3.26 \(\int x^4 \text {Li}_2(a x^2) \, dx\)

Optimal. Leaf size=73 \[ \frac {4 \tanh ^{-1}\left (\sqrt {a} x\right )}{25 a^{5/2}}-\frac {4 x}{25 a^2}+\frac {1}{5} x^5 \text {Li}_2\left (a x^2\right )-\frac {4 x^3}{75 a}+\frac {2}{25} x^5 \log \left (1-a x^2\right )-\frac {4 x^5}{125} \]

[Out]

-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
)

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Rubi [A]  time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6591, 2455, 302, 206} \[ \frac {1}{5} x^5 \text {PolyLog}\left (2,a x^2\right )-\frac {4 x}{25 a^2}+\frac {4 \tanh ^{-1}\left (\sqrt {a} x\right )}{25 a^{5/2}}-\frac {4 x^3}{75 a}+\frac {2}{25} x^5 \log \left (1-a x^2\right )-\frac {4 x^5}{125} \]

Antiderivative was successfully verified.

[In]

Int[x^4*PolyLog[2, a*x^2],x]

[Out]

(-4*x)/(25*a^2) - (4*x^3)/(75*a) - (4*x^5)/125 + (4*ArcTanh[Sqrt[a]*x])/(25*a^(5/2)) + (2*x^5*Log[1 - a*x^2])/
25 + (x^5*PolyLog[2, a*x^2])/5

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^4 \text {Li}_2\left (a x^2\right ) \, dx &=\frac {1}{5} x^5 \text {Li}_2\left (a x^2\right )+\frac {2}{5} \int x^4 \log \left (1-a x^2\right ) \, dx\\ &=\frac {2}{25} x^5 \log \left (1-a x^2\right )+\frac {1}{5} x^5 \text {Li}_2\left (a x^2\right )+\frac {1}{25} (4 a) \int \frac {x^6}{1-a x^2} \, dx\\ &=\frac {2}{25} x^5 \log \left (1-a x^2\right )+\frac {1}{5} x^5 \text {Li}_2\left (a x^2\right )+\frac {1}{25} (4 a) \int \left (-\frac {1}{a^3}-\frac {x^2}{a^2}-\frac {x^4}{a}+\frac {1}{a^3 \left (1-a x^2\right )}\right ) \, dx\\ &=-\frac {4 x}{25 a^2}-\frac {4 x^3}{75 a}-\frac {4 x^5}{125}+\frac {2}{25} x^5 \log \left (1-a x^2\right )+\frac {1}{5} x^5 \text {Li}_2\left (a x^2\right )+\frac {4 \int \frac {1}{1-a x^2} \, dx}{25 a^2}\\ &=-\frac {4 x}{25 a^2}-\frac {4 x^3}{75 a}-\frac {4 x^5}{125}+\frac {4 \tanh ^{-1}\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 \text {Li}_2\left (a x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 65, normalized size = 0.89 \[ \frac {1}{375} \left (\frac {60 \tanh ^{-1}\left (\sqrt {a} x\right )}{a^{5/2}}-\frac {60 x}{a^2}+75 x^5 \text {Li}_2\left (a x^2\right )-\frac {20 x^3}{a}+30 x^5 \log \left (1-a x^2\right )-12 x^5\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*PolyLog[2, a*x^2],x]

[Out]

((-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

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fricas [A]  time = 1.30, size = 159, normalized size = 2.18 \[ \left [\frac {75 \, a^{3} x^{5} {\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{3} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{3} x^{5} - 20 \, a^{2} x^{3} - 60 \, a x + 30 \, \sqrt {a} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right )}{375 \, a^{3}}, \frac {75 \, a^{3} x^{5} {\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{3} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{3} x^{5} - 20 \, a^{2} x^{3} - 60 \, a x - 60 \, \sqrt {-a} \arctan \left (\sqrt {-a} x\right )}{375 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*polylog(2,a*x^2),x, algorithm="fricas")

[Out]

[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)*l
og((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]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} {\rm Li}_2\left (a x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*polylog(2,a*x^2),x, algorithm="giac")

[Out]

integrate(x^4*dilog(a*x^2), x)

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maple [A]  time = 0.01, size = 58, normalized size = 0.79 \[ -\frac {4 x}{25 a^{2}}-\frac {4 x^{3}}{75 a}-\frac {4 x^{5}}{125}+\frac {4 \arctanh \left (x \sqrt {a}\right )}{25 a^{\frac {5}{2}}}+\frac {2 x^{5} \ln \left (-a \,x^{2}+1\right )}{25}+\frac {x^{5} \polylog \left (2, a \,x^{2}\right )}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*polylog(2,a*x^2),x)

[Out]

-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
)

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maxima [A]  time = 0.42, size = 80, normalized size = 1.10 \[ \frac {75 \, a^{2} x^{5} {\rm Li}_2\left (a x^{2}\right ) + 30 \, a^{2} x^{5} \log \left (-a x^{2} + 1\right ) - 12 \, a^{2} x^{5} - 20 \, a x^{3} - 60 \, x}{375 \, a^{2}} - \frac {2 \, \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right )}{25 \, a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*polylog(2,a*x^2),x, algorithm="maxima")

[Out]

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)

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mupad [B]  time = 0.41, size = 60, normalized size = 0.82 \[ \frac {x^5\,\mathrm {polylog}\left (2,a\,x^2\right )}{5}-\frac {4\,x}{25\,a^2}+\frac {2\,x^5\,\ln \left (1-a\,x^2\right )}{25}-\frac {4\,x^5}{125}-\frac {4\,x^3}{75\,a}-\frac {\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{25\,a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*polylog(2, a*x^2),x)

[Out]

(x^5*polylog(2, a*x^2))/5 - (atan(a^(1/2)*x*1i)*4i)/(25*a^(5/2)) - (4*x)/(25*a^2) + (2*x^5*log(1 - a*x^2))/25
- (4*x^5)/125 - (4*x^3)/(75*a)

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sympy [A]  time = 144.62, size = 94, normalized size = 1.29 \[ \begin {cases} - \frac {2 x^{5} \operatorname {Li}_{1}\left (a x^{2}\right )}{25} + \frac {x^{5} \operatorname {Li}_{2}\left (a x^{2}\right )}{5} - \frac {4 x^{5}}{125} - \frac {4 x^{3}}{75 a} - \frac {4 x}{25 a^{2}} - \frac {4 \log {\left (x - \sqrt {\frac {1}{a}} \right )}}{25 a^{3} \sqrt {\frac {1}{a}}} - \frac {2 \operatorname {Li}_{1}\left (a x^{2}\right )}{25 a^{3} \sqrt {\frac {1}{a}}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*polylog(2,a*x**2),x)

[Out]

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*sqrt(1/a)) - 2*polylog(1, a*x**2)/(25*a**3*sqrt(1/a)), Ne(a, 0)), (0, True
))

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