Integrand size = 9, antiderivative size = 76 \[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=-\frac {x}{16 a^3}-\frac {x^2}{32 a^2}-\frac {x^3}{48 a}-\frac {x^4}{64}-\frac {\log (1-a x)}{16 a^4}+\frac {1}{16} x^4 \log (1-a x)+\frac {1}{4} x^4 \operatorname {PolyLog}(2,a x) \]
-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)
Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86 \[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=\frac {-a x \left (12+6 a x+4 a^2 x^2+3 a^3 x^3\right )+12 \left (-1+a^4 x^4\right ) \log (1-a x)+48 a^4 x^4 \operatorname {PolyLog}(2,a x)}{192 a^4} \]
(-(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])/(192*a^4)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {7145, 25, 2842, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \operatorname {PolyLog}(2,a x) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {PolyLog}(2,a x)-\frac {1}{4} \int -x^3 \log (1-a x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \int x^3 \log (1-a x)dx+\frac {1}{4} x^4 \operatorname {PolyLog}(2,a x)\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} a \int \frac {x^4}{1-a x}dx+\frac {1}{4} x^4 \log (1-a x)\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,a x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} a \int \left (-\frac {x^3}{a}-\frac {x^2}{a^2}-\frac {x}{a^3}-\frac {1}{a^4 (a x-1)}-\frac {1}{a^4}\right )dx+\frac {1}{4} x^4 \log (1-a x)\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,a x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} a \left (-\frac {\log (1-a x)}{a^5}-\frac {x}{a^4}-\frac {x^2}{2 a^3}-\frac {x^3}{3 a^2}-\frac {x^4}{4 a}\right )+\frac {1}{4} x^4 \log (1-a x)\right )+\frac {1}{4} x^4 \operatorname {PolyLog}(2,a x)\) |
((x^4*Log[1 - a*x])/4 + (a*(-(x/a^4) - x^2/(2*a^3) - x^3/(3*a^2) - x^4/(4* a) - Log[1 - a*x]/a^5))/4)/4 + (x^4*PolyLog[2, a*x])/4
3.1.2.3.1 Defintions of rubi rules used
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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] - Simp[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]
Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbo l] :> Simp[(d*x)^(m + 1)*(PolyLog[n, a*(b*x^p)^q]/(d*(m + 1))), x] - Simp[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]
Time = 0.60 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.86
method | result | size |
meijerg | \(-\frac {\frac {a x \left (15 a^{3} x^{3}+20 a^{2} x^{2}+30 a x +60\right )}{960}+\frac {\left (-5 a^{4} x^{4}+5\right ) \ln \left (-a x +1\right )}{80}-\frac {a^{4} x^{4} \operatorname {polylog}\left (2, a x \right )}{4}}{a^{4}}\) | \(65\) |
parallelrisch | \(\frac {48 a^{4} x^{4} \operatorname {polylog}\left (2, a x \right )+12 \ln \left (-a x +1\right ) a^{4} x^{4}-3 a^{4} x^{4}-12-4 a^{3} x^{3}-6 a^{2} x^{2}-12 a x -12 \ln \left (-a x +1\right )}{192 a^{4}}\) | \(73\) |
parts | \(\frac {x^{4} \operatorname {polylog}\left (2, a x \right )}{4}+\frac {\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{4}-\frac {\left (-a x +1\right )^{4}}{16}-\ln \left (-a x +1\right ) \left (-a x +1\right )^{3}+\frac {\left (-a x +1\right )^{3}}{3}+\frac {3 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{2}-\frac {3 \left (-a x +1\right )^{2}}{4}-\ln \left (-a x +1\right ) \left (-a x +1\right )-a x +1}{4 a^{4}}\) | \(119\) |
derivativedivides | \(\frac {\frac {a^{4} x^{4} \operatorname {polylog}\left (2, a x \right )}{4}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{16}-\frac {\left (-a x +1\right )^{4}}{64}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{3}}{4}+\frac {\left (-a x +1\right )^{3}}{12}+\frac {3 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{8}-\frac {3 \left (-a x +1\right )^{2}}{16}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{4}+\frac {1}{4}-\frac {a x}{4}}{a^{4}}\) | \(120\) |
default | \(\frac {\frac {a^{4} x^{4} \operatorname {polylog}\left (2, a x \right )}{4}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{4}}{16}-\frac {\left (-a x +1\right )^{4}}{64}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{3}}{4}+\frac {\left (-a x +1\right )^{3}}{12}+\frac {3 \ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{8}-\frac {3 \left (-a x +1\right )^{2}}{16}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{4}+\frac {1}{4}-\frac {a x}{4}}{a^{4}}\) | \(120\) |
-1/a^4*(1/960*a*x*(15*a^3*x^3+20*a^2*x^2+30*a*x+60)+1/80*(-5*a^4*x^4+5)*ln (-a*x+1)-1/4*a^4*x^4*polylog(2,a*x))
Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=\frac {48 \, a^{4} x^{4} {\rm Li}_2\left (a x\right ) - 3 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 12 \, a x + 12 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )}{192 \, a^{4}} \]
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
Time = 1.40 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.76 \[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=\begin {cases} - \frac {x^{4} \operatorname {Li}_{1}\left (a x\right )}{16} + \frac {x^{4} \operatorname {Li}_{2}\left (a x\right )}{4} - \frac {x^{4}}{64} - \frac {x^{3}}{48 a} - \frac {x^{2}}{32 a^{2}} - \frac {x}{16 a^{3}} + \frac {\operatorname {Li}_{1}\left (a x\right )}{16 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
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))
Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=\frac {48 \, a^{4} x^{4} {\rm Li}_2\left (a x\right ) - 3 \, a^{4} x^{4} - 4 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 12 \, a x + 12 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right )}{192 \, a^{4}} \]
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
\[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=\int { x^{3} {\rm Li}_2\left (a x\right ) \,d x } \]
Time = 5.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.80 \[ \int x^3 \operatorname {PolyLog}(2,a x) \, dx=\frac {x^4\,\ln \left (1-a\,x\right )}{16}-\frac {\ln \left (a\,x-1\right )}{16\,a^4}-\frac {x}{16\,a^3}-\frac {x^4}{64}+\frac {x^4\,\mathrm {polylog}\left (2,a\,x\right )}{4}-\frac {x^3}{48\,a}-\frac {x^2}{32\,a^2} \]