Integrand size = 11, antiderivative size = 80 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=-\frac {8 a}{375 x^3}-\frac {8 a^2}{125 x}+\frac {8}{125} a^{5/2} \text {arctanh}\left (\sqrt {a} x\right )+\frac {4 \log \left (1-a x^2\right )}{125 x^5}-\frac {2 \operatorname {PolyLog}\left (2,a x^2\right )}{25 x^5}-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5} \]
-8/375*a/x^3-8/125*a^2/x+8/125*a^(5/2)*arctanh(x*a^(1/2))+4/125*ln(-a*x^2+ 1)/x^5-2/25*polylog(2,a*x^2)/x^5-1/5*polylog(3,a*x^2)/x^5
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.86 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=-\frac {8 a x^2+24 a^2 x^4-24 a^{5/2} x^5 \text {arctanh}\left (\sqrt {a} x\right )-12 \log \left (1-a x^2\right )+30 \operatorname {PolyLog}\left (2,a x^2\right )+75 \operatorname {PolyLog}\left (3,a x^2\right )}{375 x^5} \]
-1/375*(8*a*x^2 + 24*a^2*x^4 - 24*a^(5/2)*x^5*ArcTanh[Sqrt[a]*x] - 12*Log[ 1 - a*x^2] + 30*PolyLog[2, a*x^2] + 75*PolyLog[3, a*x^2])/x^5
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {7145, 7145, 25, 2905, 264, 264, 219}
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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2}{5} \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6}dx-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{5} \int -\frac {\log \left (1-a x^2\right )}{x^6}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{5} \left (-\frac {2}{5} \int \frac {\log \left (1-a x^2\right )}{x^6}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2}{5} \left (-\frac {2}{5} \left (-\frac {2}{5} a \int \frac {1}{x^4 \left (1-a x^2\right )}dx-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{5} \left (-\frac {2}{5} \left (-\frac {2}{5} a \left (a \int \frac {1}{x^2 \left (1-a x^2\right )}dx-\frac {1}{3 x^3}\right )-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{5} \left (-\frac {2}{5} \left (-\frac {2}{5} a \left (a \left (a \int \frac {1}{1-a x^2}dx-\frac {1}{x}\right )-\frac {1}{3 x^3}\right )-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2}{5} \left (-\frac {2}{5} \left (-\frac {2}{5} a \left (a \left (\sqrt {a} \text {arctanh}\left (\sqrt {a} x\right )-\frac {1}{x}\right )-\frac {1}{3 x^3}\right )-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (3,a x^2\right )}{5 x^5}\) |
(2*((-2*((-2*a*(-1/3*1/x^3 + a*(-x^(-1) + Sqrt[a]*ArcTanh[Sqrt[a]*x])))/5 - Log[1 - a*x^2]/(5*x^5)))/5 - PolyLog[2, a*x^2]/(5*x^5)))/5 - PolyLog[3, a*x^2]/(5*x^5)
3.1.44.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ (m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Simp[b*e*n*(p/(f*(m + 1))) Int[x^(n - 1)*((f*x)^(m + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -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]
Leaf count of result is larger than twice the leaf count of optimal. \(137\) vs. \(2(64)=128\).
Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.72
method | result | size |
meijerg | \(\frac {a^{3} \left (-\frac {16}{375 x^{3} \left (-a \right )^{\frac {3}{2}}}-\frac {16 a}{125 x \left (-a \right )^{\frac {3}{2}}}-\frac {8 x \,a^{2} \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{125 \left (-a \right )^{\frac {3}{2}} \sqrt {a \,x^{2}}}+\frac {8 \ln \left (-a \,x^{2}+1\right )}{125 x^{5} \left (-a \right )^{\frac {3}{2}} a}-\frac {4 \operatorname {polylog}\left (2, a \,x^{2}\right )}{25 x^{5} \left (-a \right )^{\frac {3}{2}} a}-\frac {2 \operatorname {polylog}\left (3, a \,x^{2}\right )}{5 x^{5} \left (-a \right )^{\frac {3}{2}} a}\right )}{2 \sqrt {-a}}\) | \(138\) |
1/2*a^3/(-a)^(1/2)*(-16/375/x^3/(-a)^(3/2)-16/125/x/(-a)^(3/2)*a-8/125*x/( -a)^(3/2)*a^2/(a*x^2)^(1/2)*(ln(1-(a*x^2)^(1/2))-ln(1+(a*x^2)^(1/2)))+8/12 5/x^5/(-a)^(3/2)*ln(-a*x^2+1)/a-4/25/x^5/(-a)^(3/2)/a*polylog(2,a*x^2)-2/5 /x^5/(-a)^(3/2)/a*polylog(3,a*x^2))
Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.88 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=\left [\frac {12 \, a^{\frac {5}{2}} x^{5} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 24 \, a^{2} x^{4} - 8 \, a x^{2} - 30 \, {\rm Li}_2\left (a x^{2}\right ) + 12 \, \log \left (-a x^{2} + 1\right ) - 75 \, {\rm polylog}\left (3, a x^{2}\right )}{375 \, x^{5}}, -\frac {24 \, \sqrt {-a} a^{2} x^{5} \arctan \left (\sqrt {-a} x\right ) + 24 \, a^{2} x^{4} + 8 \, a x^{2} + 30 \, {\rm Li}_2\left (a x^{2}\right ) - 12 \, \log \left (-a x^{2} + 1\right ) + 75 \, {\rm polylog}\left (3, a x^{2}\right )}{375 \, x^{5}}\right ] \]
[1/375*(12*a^(5/2)*x^5*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 24*a^2 *x^4 - 8*a*x^2 - 30*dilog(a*x^2) + 12*log(-a*x^2 + 1) - 75*polylog(3, a*x^ 2))/x^5, -1/375*(24*sqrt(-a)*a^2*x^5*arctan(sqrt(-a)*x) + 24*a^2*x^4 + 8*a *x^2 + 30*dilog(a*x^2) - 12*log(-a*x^2 + 1) + 75*polylog(3, a*x^2))/x^5]
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{x^{6}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=-\frac {4}{125} \, a^{\frac {5}{2}} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {24 \, a^{2} x^{4} + 8 \, a x^{2} + 30 \, {\rm Li}_2\left (a x^{2}\right ) - 12 \, \log \left (-a x^{2} + 1\right ) + 75 \, {\rm Li}_{3}(a x^{2})}{375 \, x^{5}} \]
-4/125*a^(5/2)*log((a*x - sqrt(a))/(a*x + sqrt(a))) - 1/375*(24*a^2*x^4 + 8*a*x^2 + 30*dilog(a*x^2) - 12*log(-a*x^2 + 1) + 75*polylog(3, a*x^2))/x^5
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=\int { \frac {{\rm Li}_{3}(a x^{2})}{x^{6}} \,d x } \]
Time = 5.74 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^2\right )}{x^6} \, dx=\frac {4\,\ln \left (1-a\,x^2\right )}{125\,x^5}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{5\,x^5}-\frac {8\,a^2\,x^2+\frac {8\,a}{3}}{125\,x^3}-\frac {2\,\mathrm {polylog}\left (2,a\,x^2\right )}{25\,x^5}-\frac {a^{5/2}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{125} \]