3.24 \(\int \frac {\text {Li}_2(a x^2)}{x^5} \, dx\)

Optimal. Leaf size=64 \[ -\frac {1}{8} a^2 \log \left (1-a x^2\right )+\frac {1}{4} a^2 \log (x)-\frac {\text {Li}_2\left (a x^2\right )}{4 x^4}-\frac {a}{8 x^2}+\frac {\log \left (1-a x^2\right )}{8 x^4} \]

[Out]

-1/8*a/x^2+1/4*a^2*ln(x)-1/8*a^2*ln(-a*x^2+1)+1/8*ln(-a*x^2+1)/x^4-1/4*polylog(2,a*x^2)/x^4

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Rubi [A]  time = 0.05, antiderivative size = 64, 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, 2454, 2395, 44} \[ -\frac {\text {PolyLog}\left (2,a x^2\right )}{4 x^4}-\frac {1}{8} a^2 \log \left (1-a x^2\right )+\frac {1}{4} a^2 \log (x)-\frac {a}{8 x^2}+\frac {\log \left (1-a x^2\right )}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^5,x]

[Out]

-a/(8*x^2) + (a^2*Log[x])/4 - (a^2*Log[1 - a*x^2])/8 + Log[1 - a*x^2]/(8*x^4) - PolyLog[2, a*x^2]/(4*x^4)

Rule 44

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2395

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] - Dist[(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]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

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 \frac {\text {Li}_2\left (a x^2\right )}{x^5} \, dx &=-\frac {\text {Li}_2\left (a x^2\right )}{4 x^4}-\frac {1}{2} \int \frac {\log \left (1-a x^2\right )}{x^5} \, dx\\ &=-\frac {\text {Li}_2\left (a x^2\right )}{4 x^4}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\log (1-a x)}{x^3} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{8 x^4}-\frac {\text {Li}_2\left (a x^2\right )}{4 x^4}+\frac {1}{8} a \operatorname {Subst}\left (\int \frac {1}{x^2 (1-a x)} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{8 x^4}-\frac {\text {Li}_2\left (a x^2\right )}{4 x^4}+\frac {1}{8} a \operatorname {Subst}\left (\int \left (\frac {1}{x^2}+\frac {a}{x}-\frac {a^2}{-1+a x}\right ) \, dx,x,x^2\right )\\ &=-\frac {a}{8 x^2}+\frac {1}{4} a^2 \log (x)-\frac {1}{8} a^2 \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{8 x^4}-\frac {\text {Li}_2\left (a x^2\right )}{4 x^4}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.80 \[ -\frac {-2 a^2 x^4 \log (x)+\left (a^2 x^4-1\right ) \log \left (1-a x^2\right )+2 \text {Li}_2\left (a x^2\right )+a x^2}{8 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^5,x]

[Out]

-1/8*(a*x^2 - 2*a^2*x^4*Log[x] + (-1 + a^2*x^4)*Log[1 - a*x^2] + 2*PolyLog[2, a*x^2])/x^4

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fricas [A]  time = 1.35, size = 55, normalized size = 0.86 \[ -\frac {a^{2} x^{4} \log \left (a x^{2} - 1\right ) - 2 \, a^{2} x^{4} \log \relax (x) + a x^{2} + 2 \, {\rm Li}_2\left (a x^{2}\right ) - \log \left (-a x^{2} + 1\right )}{8 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^5,x, algorithm="fricas")

[Out]

-1/8*(a^2*x^4*log(a*x^2 - 1) - 2*a^2*x^4*log(x) + a*x^2 + 2*dilog(a*x^2) - log(-a*x^2 + 1))/x^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x^{2}\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^5,x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/x^5, x)

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maple [A]  time = 0.02, size = 54, normalized size = 0.84 \[ -\frac {\polylog \left (2, a \,x^{2}\right )}{4 x^{4}}+\frac {\ln \left (-a \,x^{2}+1\right )}{8 x^{4}}-\frac {a}{8 x^{2}}+\frac {a^{2} \ln \relax (x )}{4}-\frac {a^{2} \ln \left (a \,x^{2}-1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^2)/x^5,x)

[Out]

-1/4*polylog(2,a*x^2)/x^4+1/8*ln(-a*x^2+1)/x^4-1/8*a/x^2+1/4*a^2*ln(x)-1/8*a^2*ln(a*x^2-1)

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maxima [A]  time = 0.33, size = 46, normalized size = 0.72 \[ \frac {1}{4} \, a^{2} \log \relax (x) - \frac {a x^{2} + {\left (a^{2} x^{4} - 1\right )} \log \left (-a x^{2} + 1\right ) + 2 \, {\rm Li}_2\left (a x^{2}\right )}{8 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^5,x, algorithm="maxima")

[Out]

1/4*a^2*log(x) - 1/8*(a*x^2 + (a^2*x^4 - 1)*log(-a*x^2 + 1) + 2*dilog(a*x^2))/x^4

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mupad [B]  time = 0.26, size = 53, normalized size = 0.83 \[ \frac {a^2\,\ln \relax (x)}{4}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{4\,x^4}-\frac {a^2\,\ln \left (a\,x^2-1\right )}{8}-\frac {a}{8\,x^2}+\frac {\ln \left (1-a\,x^2\right )}{8\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/x^5,x)

[Out]

(a^2*log(x))/4 - polylog(2, a*x^2)/(4*x^4) - (a^2*log(a*x^2 - 1))/8 - a/(8*x^2) + log(1 - a*x^2)/(8*x^4)

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sympy [A]  time = 7.31, size = 49, normalized size = 0.77 \[ \frac {a^{2} \log {\relax (x )}}{4} + \frac {a^{2} \operatorname {Li}_{1}\left (a x^{2}\right )}{8} - \frac {a}{8 x^{2}} - \frac {\operatorname {Li}_{1}\left (a x^{2}\right )}{8 x^{4}} - \frac {\operatorname {Li}_{2}\left (a x^{2}\right )}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x**2)/x**5,x)

[Out]

a**2*log(x)/4 + a**2*polylog(1, a*x**2)/8 - a/(8*x**2) - polylog(1, a*x**2)/(8*x**4) - polylog(2, a*x**2)/(4*x
**4)

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