3.1.1 \(\int x^4 \text {PolyLog}(2,a x) \, dx\) [1]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {6726, 2442, 45} \begin {gather*} -\frac {\log (1-a x)}{25 a^5}-\frac {x}{25 a^4}-\frac {x^2}{50 a^3}-\frac {x^3}{75 a^2}+\frac {1}{5} x^5 \text {Li}_2(a x)+\frac {1}{25} x^5 \log (1-a x)-\frac {x^4}{100 a}-\frac {x^5}{125} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*PolyLog[2, a*x],x]

[Out]

-1/25*x/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 45

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 2442

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 6726

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^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*}

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Mathematica [A]
time = 0.03, size = 73, normalized size = 0.85 \begin {gather*} \frac {-a x \left (60+30 a x+20 a^2 x^2+15 a^3 x^3+12 a^4 x^4\right )+60 \left (-1+a^5 x^5\right ) \log (1-a x)+300 a^5 x^5 \text {PolyLog}(2,a x)}{1500 a^5} \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(146\) vs. \(2(70)=140\).
time = 0.36, size = 147, normalized size = 1.71

method	result	size
meijerg	\(\frac {-\frac {a x \left (12 a^{4} x^{4}+15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{1500}-\frac {\left (-6 a^{5} x^{5}+6\right ) \ln \left (-a x +1\right )}{150}+\frac {a^{5} x^{5} \polylog \left (2, a x \right )}{5}}{a^{5}}\)	\(72\)
derivativedivides	\(\frac {\frac {a^{5} x^{5} \polylog \left (2, a x \right )}{5}-\frac {\left (-a x +1\right )^{5} \ln \left (-a x +1\right )}{25}+\frac {\left (-a x +1\right )^{5}}{125}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{5}-\frac {\left (-a x +1\right )^{4}}{20}-\frac {2 \ln \left (-a x +1\right ) \left (-a x +1\right )^{3}}{5}+\frac {2 \left (-a x +1\right )^{3}}{15}+\frac {2 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{5}-\frac {\left (-a x +1\right )^{2}}{5}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{5}+\frac {1}{5}-\frac {a x}{5}}{a^{5}}\)	\(147\)
default	\(\frac {\frac {a^{5} x^{5} \polylog \left (2, a x \right )}{5}-\frac {\left (-a x +1\right )^{5} \ln \left (-a x +1\right )}{25}+\frac {\left (-a x +1\right )^{5}}{125}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{5}-\frac {\left (-a x +1\right )^{4}}{20}-\frac {2 \ln \left (-a x +1\right ) \left (-a x +1\right )^{3}}{5}+\frac {2 \left (-a x +1\right )^{3}}{15}+\frac {2 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{5}-\frac {\left (-a x +1\right )^{2}}{5}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{5}+\frac {1}{5}-\frac {a x}{5}}{a^{5}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*polylog(2,a*x),x,method=_RETURNVERBOSE)

[Out]

1/a^5*(1/5*a^5*x^5*polylog(2,a*x)-1/25*(-a*x+1)^5*ln(-a*x+1)+1/125*(-a*x+1)^5+1/5*ln(-a*x+1)*(-a*x+1)^4-1/20*(
-a*x+1)^4-2/5*ln(-a*x+1)*(-a*x+1)^3+2/15*(-a*x+1)^3+2/5*ln(-a*x+1)*(-a*x+1)^2-1/5*(-a*x+1)^2-1/5*ln(-a*x+1)*(-
a*x+1)+1/5-1/5*a*x)

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Maxima [A]
time = 0.27, size = 72, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 0.37, size = 72, normalized size = 0.84 \begin {gather*} \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 {gather*}

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 = 3.19, size = 66, normalized size = 0.77 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.31, size = 69, normalized size = 0.80 \begin {gather*} \frac {x^5\,\ln \left (1-a\,x\right )}{25}-\frac {\ln \left (a\,x-1\right )}{25\,a^5}-\frac {x}{25\,a^4}-\frac {x^5}{125}+\frac {x^5\,\mathrm {polylog}\left (2,a\,x\right )}{5}-\frac {x^4}{100\,a}-\frac {x^3}{75\,a^2}-\frac {x^2}{50\,a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*polylog(2, a*x),x)

[Out]

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

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