3.23 \(\int \frac {\text {Li}_2(a x^2)}{x^3} \, dx\)

Optimal. Leaf size=49 \[ -\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}-\frac {1}{2} a \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{2 x^2}+a \log (x) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {6591, 2454, 2395, 36, 29, 31} \[ -\frac {\text {PolyLog}\left (2,a x^2\right )}{2 x^2}-\frac {1}{2} a \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{2 x^2}+a \log (x) \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^3,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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^3} \, dx &=-\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}-\int \frac {\log \left (1-a x^2\right )}{x^3} \, dx\\ &=-\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (1-a x)}{x^2} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{2 x^2}-\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x (1-a x)} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{2 x^2}-\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{1-a x} \, dx,x,x^2\right )\\ &=a \log (x)-\frac {1}{2} a \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{2 x^2}-\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.00 \[ -\frac {\text {Li}_2\left (a x^2\right )}{2 x^2}-\frac {1}{2} a \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{2 x^2}+a \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^3,x]

[Out]

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

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fricas [A]  time = 0.91, size = 44, normalized size = 0.90 \[ -\frac {a x^{2} \log \left (a x^{2} - 1\right ) - 2 \, a x^{2} \log \relax (x) + {\rm Li}_2\left (a x^{2}\right ) - \log \left (-a x^{2} + 1\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x^{2}\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^2)/x^3, x)

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maple [A]  time = 0.01, size = 43, normalized size = 0.88 \[ -\frac {\polylog \left (2, a \,x^{2}\right )}{2 x^{2}}+\frac {\ln \left (-a \,x^{2}+1\right )}{2 x^{2}}+a \ln \relax (x )-\frac {a \ln \left (a \,x^{2}-1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 34, normalized size = 0.69 \[ a \log \relax (x) - \frac {{\left (a x^{2} - 1\right )} \log \left (-a x^{2} + 1\right ) + {\rm Li}_2\left (a x^{2}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.21, size = 44, normalized size = 0.90 \[ \frac {3\,a\,\ln \relax (x)}{2}+\frac {\frac {\ln \left (1-a\,x^2\right )}{2}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{2}}{x^2}-\frac {a\,\ln \left (a\,x^3-x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/x^3,x)

[Out]

(3*a*log(x))/2 + (log(1 - a*x^2)/2 - polylog(2, a*x^2)/2)/x^2 - (a*log(a*x^3 - x))/2

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sympy [A]  time = 2.74, size = 37, normalized size = 0.76 \[ a \log {\relax (x )} + \frac {a \operatorname {Li}_{1}\left (a x^{2}\right )}{2} - \frac {\operatorname {Li}_{1}\left (a x^{2}\right )}{2 x^{2}} - \frac {\operatorname {Li}_{2}\left (a x^{2}\right )}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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