Integrand size = 11, antiderivative size = 88 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=-\frac {a}{108 x^4}-\frac {a^2}{54 x^2}+\frac {1}{27} a^3 \log (x)-\frac {1}{54} a^3 \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{54 x^6}-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{18 x^6}-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6} \]
-1/108*a/x^4-1/54*a^2/x^2+1/27*a^3*ln(x)-1/54*a^3*ln(-a*x^2+1)+1/54*ln(-a* x^2+1)/x^6-1/18*polylog(2,a*x^2)/x^6-1/6*polylog(3,a*x^2)/x^6
Result contains higher order function than in optimal. Order 9 vs. order 4 in optimal.
Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.34 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=\frac {G_{5,5}^{2,4}\left (-a x^2|\begin {array}{c} 1,1,1,1,4 \\ 1,3,0,0,0 \\\end {array}\right )}{2 x^6} \]
Time = 0.36 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.09, 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, 54, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{3} \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^7}dx-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int -\frac {\log \left (1-a x^2\right )}{x^7}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{6 x^6}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{3} \int \frac {\log \left (1-a x^2\right )}{x^7}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{6 x^6}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{6} \int \frac {\log \left (1-a x^2\right )}{x^8}dx^2-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{6 x^6}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{6} \left (\frac {1}{3} a \int \frac {1}{x^6 \left (1-a x^2\right )}dx^2+\frac {\log \left (1-a x^2\right )}{3 x^6}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{6 x^6}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{6} \left (\frac {1}{3} a \int \left (-\frac {a^3}{a x^2-1}+\frac {a^2}{x^2}+\frac {a}{x^4}+\frac {1}{x^6}\right )dx^2+\frac {\log \left (1-a x^2\right )}{3 x^6}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{6 x^6}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{6} \left (\frac {1}{3} a \left (a^2 \log \left (x^2\right )-a^2 \log \left (1-a x^2\right )-\frac {a}{x^2}-\frac {1}{2 x^4}\right )+\frac {\log \left (1-a x^2\right )}{3 x^6}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{6 x^6}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{6 x^6}\) |
((Log[1 - a*x^2]/(3*x^6) + (a*(-1/2*1/x^4 - a/x^2 + a^2*Log[x^2] - a^2*Log [1 - a*x^2]))/3)/6 - PolyLog[2, a*x^2]/(6*x^6))/3 - PolyLog[3, a*x^2]/(6*x ^6)
3.1.38.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.19 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.31
method | result | size |
meijerg | \(\frac {a^{3} \left (\frac {64 x^{4} a^{2}+152 a \,x^{2}+832}{1728 a^{2} x^{4}}+\frac {\left (-64 a^{3} x^{6}+64\right ) \ln \left (-a \,x^{2}+1\right )}{1728 a^{3} x^{6}}-\frac {\operatorname {polylog}\left (2, a \,x^{2}\right )}{9 a^{3} x^{6}}-\frac {\operatorname {polylog}\left (3, a \,x^{2}\right )}{3 a^{3} x^{6}}-\frac {1}{27}+\frac {2 \ln \left (x \right )}{27}+\frac {\ln \left (-a \right )}{27}-\frac {1}{2 a^{2} x^{4}}-\frac {1}{8 a \,x^{2}}\right )}{2}\) | \(115\) |
1/2*a^3*(1/1728/a^2/x^4*(64*a^2*x^4+152*a*x^2+832)+1/1728/a^3/x^6*(-64*a^3 *x^6+64)*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)-1/27+2/27*ln(x)+1/27*ln(-a)-1/2/a^2/x^4-1/8/a/x^2)
Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=-\frac {2 \, a^{3} x^{6} \log \left (a x^{2} - 1\right ) - 4 \, a^{3} x^{6} \log \left (x\right ) + 2 \, a^{2} x^{4} + a x^{2} + 6 \, {\rm Li}_2\left (a x^{2}\right ) - 2 \, \log \left (-a x^{2} + 1\right ) + 18 \, {\rm polylog}\left (3, a x^{2}\right )}{108 \, x^{6}} \]
-1/108*(2*a^3*x^6*log(a*x^2 - 1) - 4*a^3*x^6*log(x) + 2*a^2*x^4 + a*x^2 + 6*dilog(a*x^2) - 2*log(-a*x^2 + 1) + 18*polylog(3, a*x^2))/x^6
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{x^{7}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.73 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=\frac {1}{27} \, a^{3} \log \left (x\right ) - \frac {2 \, a^{2} x^{4} + a x^{2} + 2 \, {\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right ) + 6 \, {\rm Li}_2\left (a x^{2}\right ) + 18 \, {\rm Li}_{3}(a x^{2})}{108 \, x^{6}} \]
1/27*a^3*log(x) - 1/108*(2*a^2*x^4 + a*x^2 + 2*(a^3*x^6 - 1)*log(-a*x^2 + 1) + 6*dilog(a*x^2) + 18*polylog(3, a*x^2))/x^6
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=\int { \frac {{\rm Li}_{3}(a x^{2})}{x^{7}} \,d x } \]
Time = 5.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.83 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^7} \, dx=\frac {a^3\,\ln \left (x\right )}{27}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{18\,x^6}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{6\,x^6}-\frac {a^3\,\ln \left (a\,x^2-1\right )}{54}-\frac {a}{108\,x^4}+\frac {\ln \left (1-a\,x^2\right )}{54\,x^6}-\frac {a^2}{54\,x^2} \]