Integrand size = 11, antiderivative size = 73 \[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=-\frac {4 x}{25 a^2}-\frac {4 x^3}{75 a}-\frac {4 x^5}{125}+\frac {4 \text {arctanh}\left (\sqrt {a} x\right )}{25 a^{5/2}}+\frac {2}{25} x^5 \log \left (1-a x^2\right )+\frac {1}{5} x^5 \operatorname {PolyLog}\left (2,a x^2\right ) \]
-4/25*x/a^2-4/75*x^3/a-4/125*x^5+4/25*arctanh(x*a^(1/2))/a^(5/2)+2/25*x^5* ln(-a*x^2+1)+1/5*x^5*polylog(2,a*x^2)
Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\frac {1}{375} \left (-\frac {60 x}{a^2}-\frac {20 x^3}{a}-12 x^5+\frac {60 \text {arctanh}\left (\sqrt {a} x\right )}{a^{5/2}}+30 x^5 \log \left (1-a x^2\right )+75 x^5 \operatorname {PolyLog}\left (2,a x^2\right )\right ) \]
((-60*x)/a^2 - (20*x^3)/a - 12*x^5 + (60*ArcTanh[Sqrt[a]*x])/a^(5/2) + 30* x^5*Log[1 - a*x^2] + 75*x^5*PolyLog[2, a*x^2])/375
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {7145, 25, 2905, 254, 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.
\(\displaystyle \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {1}{5} x^5 \operatorname {PolyLog}\left (2,a x^2\right )-\frac {2}{5} \int -x^4 \log \left (1-a x^2\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{5} \int x^4 \log \left (1-a x^2\right )dx+\frac {1}{5} x^5 \operatorname {PolyLog}\left (2,a x^2\right )\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{5} a \int \frac {x^6}{1-a x^2}dx+\frac {1}{5} x^5 \log \left (1-a x^2\right )\right )+\frac {1}{5} x^5 \operatorname {PolyLog}\left (2,a x^2\right )\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{5} a \int \left (-\frac {x^4}{a}-\frac {x^2}{a^2}+\frac {1}{a^3 \left (1-a x^2\right )}-\frac {1}{a^3}\right )dx+\frac {1}{5} x^5 \log \left (1-a x^2\right )\right )+\frac {1}{5} x^5 \operatorname {PolyLog}\left (2,a x^2\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} \left (\frac {2}{5} a \left (\frac {\text {arctanh}\left (\sqrt {a} x\right )}{a^{7/2}}-\frac {x}{a^3}-\frac {x^3}{3 a^2}-\frac {x^5}{5 a}\right )+\frac {1}{5} x^5 \log \left (1-a x^2\right )\right )+\frac {1}{5} x^5 \operatorname {PolyLog}\left (2,a x^2\right )\) |
(2*((2*a*(-(x/a^3) - x^3/(3*a^2) - x^5/(5*a) + ArcTanh[Sqrt[a]*x]/a^(7/2)) )/5 + (x^5*Log[1 - a*x^2])/5))/5 + (x^5*PolyLog[2, a*x^2])/5
3.1.26.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
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 = 0.99 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {x^{5} \operatorname {polylog}\left (2, a \,x^{2}\right )}{5}+\frac {2 x^{5} \ln \left (-a \,x^{2}+1\right )}{25}+\frac {4 a \left (-\frac {\frac {1}{5} a^{2} x^{5}+\frac {1}{3} a \,x^{3}+x}{a^{3}}+\frac {\operatorname {arctanh}\left (x \sqrt {a}\right )}{a^{\frac {7}{2}}}\right )}{25}\) | \(63\) |
parts | \(\frac {x^{5} \operatorname {polylog}\left (2, a \,x^{2}\right )}{5}+\frac {2 x^{5} \ln \left (-a \,x^{2}+1\right )}{25}+\frac {4 a \left (-\frac {\frac {1}{5} a^{2} x^{5}+\frac {1}{3} a \,x^{3}+x}{a^{3}}+\frac {\operatorname {arctanh}\left (x \sqrt {a}\right )}{a^{\frac {7}{2}}}\right )}{25}\) | \(63\) |
