3.40 \(\int x^2 \text {Li}_3(a x^2) \, dx\)

Optimal. Leaf size=77 \[ -\frac {8 \tanh ^{-1}\left (\sqrt {a} x\right )}{27 a^{3/2}}-\frac {2}{9} x^3 \text {Li}_2\left (a x^2\right )+\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )-\frac {4}{27} x^3 \log \left (1-a x^2\right )+\frac {8 x}{27 a}+\frac {8 x^3}{81} \]

[Out]

8/27*x/a+8/81*x^3-8/27*arctanh(x*a^(1/2))/a^(3/2)-4/27*x^3*ln(-a*x^2+1)-2/9*x^3*polylog(2,a*x^2)+1/3*x^3*polyl
og(3,a*x^2)

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6591, 2455, 302, 206} \[ -\frac {2}{9} x^3 \text {PolyLog}\left (2,a x^2\right )+\frac {1}{3} x^3 \text {PolyLog}\left (3,a x^2\right )-\frac {8 \tanh ^{-1}\left (\sqrt {a} x\right )}{27 a^{3/2}}-\frac {4}{27} x^3 \log \left (1-a x^2\right )+\frac {8 x}{27 a}+\frac {8 x^3}{81} \]

Antiderivative was successfully verified.

[In]

Int[x^2*PolyLog[3, a*x^2],x]

[Out]

(8*x)/(27*a) + (8*x^3)/81 - (8*ArcTanh[Sqrt[a]*x])/(27*a^(3/2)) - (4*x^3*Log[1 - a*x^2])/27 - (2*x^3*PolyLog[2
, a*x^2])/9 + (x^3*PolyLog[3, a*x^2])/3

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2455

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] - Dist[(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]

Rule 6591

Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[((d*x)^(m + 1)*PolyLog[n
, a*(b*x^p)^q])/(d*(m + 1)), x] - Dist[(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]

Rubi steps

\begin {align*} \int x^2 \text {Li}_3\left (a x^2\right ) \, dx &=\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )-\frac {2}{3} \int x^2 \text {Li}_2\left (a x^2\right ) \, dx\\ &=-\frac {2}{9} x^3 \text {Li}_2\left (a x^2\right )+\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )-\frac {4}{9} \int x^2 \log \left (1-a x^2\right ) \, dx\\ &=-\frac {4}{27} x^3 \log \left (1-a x^2\right )-\frac {2}{9} x^3 \text {Li}_2\left (a x^2\right )+\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )-\frac {1}{27} (8 a) \int \frac {x^4}{1-a x^2} \, dx\\ &=-\frac {4}{27} x^3 \log \left (1-a x^2\right )-\frac {2}{9} x^3 \text {Li}_2\left (a x^2\right )+\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )-\frac {1}{27} (8 a) \int \left (-\frac {1}{a^2}-\frac {x^2}{a}+\frac {1}{a^2 \left (1-a x^2\right )}\right ) \, dx\\ &=\frac {8 x}{27 a}+\frac {8 x^3}{81}-\frac {4}{27} x^3 \log \left (1-a x^2\right )-\frac {2}{9} x^3 \text {Li}_2\left (a x^2\right )+\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )-\frac {8 \int \frac {1}{1-a x^2} \, dx}{27 a}\\ &=\frac {8 x}{27 a}+\frac {8 x^3}{81}-\frac {8 \tanh ^{-1}\left (\sqrt {a} x\right )}{27 a^{3/2}}-\frac {4}{27} x^3 \log \left (1-a x^2\right )-\frac {2}{9} x^3 \text {Li}_2\left (a x^2\right )+\frac {1}{3} x^3 \text {Li}_3\left (a x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 69, normalized size = 0.90 \[ \frac {1}{81} \left (-\frac {24 \tanh ^{-1}\left (\sqrt {a} x\right )}{a^{3/2}}-18 x^3 \text {Li}_2\left (a x^2\right )+27 x^3 \text {Li}_3\left (a x^2\right )-12 x^3 \log \left (1-a x^2\right )+\frac {24 x}{a}+8 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*PolyLog[3, a*x^2],x]

[Out]

((24*x)/a + 8*x^3 - (24*ArcTanh[Sqrt[a]*x])/a^(3/2) - 12*x^3*Log[1 - a*x^2] - 18*x^3*PolyLog[2, a*x^2] + 27*x^
3*PolyLog[3, a*x^2])/81

