3.1 \(\int x^4 \text{PolyLog}(2,a x) \, dx\)

Optimal. Leaf size=86 \[ \frac{1}{5} x^5 \text{PolyLog}(2,a x)-\frac{x^3}{75 a^2}-\frac{x^2}{50 a^3}-\frac{x}{25 a^4}-\frac{\log (1-a x)}{25 a^5}-\frac{x^4}{100 a}+\frac{1}{25} x^5 \log (1-a x)-\frac{x^5}{125} \]

[Out]

-x/(25*a^4) - x^2/(50*a^3) - x^3/(75*a^2) - x^4/(100*a) - x^5/125 - Log[1 - a*x]/(25*a^5) + (x^5*Log[1 - a*x])
/25 + (x^5*PolyLog[2, a*x])/5

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Rubi [A]  time = 0.0533019, antiderivative size = 86, 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} \[ \frac{1}{5} x^5 \text{PolyLog}(2,a x)-\frac{x^3}{75 a^2}-\frac{x^2}{50 a^3}-\frac{x}{25 a^4}-\frac{\log (1-a x)}{25 a^5}-\frac{x^4}{100 a}+\frac{1}{25} x^5 \log (1-a x)-\frac{x^5}{125} \]

Antiderivative was successfully verified.

[In]

Int[x^4*PolyLog[2, a*x],x]

[Out]

-x/(25*a^4) - x^2/(50*a^3) - x^3/(75*a^2) - x^4/(100*a) - x^5/125 - Log[1 - a*x]/(25*a^5) + (x^5*Log[1 - a*x])
/25 + (x^5*PolyLog[2, a*x])/5

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^4 \text{Li}_2(a x) \, dx &=\frac{1}{5} x^5 \text{Li}_2(a x)+\frac{1}{5} \int x^4 \log (1-a x) \, dx\\ &=\frac{1}{25} x^5 \log (1-a x)+\frac{1}{5} x^5 \text{Li}_2(a x)+\frac{1}{25} a \int \frac{x^5}{1-a x} \, dx\\ &=\frac{1}{25} x^5 \log (1-a x)+\frac{1}{5} x^5 \text{Li}_2(a x)+\frac{1}{25} a \int \left (-\frac{1}{a^5}-\frac{x}{a^4}-\frac{x^2}{a^3}-\frac{x^3}{a^2}-\frac{x^4}{a}-\frac{1}{a^5 (-1+a x)}\right ) \, dx\\ &=-\frac{x}{25 a^4}-\frac{x^2}{50 a^3}-\frac{x^3}{75 a^2}-\frac{x^4}{100 a}-\frac{x^5}{125}-\frac{\log (1-a x)}{25 a^5}+\frac{1}{25} x^5 \log (1-a x)+\frac{1}{5} x^5 \text{Li}_2(a x)\\ \end{align*}

Mathematica [A]  time = 0.04193, size = 73, normalized size = 0.85 \[ \frac{300 a^5 x^5 \text{PolyLog}(2,a x)-a x \left (12 a^4 x^4+15 a^3 x^3+20 a^2 x^2+30 a x+60\right )+60 \left (a^5 x^5-1\right ) \log (1-a x)}{1500 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*PolyLog[2, a*x],x]

[Out]

(-(a*x*(60 + 30*a*x + 20*a^2*x^2 + 15*a^3*x^3 + 12*a^4*x^4)) + 60*(-1 + a^5*x^5)*Log[1 - a*x] + 300*a^5*x^5*Po
lyLog[2, a*x])/(1500*a^5)

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Maple [A]  time = 0.046, size = 76, normalized size = 0.9 \begin{align*} -{\frac{x}{25\,{a}^{4}}}-{\frac{\ln \left ( -ax+1 \right ) }{25\,{a}^{5}}}-{\frac{{x}^{2}}{50\,{a}^{3}}}+{\frac{{x}^{5}{\it polylog} \left ( 2,ax \right ) }{5}}+{\frac{{x}^{5}\ln \left ( -ax+1 \right ) }{25}}+{\frac{137}{1500\,{a}^{5}}}-{\frac{{x}^{4}}{100\,a}}-{\frac{{x}^{3}}{75\,{a}^{2}}}-{\frac{{x}^{5}}{125}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*polylog(2,a*x),x)

[Out]

-1/25*x/a^4-1/25*ln(-a*x+1)/a^5-1/50*x^2/a^3+1/5*x^5*polylog(2,a*x)+1/25*x^5*ln(-a*x+1)+137/1500/a^5-1/100*x^4
/a-1/75*x^3/a^2-1/125*x^5

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Maxima [A]  time = 0.964112, size = 97, normalized size = 1.13 \begin{align*} \frac{300 \, a^{5} x^{5}{\rm Li}_2\left (a x\right ) - 12 \, a^{5} x^{5} - 15 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 30 \, a^{2} x^{2} - 60 \, a x + 60 \,{\left (a^{5} x^{5} - 1\right )} \log \left (-a x + 1\right )}{1500 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*polylog(2,a*x),x, algorithm="maxima")

[Out]

1/1500*(300*a^5*x^5*dilog(a*x) - 12*a^5*x^5 - 15*a^4*x^4 - 20*a^3*x^3 - 30*a^2*x^2 - 60*a*x + 60*(a^5*x^5 - 1)
*log(-a*x + 1))/a^5

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Fricas [A]  time = 2.68893, size = 177, normalized size = 2.06 \begin{align*} \frac{300 \, a^{5} x^{5}{\rm Li}_2\left (a x\right ) - 12 \, a^{5} x^{5} - 15 \, a^{4} x^{4} - 20 \, a^{3} x^{3} - 30 \, a^{2} x^{2} - 60 \, a x + 60 \,{\left (a^{5} x^{5} - 1\right )} \log \left (-a x + 1\right )}{1500 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*polylog(2,a*x),x, algorithm="fricas")

[Out]

1/1500*(300*a^5*x^5*dilog(a*x) - 12*a^5*x^5 - 15*a^4*x^4 - 20*a^3*x^3 - 30*a^2*x^2 - 60*a*x + 60*(a^5*x^5 - 1)
*log(-a*x + 1))/a^5

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Sympy [A]  time = 15.264, size = 66, normalized size = 0.77 \begin{align*} \begin{cases} - \frac{x^{5} \operatorname{Li}_{1}\left (a x\right )}{25} + \frac{x^{5} \operatorname{Li}_{2}\left (a x\right )}{5} - \frac{x^{5}}{125} - \frac{x^{4}}{100 a} - \frac{x^{3}}{75 a^{2}} - \frac{x^{2}}{50 a^{3}} - \frac{x}{25 a^{4}} + \frac{\operatorname{Li}_{1}\left (a x\right )}{25 a^{5}} & \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**4*polylog(2,a*x),x)

[Out]

Piecewise((-x**5*polylog(1, a*x)/25 + x**5*polylog(2, a*x)/5 - x**5/125 - x**4/(100*a) - x**3/(75*a**2) - x**2
/(50*a**3) - x/(25*a**4) + polylog(1, a*x)/(25*a**5), Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm Li}_2\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*polylog(2,a*x),x, algorithm="giac")

[Out]

integrate(x^4*dilog(a*x), x)