Integrand size = 11, antiderivative size = 78 \[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\frac {x^2}{16 a}+\frac {x^4}{32}+\frac {\log \left (1-a x^2\right )}{16 a^2}-\frac {1}{16} x^4 \log \left (1-a x^2\right )-\frac {1}{8} x^4 \operatorname {PolyLog}\left (2,a x^2\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right ) \]
1/16*x^2/a+1/32*x^4+1/16*ln(-a*x^2+1)/a^2-1/16*x^4*ln(-a*x^2+1)-1/8*x^4*po lylog(2,a*x^2)+1/4*x^4*polylog(3,a*x^2)
Time = 0.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\frac {2 a x^2+a^2 x^4+2 \log \left (1-a x^2\right )-2 a^2 x^4 \log \left (1-a x^2\right )-4 a^2 x^4 \operatorname {PolyLog}\left (2,a x^2\right )+8 a^2 x^4 \operatorname {PolyLog}\left (3,a x^2\right )}{32 a^2} \]
(2*a*x^2 + a^2*x^4 + 2*Log[1 - a*x^2] - 2*a^2*x^4*Log[1 - a*x^2] - 4*a^2*x ^4*PolyLog[2, a*x^2] + 8*a^2*x^4*PolyLog[3, a*x^2])/(32*a^2)
Time = 0.36 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19, 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, 49, 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.
\(\displaystyle \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )-\frac {1}{2} \int x^3 \operatorname {PolyLog}\left (2,a x^2\right )dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -x^3 \log \left (1-a x^2\right )dx-\frac {1}{4} x^4 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int x^3 \log \left (1-a x^2\right )dx-\frac {1}{4} x^4 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{4} \int x^2 \log \left (1-a x^2\right )dx^2-\frac {1}{4} x^4 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {1}{2} a \int \frac {x^4}{1-a x^2}dx^2-\frac {1}{2} x^4 \log \left (1-a x^2\right )\right )-\frac {1}{4} x^4 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {1}{2} a \int \left (-\frac {x^2}{a}-\frac {1}{a^2 \left (a x^2-1\right )}-\frac {1}{a^2}\right )dx^2-\frac {1}{2} x^4 \log \left (1-a x^2\right )\right )-\frac {1}{4} x^4 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (-\frac {1}{2} a \left (-\frac {\log \left (1-a x^2\right )}{a^3}-\frac {x^2}{a^2}-\frac {x^4}{2 a}\right )-\frac {1}{2} x^4 \log \left (1-a x^2\right )\right )-\frac {1}{4} x^4 \operatorname {PolyLog}\left (2,a x^2\right )\right )+\frac {1}{4} x^4 \operatorname {PolyLog}\left (3,a x^2\right )\) |
((-1/2*(x^4*Log[1 - a*x^2]) - (a*(-(x^2/a^2) - x^4/(2*a) - Log[1 - a*x^2]/ a^3))/2)/4 - (x^4*PolyLog[2, a*x^2])/4)/2 + (x^4*PolyLog[3, a*x^2])/4
3.1.33.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[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 = 0.06 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92
method | result | size |
meijerg | \(-\frac {-\frac {a \,x^{2} \left (3 a \,x^{2}+6\right )}{48}-\frac {\left (-3 x^{4} a^{2}+3\right ) \ln \left (-a \,x^{2}+1\right )}{24}+\frac {a^{2} x^{4} \operatorname {polylog}\left (2, a \,x^{2}\right )}{4}-\frac {a^{2} x^{4} \operatorname {polylog}\left (3, a \,x^{2}\right )}{2}}{2 a^{2}}\) | \(72\) |
-1/2/a^2*(-1/48*a*x^2*(3*a*x^2+6)-1/24*(-3*a^2*x^4+3)*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 = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=-\frac {4 \, a^{2} x^{4} {\rm Li}_2\left (a x^{2}\right ) - 8 \, a^{2} x^{4} {\rm polylog}\left (3, a x^{2}\right ) - a^{2} x^{4} - 2 \, a x^{2} + 2 \, {\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right )}{32 \, a^{2}} \]
-1/32*(4*a^2*x^4*dilog(a*x^2) - 8*a^2*x^4*polylog(3, a*x^2) - a^2*x^4 - 2* a*x^2 + 2*(a^2*x^4 - 1)*log(-a*x^2 + 1))/a^2
\[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\int x^{3} \operatorname {Li}_{3}\left (a x^{2}\right )\, dx \]
Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=-\frac {4 \, a^{2} x^{4} {\rm Li}_2\left (a x^{2}\right ) - 8 \, a^{2} x^{4} {\rm Li}_{3}(a x^{2}) - a^{2} x^{4} - 2 \, a x^{2} + 2 \, {\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right )}{32 \, a^{2}} \]
-1/32*(4*a^2*x^4*dilog(a*x^2) - 8*a^2*x^4*polylog(3, a*x^2) - a^2*x^4 - 2* a*x^2 + 2*(a^2*x^4 - 1)*log(-a*x^2 + 1))/a^2
\[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\int { x^{3} {\rm Li}_{3}(a x^{2}) \,d x } \]
Time = 4.89 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.83 \[ \int x^3 \operatorname {PolyLog}\left (3,a x^2\right ) \, dx=\frac {x^4\,\mathrm {polylog}\left (3,a\,x^2\right )}{4}-\frac {x^4\,\mathrm {polylog}\left (2,a\,x^2\right )}{8}+\frac {\ln \left (a\,x^2-1\right )}{16\,a^2}-\frac {x^4\,\ln \left (1-a\,x^2\right )}{16}+\frac {x^4}{32}+\frac {x^2}{16\,a} \]