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fricas [C]  time = 0.65, size = 173, normalized size = 2.25 \[ \left [-\frac {18 \, a^{2} x^{3} {\rm Li}_2\left (a x^{2}\right ) + 12 \, a^{2} x^{3} \log \left (-a x^{2} + 1\right ) - 27 \, a^{2} x^{3} {\rm polylog}\left (3, a x^{2}\right ) - 8 \, a^{2} x^{3} - 24 \, a x - 12 \, \sqrt {a} \log \left (\frac {a x^{2} - 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right )}{81 \, a^{2}}, -\frac {18 \, a^{2} x^{3} {\rm Li}_2\left (a x^{2}\right ) + 12 \, a^{2} x^{3} \log \left (-a x^{2} + 1\right ) - 27 \, a^{2} x^{3} {\rm polylog}\left (3, a x^{2}\right ) - 8 \, a^{2} x^{3} - 24 \, a x - 24 \, \sqrt {-a} \arctan \left (\sqrt {-a} x\right )}{81 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*polylog(3,a*x^2),x, algorithm="fricas")

[Out]

[-1/81*(18*a^2*x^3*dilog(a*x^2) + 12*a^2*x^3*log(-a*x^2 + 1) - 27*a^2*x^3*polylog(3, a*x^2) - 8*a^2*x^3 - 24*a
*x - 12*sqrt(a)*log((a*x^2 - 2*sqrt(a)*x + 1)/(a*x^2 - 1)))/a^2, -1/81*(18*a^2*x^3*dilog(a*x^2) + 12*a^2*x^3*l
og(-a*x^2 + 1) - 27*a^2*x^3*polylog(3, a*x^2) - 8*a^2*x^3 - 24*a*x - 24*sqrt(-a)*arctan(sqrt(-a)*x))/a^2]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Li}_{3}(a x^{2})\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*polylog(3,a*x^2),x, algorithm="giac")

[Out]

integrate(x^2*polylog(3, a*x^2), x)

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maple [B]  time = 0.16, size = 136, normalized size = 1.77 \[ \frac {\frac {2 x \left (-a \right )^{\frac {5}{2}} \left (40 a \,x^{2}+120\right )}{405 a^{2}}+\frac {8 x \left (-a \right )^{\frac {5}{2}} \left (\ln \left (1-\sqrt {a \,x^{2}}\right )-\ln \left (1+\sqrt {a \,x^{2}}\right )\right )}{27 a^{2} \sqrt {a \,x^{2}}}-\frac {8 x^{3} \left (-a \right )^{\frac {5}{2}} \ln \left (-a \,x^{2}+1\right )}{27 a}-\frac {4 x^{3} \left (-a \right )^{\frac {5}{2}} \polylog \left (2, a \,x^{2}\right )}{9 a}+\frac {2 x^{3} \left (-a \right )^{\frac {5}{2}} \polylog \left (3, a \,x^{2}\right )}{3 a}}{2 a \sqrt {-a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*polylog(3,a*x^2),x)

[Out]

1/2/a/(-a)^(1/2)*(2/405*x*(-a)^(5/2)*(40*a*x^2+120)/a^2+8/27*x*(-a)^(5/2)/a^2/(a*x^2)^(1/2)*(ln(1-(a*x^2)^(1/2
))-ln(1+(a*x^2)^(1/2)))-8/27*x^3*(-a)^(5/2)/a*ln(-a*x^2+1)-4/9*x^3*(-a)^(5/2)*polylog(2,a*x^2)/a+2/3*x^3*(-a)^
(5/2)/a*polylog(3,a*x^2))

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maxima [A]  time = 0.42, size = 81, normalized size = 1.05 \[ -\frac {18 \, a x^{3} {\rm Li}_2\left (a x^{2}\right ) + 12 \, a x^{3} \log \left (-a x^{2} + 1\right ) - 27 \, a x^{3} {\rm Li}_{3}(a x^{2}) - 8 \, a x^{3} - 24 \, x}{81 \, a} + \frac {4 \, \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right )}{27 \, a^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*polylog(3,a*x^2),x, algorithm="maxima")

[Out]

-1/81*(18*a*x^3*dilog(a*x^2) + 12*a*x^3*log(-a*x^2 + 1) - 27*a*x^3*polylog(3, a*x^2) - 8*a*x^3 - 24*x)/a + 4/2
7*log((a*x - sqrt(a))/(a*x + sqrt(a)))/a^(3/2)

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mupad [B]  time = 0.49, size = 64, normalized size = 0.83 \[ \frac {x^3\,\mathrm {polylog}\left (3,a\,x^2\right )}{3}-\frac {2\,x^3\,\mathrm {polylog}\left (2,a\,x^2\right )}{9}+\frac {8\,x}{27\,a}-\frac {4\,x^3\,\ln \left (1-a\,x^2\right )}{27}+\frac {8\,x^3}{81}+\frac {\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{27\,a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*polylog(3, a*x^2),x)

[Out]

(atan(a^(1/2)*x*1i)*8i)/(27*a^(3/2)) - (2*x^3*polylog(2, a*x^2))/9 + (x^3*polylog(3, a*x^2))/3 + (8*x)/(27*a)
- (4*x^3*log(1 - a*x^2))/27 + (8*x^3)/81

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {Li}_{3}\left (a x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*polylog(3,a*x**2),x)

[Out]

Integral(x**2*polylog(3, a*x**2), x)

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