meijerg | \(-\frac {-\frac {2 x \left (-a \right )^{\frac {7}{2}} \left (84 x^{4} a^{2}+140 a \,x^{2}+420\right )}{2625 a^{3}}-\frac {4 x \left (-a \right )^{\frac {7}{2}} \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{25 a^{3} \sqrt {a \,x^{2}}}+\frac {4 x^{5} \left (-a \right )^{\frac {7}{2}} \ln \left (-a \,x^{2}+1\right )}{25 a}+\frac {2 x^{5} \left (-a \right )^{\frac {7}{2}} \operatorname {polylog}\left (2, a \,x^{2}\right )}{5 a}}{2 a^{2} \sqrt {-a}}\) | \(124\) |
1/5*x^5*polylog(2,a*x^2)+2/25*x^5*ln(-a*x^2+1)+4/25*a*(-1/a^3*(1/5*a^2*x^5 +1/3*a*x^3+x)+1/a^(7/2)*arctanh(x*a^(1/2)))
Time = 0.26 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.18 \[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\left [\frac {75 \, a^{3} x^{5} {\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{3} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{3} x^{5} - 20 \, a^{2} x^{3} - 60 \, a x + 30 \, \sqrt {a} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right )}{375 \, a^{3}}, \frac {75 \, a^{3} x^{5} {\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{3} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{3} x^{5} - 20 \, a^{2} x^{3} - 60 \, a x - 60 \, \sqrt {-a} \arctan \left (\sqrt {-a} x\right )}{375 \, a^{3}}\right ] \]
[1/375*(75*a^3*x^5*dilog(a*x^2) + 30*a^3*x^5*log(-a*x^2 + 1) - 12*a^3*x^5 - 20*a^2*x^3 - 60*a*x + 30*sqrt(a)*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)))/a^3, 1/375*(75*a^3*x^5*dilog(a*x^2) + 30*a^3*x^5*log(-a*x^2 + 1) - 12 *a^3*x^5 - 20*a^2*x^3 - 60*a*x - 60*sqrt(-a)*arctan(sqrt(-a)*x))/a^3]
Time = 60.57 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.29 \[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\begin {cases} - \frac {2 x^{5} \operatorname {Li}_{1}\left (a x^{2}\right )}{25} + \frac {x^{5} \operatorname {Li}_{2}\left (a x^{2}\right )}{5} - \frac {4 x^{5}}{125} - \frac {4 x^{3}}{75 a} - \frac {4 x}{25 a^{2}} - \frac {4 \log {\left (x - \sqrt {\frac {1}{a}} \right )}}{25 a^{3} \sqrt {\frac {1}{a}}} - \frac {2 \operatorname {Li}_{1}\left (a x^{2}\right )}{25 a^{3} \sqrt {\frac {1}{a}}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((-2*x**5*polylog(1, a*x**2)/25 + x**5*polylog(2, a*x**2)/5 - 4*x **5/125 - 4*x**3/(75*a) - 4*x/(25*a**2) - 4*log(x - sqrt(1/a))/(25*a**3*sq rt(1/a)) - 2*polylog(1, a*x**2)/(25*a**3*sqrt(1/a)), Ne(a, 0)), (0, True))
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10 \[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\frac {75 \, a^{2} x^{5} {\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{2} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{2} x^{5} - 20 \, a x^{3} - 60 \, x}{375 \, a^{2}} - \frac {2 \, \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right )}{25 \, a^{\frac {5}{2}}} \]
1/375*(75*a^2*x^5*dilog(a*x^2) + 30*a^2*x^5*log(-a*x^2 + 1) - 12*a^2*x^5 - 20*a*x^3 - 60*x)/a^2 - 2/25*log((a*x - sqrt(a))/(a*x + sqrt(a)))/a^(5/2)
\[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\int { x^{4} {\rm Li}_2\left (a x^{2}\right ) \,d x } \]
Time = 5.14 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int x^4 \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\frac {x^5\,\mathrm {polylog}\left (2,a\,x^2\right )}{5}-\frac {4\,x}{25\,a^2}+\frac {2\,x^5\,\ln \left (1-a\,x^2\right )}{25}-\frac {4\,x^5}{125}-\frac {4\,x^3}{75\,a}-\frac {\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{25\,a^{5/2}} \]