Integrand size = 11, antiderivative size = 66 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=-\frac {4 a}{75 x^3}-\frac {4 a^2}{25 x}+\frac {4}{25} a^{5/2} \text {arctanh}\left (\sqrt {a} x\right )+\frac {2 \log \left (1-a x^2\right )}{25 x^5}-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5} \]
-4/75*a/x^3-4/25*a^2/x+4/25*a^(5/2)*arctanh(x*a^(1/2))+2/25*ln(-a*x^2+1)/x ^5-1/5*polylog(2,a*x^2)/x^5
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.71 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=-\frac {4 a x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},a x^2\right )-6 \log \left (1-a x^2\right )+15 \operatorname {PolyLog}\left (2,a x^2\right )}{75 x^5} \]
-1/75*(4*a*x^2*Hypergeometric2F1[-3/2, 1, -1/2, a*x^2] - 6*Log[1 - a*x^2] + 15*PolyLog[2, a*x^2])/x^5
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2}{5} \int -\frac {\log \left (1-a x^2\right )}{x^6}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2}{5} \int \frac {\log \left (1-a x^2\right )}{x^6}dx-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle -\frac {2}{5} \left (-\frac {2}{5} a \int \frac {1}{x^4 \left (1-a x^2\right )}dx-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {2}{5} \left (-\frac {2}{5} a \left (a \int \frac {1}{x^2 \left (1-a x^2\right )}dx-\frac {1}{3 x^3}\right )-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {2}{5} \left (-\frac {2}{5} a \left (a \left (a \int \frac {1}{1-a x^2}dx-\frac {1}{x}\right )-\frac {1}{3 x^3}\right )-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2}{5} \left (-\frac {2}{5} a \left (a \left (\sqrt {a} \text {arctanh}\left (\sqrt {a} x\right )-\frac {1}{x}\right )-\frac {1}{3 x^3}\right )-\frac {\log \left (1-a x^2\right )}{5 x^5}\right )-\frac {\operatorname {PolyLog}\left (2,a x^2\right )}{5 x^5}\) |
(-2*((-2*a*(-1/3*1/x^3 + a*(-x^(-1) + Sqrt[a]*ArcTanh[Sqrt[a]*x])))/5 - Lo g[1 - a*x^2]/(5*x^5)))/5 - PolyLog[2, a*x^2]/(5*x^5)
3.1.31.3.1 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]
Time = 1.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.80
method | result | size |
default | \(-\frac {\operatorname {polylog}\left (2, a \,x^{2}\right )}{5 x^{5}}+\frac {2 \ln \left (-a \,x^{2}+1\right )}{25 x^{5}}+\frac {4 a \left (a^{\frac {3}{2}} \operatorname {arctanh}\left (x \sqrt {a}\right )-\frac {1}{3 x^{3}}-\frac {a}{x}\right )}{25}\) | \(53\) |
parts | \(-\frac {\operatorname {polylog}\left (2, a \,x^{2}\right )}{5 x^{5}}+\frac {2 \ln \left (-a \,x^{2}+1\right )}{25 x^{5}}+\frac {4 a \left (a^{\frac {3}{2}} \operatorname {arctanh}\left (x \sqrt {a}\right )-\frac {1}{3 x^{3}}-\frac {a}{x}\right )}{25}\) | \(53\) |
meijerg | \(\frac {a^{3} \left (-\frac {8}{75 x^{3} \left (-a \right )^{\frac {3}{2}}}-\frac {8 a}{25 x \left (-a \right )^{\frac {3}{2}}}-\frac {4 x \,a^{2} \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{25 \left (-a \right )^{\frac {3}{2}} \sqrt {a \,x^{2}}}+\frac {4 \ln \left (-a \,x^{2}+1\right )}{25 x^{5} \left (-a \right )^{\frac {3}{2}} a}-\frac {2 \operatorname {polylog}\left (2, a \,x^{2}\right )}{5 x^{5} \left (-a \right )^{\frac {3}{2}} a}\right )}{2 \sqrt {-a}}\) | \(118\) |
-1/5*polylog(2,a*x^2)/x^5+2/25*ln(-a*x^2+1)/x^5+4/25*a*(a^(3/2)*arctanh(x* a^(1/2))-1/3/x^3-a/x)
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.00 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=\left [\frac {6 \, a^{\frac {5}{2}} x^{5} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 12 \, a^{2} x^{4} - 4 \, a x^{2} - 15 \, {\rm Li}_2\left (a x^{2}\right ) + 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}}, -\frac {12 \, \sqrt {-a} a^{2} x^{5} \arctan \left (\sqrt {-a} x\right ) + 12 \, a^{2} x^{4} + 4 \, a x^{2} + 15 \, {\rm Li}_2\left (a x^{2}\right ) - 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}}\right ] \]
[1/75*(6*a^(5/2)*x^5*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 12*a^2*x ^4 - 4*a*x^2 - 15*dilog(a*x^2) + 6*log(-a*x^2 + 1))/x^5, -1/75*(12*sqrt(-a )*a^2*x^5*arctan(sqrt(-a)*x) + 12*a^2*x^4 + 4*a*x^2 + 15*dilog(a*x^2) - 6* log(-a*x^2 + 1))/x^5]
Timed out. \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=-\frac {2}{25} \, a^{\frac {5}{2}} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {12 \, a^{2} x^{4} + 4 \, a x^{2} + 15 \, {\rm Li}_2\left (a x^{2}\right ) - 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}} \]
-2/25*a^(5/2)*log((a*x - sqrt(a))/(a*x + sqrt(a))) - 1/75*(12*a^2*x^4 + 4* a*x^2 + 15*dilog(a*x^2) - 6*log(-a*x^2 + 1))/x^5
\[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=\int { \frac {{\rm Li}_2\left (a x^{2}\right )}{x^{6}} \,d x } \]
Time = 5.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int \frac {\operatorname {PolyLog}\left (2,a x^2\right )}{x^6} \, dx=\frac {2\,\ln \left (1-a\,x^2\right )}{25\,x^5}-\frac {4\,a^2\,x^2+\frac {4\,a}{3}}{25\,x^3}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{5\,x^5}-\frac {a^{5/2}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{25} \]