out]

(2*a*x + a^2*x^2 + 2*Log[1 - a*x] - 2*a^2*x^2*Log[1 - a*x] - 4*a^2*x^2*PolyLog[2, a*x] + 8*a^2*x^2*PolyLog[3,
a*x])/(16*a^2)

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fricas [C]  time = 0.77, size = 61, normalized size = 0.90 \[ -\frac {4 \, a^{2} x^{2} {\rm Li}_2\left (a x\right ) - 8 \, a^{2} x^{2} {\rm polylog}\left (3, a x\right ) - a^{2} x^{2} - 2 \, a x + 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}{16 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a^2*x^2*dilog(a*x) - 8*a^2*x^2*polylog(3, 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm Li}_{3}(a x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*polylog(3, a*x), x)

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maple [A]  time = 0.13, size = 62, normalized size = 0.91 \[ -\frac {-\frac {a x \left (3 a x +6\right )}{48}-\frac {\left (-3 a^{2} x^{2}+3\right ) \ln \left (-a x +1\right )}{24}+\frac {a^{2} x^{2} \polylog \left (2, a x \right )}{4}-\frac {a^{2} x^{2} \polylog \left (3, a x \right )}{2}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^2*(-1/48*a*x*(3*a*x+6)-1/24*(-3*a^2*x^2+3)*ln(-a*x+1)+1/4*a^2*x^2*polylog(2,a*x)-1/2*a^2*x^2*polylog(3,a*
x))

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maxima [A]  time = 0.33, size = 61, normalized size = 0.90 \[ -\frac {4 \, a^{2} x^{2} {\rm Li}_2\left (a x\right ) - 8 \, a^{2} x^{2} {\rm Li}_{3}(a x) - a^{2} x^{2} - 2 \, a x + 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}{16 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a^2*x^2*dilog(a*x) - 8*a^2*x^2*polylog(3, 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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mupad [B]  time = 0.90, size = 55, normalized size = 0.81 \[ \frac {\ln \left (a\,x-1\right )}{8\,a^2}-\frac {x^2\,\ln \left (1-a\,x\right )}{8}+\frac {x}{8\,a}+\frac {x^2}{16}-\frac {x^2\,\mathrm {polylog}\left (2,a\,x\right )}{4}+\frac {x^2\,\mathrm {polylog}\left (3,a\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(3, a*x),x)

[Out]

log(a*x - 1)/(8*a^2) - (x^2*log(1 - a*x))/8 + x/(8*a) + x^2/16 - (x^2*polylog(2, a*x))/4 + (x^2*polylog(3, a*x
))/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*polylog(3, a*x), x)

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