3.1.38 \(\int \frac {\text {PolyLog}(3,a x^2)}{x^7} \, dx\) [38]

Optimal. Leaf size=88 \[ -\frac {a}{108 x^4}-\frac {a^2}{54 x^2}+\frac {1}{27} a^3 \log (x)-\frac {1}{54} a^3 \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{54 x^6}-\frac {\text {PolyLog}\left (2,a x^2\right )}{18 x^6}-\frac {\text {PolyLog}\left (3,a x^2\right )}{6 x^6} \]

[Out]

-1/108*a/x^4-1/54*a^2/x^2+1/27*a^3*ln(x)-1/54*a^3*ln(-a*x^2+1)+1/54*ln(-a*x^2+1)/x^6-1/18*polylog(2,a*x^2)/x^6
-1/6*polylog(3,a*x^2)/x^6

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Rubi [A]
time = 0.05, antiderivative size = 88, 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 = {6726, 2504, 2442, 46} \begin {gather*} -\frac {1}{54} a^3 \log \left (1-a x^2\right )+\frac {1}{27} a^3 \log (x)-\frac {a^2}{54 x^2}-\frac {\text {Li}_2\left (a x^2\right )}{18 x^6}-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}-\frac {a}{108 x^4}+\frac {\log \left (1-a x^2\right )}{54 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/x^7,x]

[Out]

-1/108*a/x^4 - a^2/(54*x^2) + (a^3*Log[x])/27 - (a^3*Log[1 - a*x^2])/54 + Log[1 - a*x^2]/(54*x^6) - PolyLog[2,
a*x^2]/(18*x^6) - PolyLog[3, a*x^2]/(6*x^6)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2442

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 2504

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 6726

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}_3\left (a x^2\right )}{x^7} \, dx &=-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}+\frac {1}{3} \int \frac {\text {Li}_2\left (a x^2\right )}{x^7} \, dx\\ &=-\frac {\text {Li}_2\left (a x^2\right )}{18 x^6}-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}-\frac {1}{9} \int \frac {\log \left (1-a x^2\right )}{x^7} \, dx\\ &=-\frac {\text {Li}_2\left (a x^2\right )}{18 x^6}-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}-\frac {1}{18} \text {Subst}\left (\int \frac {\log (1-a x)}{x^4} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{54 x^6}-\frac {\text {Li}_2\left (a x^2\right )}{18 x^6}-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}+\frac {1}{54} a \text {Subst}\left (\int \frac {1}{x^3 (1-a x)} \, dx,x,x^2\right )\\ &=\frac {\log \left (1-a x^2\right )}{54 x^6}-\frac {\text {Li}_2\left (a x^2\right )}{18 x^6}-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}+\frac {1}{54} a \text {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}{108 x^4}-\frac {a^2}{54 x^2}+\frac {1}{27} a^3 \log (x)-\frac {1}{54} a^3 \log \left (1-a x^2\right )+\frac {\log \left (1-a x^2\right )}{54 x^6}-\frac {\text {Li}_2\left (a x^2\right )}{18 x^6}-\frac {\text {Li}_3\left (a x^2\right )}{6 x^6}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 4 in optimal.
time = 0.01, size = 30, normalized size = 0.34 \begin {gather*} \frac {G_{5,5}^{2,4}\left (-a x^2|\begin {array}{c} 1,1,1,1,4 \\ 1,3,0,0,0 \\\end {array}\right )}{2 x^6} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^2]/x^7,x]

[Out]

MeijerG[{{1, 1, 1, 1}, {4}}, {{1, 3}, {0, 0, 0}}, -(a*x^2)]/(2*x^6)

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Maple [A]
time = 0.10, size = 115, normalized size = 1.31

method	result	size
meijerg	\(\frac {a^{3} \left (\frac {64 a^{2} x^{4}+152 a \,x^{2}+832}{1728 a^{2} x^{4}}+\frac {\left (-64 a^{3} x^{6}+64\right ) \ln \left (-a \,x^{2}+1\right )}{1728 a^{3} x^{6}}-\frac {\polylog \left (2, a \,x^{2}\right )}{9 a^{3} x^{6}}-\frac {\polylog \left (3, a \,x^{2}\right )}{3 a^{3} x^{6}}-\frac {1}{27}+\frac {2 \ln \left (x \right )}{27}+\frac {\ln \left (-a \right )}{27}-\frac {1}{2 a^{2} x^{4}}-\frac {1}{8 a \,x^{2}}\right )}{2}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/x^7,x,method=_RETURNVERBOSE)

[Out]

1/2*a^3*(1/1728/a^2/x^4*(64*a^2*x^4+152*a*x^2+832)+1/1728/a^3/x^6*(-64*a^3*x^6+64)*ln(-a*x^2+1)-1/9/a^3/x^6*po
lylog(2,a*x^2)-1/3/a^3/x^6*polylog(3,a*x^2)-1/27+2/27*ln(x)+1/27*ln(-a)-1/2/a^2/x^4-1/8/a/x^2)

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Maxima [A]
time = 0.27, size = 64, normalized size = 0.73 \begin {gather*} \frac {1}{27} \, a^{3} \log \left (x\right ) - \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 ) + 18 \, {\rm Li}_{3}(a x^{2})}{108 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^7,x, algorithm="maxima")

[Out]

1/27*a^3*log(x) - 1/108*(2*a^2*x^4 + a*x^2 + 2*(a^3*x^6 - 1)*log(-a*x^2 + 1) + 6*dilog(a*x^2) + 18*polylog(3,
a*x^2))/x^6

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Fricas [A]
time = 0.46, size = 73, normalized size = 0.83 \begin {gather*} -\frac {2 \, a^{3} x^{6} \log \left (a x^{2} - 1\right ) - 4 \, a^{3} x^{6} \log \left (x\right ) + 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 ) + 18 \, {\rm polylog}\left (3, a x^{2}\right )}{108 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^7,x, algorithm="fricas")

[Out]

-1/108*(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) +
18*polylog(3, a*x^2))/x^6

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/x**7,x)

[Out]

Integral(polylog(3, a*x**2)/x**7, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/x^7,x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/x^7, x)

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Mupad [B]
time = 1.02, size = 73, normalized size = 0.83 \begin {gather*} \frac {a^3\,\ln \left (x\right )}{27}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{18\,x^6}-\frac {\mathrm {polylog}\left (3,a\,x^2\right )}{6\,x^6}-\frac {a^3\,\ln \left (a\,x^2-1\right )}{54}-\frac {a}{108\,x^4}+\frac {\ln \left (1-a\,x^2\right )}{54\,x^6}-\frac {a^2}{54\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/x^7,x)

[Out]

(a^3*log(x))/27 - polylog(2, a*x^2)/(18*x^6) - polylog(3, a*x^2)/(6*x^6) - (a^3*log(a*x^2 - 1))/54 - a/(108*x^
4) + log(1 - a*x^2)/(54*x^6) - a^2/(54*x^2)

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