Integrand size = 11, antiderivative size = 78 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=-\frac {a}{16 x^2}+\frac {1}{8} a^2 \log (x)-\frac {1}{16} a^2 \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{16 x^4}-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{8 x^4}-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4} \] Output:
-1/16*a/x^2+1/8*a^2*ln(x)-1/16*a^2*ln(-a*x^2+1)+1/16*ln(-a*x^2+1)/x^4-1/8* polylog(2,a*x^2)/x^4-1/4*polylog(3,a*x^2)/x^4
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 = 0.38 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=\frac {G_{5,5}^{2,4}\left (-a x^2|\begin {array}{c} 1,1,1,1,3 \\ 1,2,0,0,0 \\\end {array}\right )}{2 x^4} \] Input:
Integrate[PolyLog[3, a*x^2]/x^5,x]
Output:
MeijerG[{{1, 1, 1, 1}, {3}}, {{1, 2}, {0, 0, 0}}, -(a*x^2)]/(2*x^4)
Time = 0.36 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{2} \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^5}dx-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {\log \left (1-a x^2\right )}{x^5}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{4 x^4}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {\log \left (1-a x^2\right )}{x^5}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{4 x^4}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{4} \int \frac {\log \left (1-a x^2\right )}{x^6}dx^2-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{4 x^4}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} a \int \frac {1}{x^4 \left (1-a x^2\right )}dx^2+\frac {\log \left (1-a x^2\right )}{2 x^4}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{4 x^4}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} a \int \left (-\frac {a^2}{a x^2-1}+\frac {a}{x^2}+\frac {1}{x^4}\right )dx^2+\frac {\log \left (1-a x^2\right )}{2 x^4}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{4 x^4}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} a \left (a \log \left (x^2\right )-a \log \left (1-a x^2\right )-\frac {1}{x^2}\right )+\frac {\log \left (1-a x^2\right )}{2 x^4}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{4 x^4}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{4 x^4}\) |
Input:
Int[PolyLog[3, a*x^2]/x^5,x]
Output:
((Log[1 - a*x^2]/(2*x^4) + (a*(-x^(-2) + a*Log[x^2] - a*Log[1 - a*x^2]))/2 )/4 - PolyLog[2, a*x^2]/(4*x^4))/2 - PolyLog[3, a*x^2]/(4*x^4)
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 = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.26
method | result | size |
meijerg | \(-\frac {a^{2} \left (\frac {1}{a \,x^{2}}+\frac {3}{16}-\frac {\ln \left (x \right )}{4}-\frac {\ln \left (-a \right )}{8}-\frac {81 a \,x^{2}+378}{432 a \,x^{2}}-\frac {\left (-27 a^{2} x^{4}+27\right ) \ln \left (-a \,x^{2}+1\right )}{216 a^{2} x^{4}}+\frac {\operatorname {polylog}\left (2, a \,x^{2}\right )}{4 a^{2} x^{4}}+\frac {\operatorname {polylog}\left (3, a \,x^{2}\right )}{2 a^{2} x^{4}}\right )}{2}\) | \(98\) |
Input:
int(polylog(3,a*x^2)/x^5,x,method=_RETURNVERBOSE)
Output:
-1/2*a^2*(1/a/x^2+3/16-1/4*ln(x)-1/8*ln(-a)-1/432/a/x^2*(81*a*x^2+378)-1/2 16/a^2/x^4*(-27*a^2*x^4+27)*ln(-a*x^2+1)+1/4/a^2/x^4*polylog(2,a*x^2)+1/2/ a^2/x^4*polylog(3,a*x^2))
Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=-\frac {a^{2} x^{4} \log \left (a x^{2} - 1\right ) - 2 \, a^{2} x^{4} \log \left (x\right ) + a x^{2} + 2 \, {\rm Li}_2\left (a x^{2}\right ) - \log \left (-a x^{2} + 1\right ) + 4 \, {\rm polylog}\left (3, a x^{2}\right )}{16 \, x^{4}} \] Input:
integrate(polylog(3,a*x^2)/x^5,x, algorithm="fricas")
Output:
-1/16*(a^2*x^4*log(a*x^2 - 1) - 2*a^2*x^4*log(x) + a*x^2 + 2*dilog(a*x^2) - log(-a*x^2 + 1) + 4*polylog(3, a*x^2))/x^4
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{x^{5}}\, dx \] Input:
integrate(polylog(3,a*x**2)/x**5,x)
Output:
Integral(polylog(3, a*x**2)/x**5, x)
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=\frac {1}{8} \, a^{2} \log \left (x\right ) - \frac {a x^{2} + {\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right ) + 2 \, {\rm Li}_2\left (a x^{2}\right ) + 4 \, {\rm Li}_{3}(a x^{2})}{16 \, x^{4}} \] Input:
integrate(polylog(3,a*x^2)/x^5,x, algorithm="maxima")
Output:
1/8*a^2*log(x) - 1/16*(a*x^2 + (a^2*x^4 - 1)*log(-a*x^2 + 1) + 2*dilog(a*x ^2) + 4*polylog(3, a*x^2))/x^4
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=\int { \frac {{\rm Li}_{3}(a x^{2})}{x^{5}} \,d x } \] Input:
integrate(polylog(3,a*x^2)/x^5,x, algorithm="giac")
Output:
integrate(polylog(3, a*x^2)/x^5, x)
Time = 0.52 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=\frac {a^2\,\ln \left (x\right )}{8}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{8\,x^4}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{4\,x^4}-\frac {a^2\,\ln \left (a\,x^2-1\right )}{16}-\frac {a}{16\,x^2}+\frac {\ln \left (1-a\,x^2\right )}{16\,x^4} \] Input:
int(polylog(3, a*x^2)/x^5,x)
Output:
(a^2*log(x))/8 - polylog(2, a*x^2)/(8*x^4) - polylog(3, a*x^2)/(4*x^4) - ( a^2*log(a*x^2 - 1))/16 - a/(16*x^2) + log(1 - a*x^2)/(16*x^4)
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^5} \, dx=\int \frac {\mathit {polylog}\left (3, a \,x^{2}\right )}{x^{5}}d x \] Input:
int(polylog(3,a*x^2)/x^5,x)
Output:
int(polylog(3,a*x**2)/x**5,x)