3.25 \(\int \frac {\text {Li}_2(a x^2)}{x^7} \, dx\)

Optimal. Leaf size=74 \[ -\frac {1}{18} a^3 \log \left (1-a x^2\right )+\frac {1}{9} a^3 \log (x)-\frac {a^2}{18 x^2}-\frac {\text {Li}_2\left (a x^2\right )}{6 x^6}-\frac {a}{36 x^4}+\frac {\log \left (1-a x^2\right )}{18 x^6} \]

[Out]

-1/36*a/x^4-1/18*a^2/x^2+1/9*a^3*ln(x)-1/18*a^3*ln(-a*x^2+1)+1/18*ln(-a*x^2+1)/x^6-1/6*polylog(2,a*x^2)/x^6

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Rubi [A]  time = 0.05, antiderivative size = 74, 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} \[ -\frac {\text {PolyLog}\left (2,a x^2\right )}{6 x^6}-\frac {a^2}{18 x^2}-\frac {1}{18} a^3 \log \left (1-a x^2\right )+\frac {1}{9} a^3 \log (x)-\frac {a}{36 x^4}+\frac {\log \left (1-a x^2\right )}{18 x^6} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^7,x]

[Out]

-a/(36*x^4) - a^2/(18*x^2) + (a^3*Log[x])/9 - (a^3*Log[1 - a*x^2])/18 + Log[1 - a*x^2]/(18*x^6) - PolyLog[2, a
*x^2]/(6*x^6)

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^7} \, dx &=-\frac {\text {Li}_2\left (a x^2\right )}{6 x^6}-\frac {1}{3} \int \frac {\log \left (1-a x^2\right )}{x^7} \, dx\\ &=-\frac {\text {Li}_2\left (a x^2\right )}{6 x^6}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (1-a x)}{x^4} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{18 x^6}-\frac {\text {Li}_2\left (a x^2\right )}{6 x^6}+\frac {1}{18} a \operatorname {Subst}\left (\int \frac {1}{x^3 (1-a x)} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{18 x^6}-\frac {\text {Li}_2\left (a x^2\right )}{6 x^6}+\frac {1}{18} a \operatorname {Subst}\left (\int \left (\frac {1}{x^3}+\frac {a}{x^2}+\frac {a^2}{x}-\frac {a^3}{-1+a x}\right ) \, dx,x,x^2\right )\\ &=-\frac {a}{36 x^4}-\frac {a^2}{18 x^2}+\frac {1}{9} a^3 \log (x)-\frac {1}{18} a^3 \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{18 x^6}-\frac {\text {Li}_2\left (a x^2\right )}{6 x^6}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 60, normalized size = 0.81 \[ -\frac {-4 a^3 x^6 \log (x)+2 \left (a^3 x^6-1\right ) \log \left (1-a x^2\right )+6 \text {Li}_2\left (a x^2\right )+a x^2 \left (2 a x^2+1\right )}{36 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^7,x]

[Out]

-1/36*(a*x^2*(1 + 2*a*x^2) - 4*a^3*x^6*Log[x] + 2*(-1 + a^3*x^6)*Log[1 - a*x^2] + 6*PolyLog[2, a*x^2])/x^6

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fricas [A]  time = 0.45, size = 64, normalized size = 0.86 \[ -\frac {2 \, a^{3} x^{6} \log \left (a x^{2} - 1\right ) - 4 \, a^{3} x^{6} \log \relax (x) + 2 \, a^{2} x^{4} + a x^{2} + 6 \, {\rm Li}_2\left (a x^{2}\right ) - 2 \, \log \left (-a x^{2} + 1\right )}{36 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^7,x, algorithm="fricas")

[Out]

-1/36*(2*a^3*x^6*log(a*x^2 - 1) - 4*a^3*x^6*log(x) + 2*a^2*x^4 + a*x^2 + 6*dilog(a*x^2) - 2*log(-a*x^2 + 1))/x
^6

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x^{2}\right )}{x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^7,x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/x^7, x)

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maple [A]  time = 0.01, size = 62, normalized size = 0.84 \[ -\frac {\polylog \left (2, a \,x^{2}\right )}{6 x^{6}}+\frac {\ln \left (-a \,x^{2}+1\right )}{18 x^{6}}-\frac {a}{36 x^{4}}-\frac {a^{2}}{18 x^{2}}+\frac {a^{3} \ln \relax (x )}{9}-\frac {a^{3} \ln \left (a \,x^{2}-1\right )}{18} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^2)/x^7,x)

[Out]

-1/6*polylog(2,a*x^2)/x^6+1/18*ln(-a*x^2+1)/x^6-1/36*a/x^4-1/18*a^2/x^2+1/9*a^3*ln(x)-1/18*a^3*ln(a*x^2-1)

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maxima [A]  time = 0.32, size = 55, normalized size = 0.74 \[ \frac {1}{9} \, a^{3} \log \relax (x) - \frac {2 \, a^{2} x^{4} + a x^{2} + 2 \, {\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right ) + 6 \, {\rm Li}_2\left (a x^{2}\right )}{36 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^7,x, algorithm="maxima")

[Out]

1/9*a^3*log(x) - 1/36*(2*a^2*x^4 + a*x^2 + 2*(a^3*x^6 - 1)*log(-a*x^2 + 1) + 6*dilog(a*x^2))/x^6

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mupad [B]  time = 0.28, size = 61, normalized size = 0.82 \[ \frac {a^3\,\ln \relax (x)}{9}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{6\,x^6}-\frac {a^3\,\ln \left (a\,x^2-1\right )}{18}-\frac {a}{36\,x^4}+\frac {\ln \left (1-a\,x^2\right )}{18\,x^6}-\frac {a^2}{18\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/x^7,x)

[Out]

(a^3*log(x))/9 - polylog(2, a*x^2)/(6*x^6) - (a^3*log(a*x^2 - 1))/18 - a/(36*x^4) + log(1 - a*x^2)/(18*x^6) -
a^2/(18*x^2)

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sympy [A]  time = 18.53, size = 58, normalized size = 0.78 \[ \frac {a^{3} \log {\relax (x )}}{9} + \frac {a^{3} \operatorname {Li}_{1}\left (a x^{2}\right )}{18} - \frac {a^{2}}{18 x^{2}} - \frac {a}{36 x^{4}} - \frac {\operatorname {Li}_{1}\left (a x^{2}\right )}{18 x^{6}} - \frac {\operatorname {Li}_{2}\left (a x^{2}\right )}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x**2)/x**7,x)

[Out]

a**3*log(x)/9 + a**3*polylog(1, a*x**2)/18 - a**2/(18*x**2) - a/(36*x**4) - polylog(1, a*x**2)/(18*x**6) - pol
ylog(2, a*x**2)/(6*x**6)

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