3.1.2 \(\int x^3 \text {PolyLog}(2,a x) \, dx\) [2]

Optimal. Leaf size=76 \[ -\frac {x}{16 a^3}-\frac {x^2}{32 a^2}-\frac {x^3}{48 a}-\frac {x^4}{64}-\frac {\log (1-a x)}{16 a^4}+\frac {1}{16} x^4 \log (1-a x)+\frac {1}{4} x^4 \text {PolyLog}(2,a x) \]

[Out]

-1/16*x/a^3-1/32*x^2/a^2-1/48*x^3/a-1/64*x^4-1/16*ln(-a*x+1)/a^4+1/16*x^4*ln(-a*x+1)+1/4*x^4*polylog(2,a*x)

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Rubi [A]
time = 0.03, antiderivative size = 76, 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)}{16 a^4}-\frac {x}{16 a^3}-\frac {x^2}{32 a^2}+\frac {1}{4} x^4 \text {Li}_2(a x)+\frac {1}{16} x^4 \log (1-a x)-\frac {x^3}{48 a}-\frac {x^4}{64} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*PolyLog[2, a*x],x]

[Out]

-1/16*x/a^3 - x^2/(32*a^2) - x^3/(48*a) - x^4/64 - Log[1 - a*x]/(16*a^4) + (x^4*Log[1 - a*x])/16 + (x^4*PolyLo
g[2, a*x])/4

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^3 \text {Li}_2(a x) \, dx &=\frac {1}{4} x^4 \text {Li}_2(a x)+\frac {1}{4} \int x^3 \log (1-a x) \, dx\\ &=\frac {1}{16} x^4 \log (1-a x)+\frac {1}{4} x^4 \text {Li}_2(a x)+\frac {1}{16} a \int \frac {x^4}{1-a x} \, dx\\ &=\frac {1}{16} x^4 \log (1-a x)+\frac {1}{4} x^4 \text {Li}_2(a x)+\frac {1}{16} 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}{16 a^3}-\frac {x^2}{32 a^2}-\frac {x^3}{48 a}-\frac {x^4}{64}-\frac {\log (1-a x)}{16 a^4}+\frac {1}{16} x^4 \log (1-a x)+\frac {1}{4} x^4 \text {Li}_2(a x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 65, normalized size = 0.86 \begin {gather*} \frac {-a x \left (12+6 a x+4 a^2 x^2+3 a^3 x^3\right )+12 \left (-1+a^4 x^4\right ) \log (1-a x)+48 a^4 x^4 \text {PolyLog}(2,a x)}{192 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*PolyLog[2, a*x],x]

[Out]

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

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Maple [A]
time = 0.36, size = 120, normalized size = 1.58

method	result	size
meijerg	\(-\frac {\frac {a x \left (15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{960}+\frac {\left (-5 a^{4} x^{4}+5\right ) \ln \left (-a x +1\right )}{80}-\frac {a^{4} x^{4} \polylog \left (2, a x \right )}{4}}{a^{4}}\)	\(65\)
derivativedivides	\(\frac {\frac {a^{4} x^{4} \polylog \left (2, a x \right )}{4}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{16}-\frac {\left (-a x +1\right )^{4}}{64}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{3}}{4}+\frac {\left (-a x +1\right )^{3}}{12}+\frac {3 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{8}-\frac {3 \left (-a x +1\right )^{2}}{16}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{4}+\frac {1}{4}-\frac {a x}{4}}{a^{4}}\)	\(120\)
default	\(\frac {\frac {a^{4} x^{4} \polylog \left (2, a x \right )}{4}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{16}-\frac {\left (-a x +1\right )^{4}}{64}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{3}}{4}+\frac {\left (-a x +1\right )^{3}}{12}+\frac {3 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{8}-\frac {3 \left (-a x +1\right )^{2}}{16}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{4}+\frac {1}{4}-\frac {a x}{4}}{a^{4}}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(2,a*x),x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4*a^4*x^4*polylog(2,a*x)+1/16*ln(-a*x+1)*(-a*x+1)^4-1/64*(-a*x+1)^4-1/4*ln(-a*x+1)*(-a*x+1)^3+1/12*(-
a*x+1)^3+3/8*ln(-a*x+1)*(-a*x+1)^2-3/16*(-a*x+1)^2-1/4*ln(-a*x+1)*(-a*x+1)+1/4-1/4*a*x)

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Maxima [A]
time = 0.26, size = 64, normalized size = 0.84 \begin {gather*} \frac {48 \, a^{4} x^{4} {\rm Li}_2\left (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 )}{192 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(2,a*x),x, algorithm="maxima")

[Out]

1/192*(48*a^4*x^4*dilog(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 [A]
time = 0.35, size = 64, normalized size = 0.84 \begin {gather*} \frac {48 \, a^{4} x^{4} {\rm Li}_2\left (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 )}{192 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*polylog(2,a*x),x, algorithm="fricas")

[Out]

1/192*(48*a^4*x^4*dilog(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 [A]
time = 1.73, size = 58, normalized size = 0.76 \begin {gather*} \begin {cases} - \frac {x^{4} \operatorname {Li}_{1}\left (a x\right )}{16} + \frac {x^{4} \operatorname {Li}_{2}\left (a x\right )}{4} - \frac {x^{4}}{64} - \frac {x^{3}}{48 a} - \frac {x^{2}}{32 a^{2}} - \frac {x}{16 a^{3}} + \frac {\operatorname {Li}_{1}\left (a x\right )}{16 a^{4}} & \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**3*polylog(2,a*x),x)

[Out]

Piecewise((-x**4*polylog(1, a*x)/16 + x**4*polylog(2, a*x)/4 - x**4/64 - x**3/(48*a) - x**2/(32*a**2) - x/(16*
a**3) + polylog(1, a*x)/(16*a**4), 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^3*polylog(2,a*x),x, algorithm="giac")

[Out]

integrate(x^3*dilog(a*x), x)

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Mupad [B]
time = 0.25, size = 61, normalized size = 0.80 \begin {gather*} \frac {x^4\,\ln \left (1-a\,x\right )}{16}-\frac {\ln \left (a\,x-1\right )}{16\,a^4}-\frac {x}{16\,a^3}-\frac {x^4}{64}+\frac {x^4\,\mathrm {polylog}\left (2,a\,x\right )}{4}-\frac {x^3}{48\,a}-\frac {x^2}{32\,a^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*polylog(2, a*x),x)

[Out]

(x^4*log(1 - a*x))/16 - log(a*x - 1)/(16*a^4) - x/(16*a^3) - x^4/64 + (x^4*polylog(2, a*x))/4 - x^3/(48*a) - x
^2/(32*a^2)

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