Integrand size = 11, antiderivative size = 88 \[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\frac {x^2}{54 a^2}+\frac {x^4}{108 a}+\frac {x^6}{162}+\frac {\log \left (1-a x^2\right )}{54 a^3}-\frac {1}{54} x^6 \log \left (1-a x^2\right )-\frac {1}{18} x^6 \operatorname {PolyLog}\left (2,a x^2\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right ) \]
1/54*x^2/a^2+1/108*x^4/a+1/162*x^6+1/54*ln(-a*x^2+1)/a^3-1/54*x^6*ln(-a*x^ 2+1)-1/18*x^6*polylog(2,a*x^2)+1/6*x^6*polylog(3,a*x^2)
Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\frac {6 a x^2+3 a^2 x^4+2 a^3 x^6+6 \log \left (1-a x^2\right )-6 a^3 x^6 \log \left (1-a x^2\right )-18 a^3 x^6 \operatorname {PolyLog}\left (2,a x^2\right )+54 a^3 x^6 \operatorname {PolyLog}\left (3,a x^2\right )}{324 a^3} \]
(6*a*x^2 + 3*a^2*x^4 + 2*a^3*x^6 + 6*Log[1 - a*x^2] - 6*a^3*x^6*Log[1 - a* x^2] - 18*a^3*x^6*PolyLog[2, a*x^2] + 54*a^3*x^6*PolyLog[3, a*x^2])/(324*a ^3)
Time = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {7145, 7145, 25, 2904, 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.
\(\displaystyle \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )-\frac {1}{3} \int x^5 \operatorname {PolyLog}\left (2,a x^2\right )dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int -x^5 \log \left (1-a x^2\right )dx-\frac {1}{6} x^6 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{3} \int x^5 \log \left (1-a x^2\right )dx-\frac {1}{6} x^6 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{6} \int x^4 \log \left (1-a x^2\right )dx^2-\frac {1}{6} x^6 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{6} \left (-\frac {1}{3} a \int \frac {x^6}{1-a x^2}dx^2-\frac {1}{3} x^6 \log \left (1-a x^2\right )\right )-\frac {1}{6} x^6 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{6} \left (-\frac {1}{3} a \int \left (-\frac {x^4}{a}-\frac {x^2}{a^2}-\frac {1}{a^3 \left (a x^2-1\right )}-\frac {1}{a^3}\right )dx^2-\frac {1}{3} x^6 \log \left (1-a x^2\right )\right )-\frac {1}{6} x^6 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{6} \left (-\frac {1}{3} a \left (-\frac {\log \left (1-a x^2\right )}{a^4}-\frac {x^2}{a^3}-\frac {x^4}{2 a^2}-\frac {x^6}{3 a}\right )-\frac {1}{3} x^6 \log \left (1-a x^2\right )\right )-\frac {1}{6} x^6 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{6} x^6 \operatorname {PolyLog}\left (3,a x^2\right )\) |
((-1/3*(x^6*Log[1 - a*x^2]) - (a*(-(x^2/a^3) - x^4/(2*a^2) - x^6/(3*a) - L og[1 - a*x^2]/a^4))/3)/6 - (x^6*PolyLog[2, a*x^2])/6)/3 + (x^6*PolyLog[3, a*x^2])/6
3.1.32.3.1 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[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
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 = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.91
method | result | size |
meijerg | \(\frac {\frac {a \,x^{2} \left (4 x^{4} a^{2}+6 a \,x^{2}+12\right )}{324}+\frac {\left (-4 a^{3} x^{6}+4\right ) \ln \left (-a \,x^{2}+1\right )}{108}-\frac {a^{3} x^{6} \operatorname {polylog}\left (2, a \,x^{2}\right )}{9}+\frac {a^{3} x^{6} \operatorname {polylog}\left (3, a \,x^{2}\right )}{3}}{2 a^{3}}\) | \(80\) |
1/2/a^3*(1/324*a*x^2*(4*a^2*x^4+6*a*x^2+12)+1/108*(-4*a^3*x^6+4)*ln(-a*x^2 +1)-1/9*a^3*x^6*polylog(2,a*x^2)+1/3*a^3*x^6*polylog(3,a*x^2))
Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=-\frac {18 \, a^{3} x^{6} {\rm Li}_2\left (a x^{2}\right ) - 54 \, a^{3} x^{6} {\rm polylog}\left (3, a x^{2}\right ) - 2 \, a^{3} x^{6} - 3 \, a^{2} x^{4} - 6 \, a x^{2} + 6 \, {\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right )}{324 \, a^{3}} \]
-1/324*(18*a^3*x^6*dilog(a*x^2) - 54*a^3*x^6*polylog(3, a*x^2) - 2*a^3*x^6 - 3*a^2*x^4 - 6*a*x^2 + 6*(a^3*x^6 - 1)*log(-a*x^2 + 1))/a^3
\[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\int x^{5} \operatorname {Li}_{3}\left (a x^{2}\right )\, dx \]
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=-\frac {18 \, a^{3} x^{6} {\rm Li}_2\left (a x^{2}\right ) - 54 \, a^{3} x^{6} {\rm Li}_{3}(a x^{2}) - 2 \, a^{3} x^{6} - 3 \, a^{2} x^{4} - 6 \, a x^{2} + 6 \, {\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right )}{324 \, a^{3}} \]
-1/324*(18*a^3*x^6*dilog(a*x^2) - 54*a^3*x^6*polylog(3, a*x^2) - 2*a^3*x^6 - 3*a^2*x^4 - 6*a*x^2 + 6*(a^3*x^6 - 1)*log(-a*x^2 + 1))/a^3
\[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\int { x^{5} {\rm Li}_{3}(a x^{2}) \,d x } \]
Time = 4.98 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83 \[ \int x^5 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\frac {x^6\,\mathrm {polylog}\left (3,a\,x^2\right )}{6}-\frac {x^6\,\mathrm {polylog}\left (2,a\,x^2\right )}{18}+\frac {\ln \left (a\,x^2-1\right )}{54\,a^3}-\frac {x^6\,\ln \left (1-a\,x^2\right )}{54}+\frac {x^6}{162}+\frac {x^2}{54\,a^2}+\frac {x^4}{108\,a} \]