Integrand size = 9, antiderivative size = 78 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=-\frac {a}{48 x^3}-\frac {a^2}{32 x^2}-\frac {a^3}{16 x}+\frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)+\frac {\log (1-a x)}{16 x^4}-\frac {\operatorname {PolyLog}(2,a x)}{4 x^4} \]
-1/48*a/x^3-1/32*a^2/x^2-1/16*a^3/x+1/16*a^4*ln(x)-1/16*a^4*ln(-a*x+1)+1/1 6*ln(-a*x+1)/x^4-1/4*polylog(2,a*x)/x^4
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=-\frac {a x \left (2+3 a x+6 a^2 x^2\right )-6 a^4 x^4 \log (x)+6 \left (-1+a^4 x^4\right ) \log (1-a x)+24 \operatorname {PolyLog}(2,a x)}{96 x^4} \]
-1/96*(a*x*(2 + 3*a*x + 6*a^2*x^2) - 6*a^4*x^4*Log[x] + 6*(-1 + a^4*x^4)*L og[1 - a*x] + 24*PolyLog[2, a*x])/x^4
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01, 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, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{4} \int -\frac {\log (1-a x)}{x^5}dx-\frac {\operatorname {PolyLog}(2,a x)}{4 x^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {\log (1-a x)}{x^5}dx-\frac {\operatorname {PolyLog}(2,a x)}{4 x^4}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} a \int \frac {1}{x^4 (1-a x)}dx+\frac {\log (1-a x)}{4 x^4}\right )-\frac {\operatorname {PolyLog}(2,a x)}{4 x^4}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} a \int \left (-\frac {a^4}{a x-1}+\frac {a^3}{x}+\frac {a^2}{x^2}+\frac {a}{x^3}+\frac {1}{x^4}\right )dx+\frac {\log (1-a x)}{4 x^4}\right )-\frac {\operatorname {PolyLog}(2,a x)}{4 x^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\frac {1}{4} a \left (a^3 \log (x)-a^3 \log (1-a x)-\frac {a^2}{x}-\frac {a}{2 x^2}-\frac {1}{3 x^3}\right )+\frac {\log (1-a x)}{4 x^4}\right )-\frac {\operatorname {PolyLog}(2,a x)}{4 x^4}\) |
(Log[1 - a*x]/(4*x^4) + (a*(-1/3*1/x^3 - a/(2*x^2) - a^2/x + a^3*Log[x] - a^3*Log[1 - a*x]))/4)/4 - PolyLog[2, a*x]/(4*x^4)
3.1.10.3.1 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[((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 = 0.80 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(\frac {6 \ln \left (x \right ) x^{4} a^{4}-6 \ln \left (-a x +1\right ) a^{4} x^{4}-6 a^{4} x^{4}-6 a^{3} x^{3}-3 a^{2} x^{2}-2 a x -24 \operatorname {polylog}\left (2, a x \right )+6 \ln \left (-a x +1\right )}{96 x^{4}}\) | \(76\) |
parts | \(-\frac {\operatorname {polylog}\left (2, a x \right )}{4 x^{4}}-\frac {a^{4} \left (\frac {1}{12 a^{3} x^{3}}+\frac {1}{4 a x}-\frac {\ln \left (-a x \right )}{4}+\frac {1}{8 a^{2} x^{2}}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right ) \left (\left (-a x +1\right )^{3}-4 \left (-a x +1\right )^{2}+2-6 a x \right )}{4 a^{4} x^{4}}\right )}{4}\) | \(94\) |
derivativedivides | \(a^{4} \left (-\frac {\operatorname {polylog}\left (2, a x \right )}{4 a^{4} x^{4}}-\frac {1}{48 a^{3} x^{3}}+\frac {\ln \left (-a x \right )}{16}-\frac {1}{16 a x}-\frac {1}{32 a^{2} x^{2}}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right ) \left (\left (-a x +1\right )^{3}-4 \left (-a x +1\right )^{2}+2-6 a x \right )}{16 a^{4} x^{4}}\right )\) | \(95\) |
default | \(a^{4} \left (-\frac {\operatorname {polylog}\left (2, a x \right )}{4 a^{4} x^{4}}-\frac {1}{48 a^{3} x^{3}}+\frac {\ln \left (-a x \right )}{16}-\frac {1}{16 a x}-\frac {1}{32 a^{2} x^{2}}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right ) \left (\left (-a x +1\right )^{3}-4 \left (-a x +1\right )^{2}+2-6 a x \right )}{16 a^{4} x^{4}}\right )\) | \(95\) |
meijerg | \(-a^{4} \left (-\frac {225 a^{3} x^{3}+350 a^{2} x^{2}+675 a x +2250}{7200 a^{3} x^{3}}-\frac {\left (-25 a^{4} x^{4}+25\right ) \ln \left (-a x +1\right )}{400 a^{4} x^{4}}+\frac {\operatorname {polylog}\left (2, a x \right )}{4 a^{4} x^{4}}+\frac {1}{32}-\frac {\ln \left (x \right )}{16}-\frac {\ln \left (-a \right )}{16}+\frac {1}{3 a^{3} x^{3}}+\frac {1}{8 a^{2} x^{2}}+\frac {1}{9 a x}\right )\) | \(110\) |
1/96*(6*ln(x)*x^4*a^4-6*ln(-a*x+1)*a^4*x^4-6*a^4*x^4-6*a^3*x^3-3*a^2*x^2-2 *a*x-24*polylog(2,a*x)+6*ln(-a*x+1))/x^4
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=-\frac {6 \, a^{4} x^{4} \log \left (a x - 1\right ) - 6 \, a^{4} x^{4} \log \left (x\right ) + 6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x + 24 \, {\rm Li}_2\left (a x\right ) - 6 \, \log \left (-a x + 1\right )}{96 \, x^{4}} \]
-1/96*(6*a^4*x^4*log(a*x - 1) - 6*a^4*x^4*log(x) + 6*a^3*x^3 + 3*a^2*x^2 + 2*a*x + 24*dilog(a*x) - 6*log(-a*x + 1))/x^4
Time = 1.80 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=\frac {a^{4} \log {\left (x \right )}}{16} + \frac {a^{4} \operatorname {Li}_{1}\left (a x\right )}{16} - \frac {a^{3}}{16 x} - \frac {a^{2}}{32 x^{2}} - \frac {a}{48 x^{3}} - \frac {\operatorname {Li}_{1}\left (a x\right )}{16 x^{4}} - \frac {\operatorname {Li}_{2}\left (a x\right )}{4 x^{4}} \]
a**4*log(x)/16 + a**4*polylog(1, a*x)/16 - a**3/(16*x) - a**2/(32*x**2) - a/(48*x**3) - polylog(1, a*x)/(16*x**4) - polylog(2, a*x)/(4*x**4)
Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=\frac {1}{16} \, a^{4} \log \left (x\right ) - \frac {6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x + 6 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right ) + 24 \, {\rm Li}_2\left (a x\right )}{96 \, x^{4}} \]
1/16*a^4*log(x) - 1/96*(6*a^3*x^3 + 3*a^2*x^2 + 2*a*x + 6*(a^4*x^4 - 1)*lo g(-a*x + 1) + 24*dilog(a*x))/x^4
\[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=\int { \frac {{\rm Li}_2\left (a x\right )}{x^{5}} \,d x } \]
Time = 5.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.77 \[ \int \frac {\operatorname {PolyLog}(2,a x)}{x^5} \, dx=\frac {\ln \left (1-a\,x\right )}{16\,x^4}-\frac {\mathrm {polylog}\left (2,a\,x\right )}{4\,x^4}-\frac {a^3\,x^2+\frac {a^2\,x}{2}+\frac {a}{3}}{16\,x^3}-\frac {a^4\,\mathrm {atan}\left (a\,x\,2{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{8} \]