3.1.10 \(\int \frac {\text {PolyLog}(2,a x)}{x^5} \, dx\) [10]

Optimal. Leaf size=78 \[ -\frac {a}{48 x^3}-\frac {a^2}{32 x^2}-\frac {a^3}{16 x}+\frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)+\frac {\log (1-a x)}{16 x^4}-\frac {\text {PolyLog}(2,a x)}{4 x^4} \]

[Out]

-1/48*a/x^3-1/32*a^2/x^2-1/16*a^3/x+1/16*a^4*ln(x)-1/16*a^4*ln(-a*x+1)+1/16*ln(-a*x+1)/x^4-1/4*polylog(2,a*x)/
x^4

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Rubi [A]
time = 0.03, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6726, 2442, 46} \begin {gather*} \frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)-\frac {a^3}{16 x}-\frac {a^2}{32 x^2}-\frac {\text {Li}_2(a x)}{4 x^4}+\frac {\log (1-a x)}{16 x^4}-\frac {a}{48 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x]/x^5,x]

[Out]

-1/48*a/x^3 - a^2/(32*x^2) - a^3/(16*x) + (a^4*Log[x])/16 - (a^4*Log[1 - a*x])/16 + Log[1 - a*x]/(16*x^4) - Po
lyLog[2, a*x]/(4*x^4)

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 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}_2(a x)}{x^5} \, dx &=-\frac {\text {Li}_2(a x)}{4 x^4}-\frac {1}{4} \int \frac {\log (1-a x)}{x^5} \, dx\\ &=\frac {\log (1-a x)}{16 x^4}-\frac {\text {Li}_2(a x)}{4 x^4}+\frac {1}{16} a \int \frac {1}{x^4 (1-a x)} \, dx\\ &=\frac {\log (1-a x)}{16 x^4}-\frac {\text {Li}_2(a x)}{4 x^4}+\frac {1}{16} a \int \left (\frac {1}{x^4}+\frac {a}{x^3}+\frac {a^2}{x^2}+\frac {a^3}{x}-\frac {a^4}{-1+a x}\right ) \, dx\\ &=-\frac {a}{48 x^3}-\frac {a^2}{32 x^2}-\frac {a^3}{16 x}+\frac {1}{16} a^4 \log (x)-\frac {1}{16} a^4 \log (1-a x)+\frac {\log (1-a x)}{16 x^4}-\frac {\text {Li}_2(a x)}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 60, normalized size = 0.77 \begin {gather*} -\frac {a x \left (2+3 a x+6 a^2 x^2\right )-6 a^4 x^4 \log (x)+6 \left (-1+a^4 x^4\right ) \log (1-a x)+24 \text {PolyLog}(2,a x)}{96 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x]/x^5,x]

[Out]

-1/96*(a*x*(2 + 3*a*x + 6*a^2*x^2) - 6*a^4*x^4*Log[x] + 6*(-1 + a^4*x^4)*Log[1 - a*x] + 24*PolyLog[2, a*x])/x^
4

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Maple [A]
time = 0.44, size = 95, normalized size = 1.22

method	result	size
derivativedivides	\(a^{4} \left (-\frac {\polylog \left (2, a x \right )}{4 a^{4} x^{4}}-\frac {1}{48 a^{3} x^{3}}+\frac {\ln \left (-a x \right )}{16}-\frac {1}{16 a x}-\frac {1}{32 a^{2} x^{2}}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right ) \left (\left (-a x +1\right )^{3}-4 \left (-a x +1\right )^{2}+2-6 a x \right )}{16 a^{4} x^{4}}\right )\)	\(95\)
default	\(a^{4} \left (-\frac {\polylog \left (2, a x \right )}{4 a^{4} x^{4}}-\frac {1}{48 a^{3} x^{3}}+\frac {\ln \left (-a x \right )}{16}-\frac {1}{16 a x}-\frac {1}{32 a^{2} x^{2}}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right ) \left (\left (-a x +1\right )^{3}-4 \left (-a x +1\right )^{2}+2-6 a x \right )}{16 a^{4} x^{4}}\right )\)	\(95\)
meijerg	\(-a^{4} \left (-\frac {225 a^{3} x^{3}+350 a^{2} x^{2}+675 a x +2250}{7200 a^{3} x^{3}}-\frac {\left (-25 a^{4} x^{4}+25\right ) \ln \left (-a x +1\right )}{400 a^{4} x^{4}}+\frac {\polylog \left (2, a x \right )}{4 a^{4} x^{4}}+\frac {1}{32}-\frac {\ln \left (x \right )}{16}-\frac {\ln \left (-a \right )}{16}+\frac {1}{3 a^{3} x^{3}}+\frac {1}{8 a^{2} x^{2}}+\frac {1}{9 a x}\right )\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*(-1/4/a^4/x^4*polylog(2,a*x)-1/48/a^3/x^3+1/16*ln(-a*x)-1/16/a/x-1/32/a^2/x^2-1/16*ln(-a*x+1)*(-a*x+1)*((-
a*x+1)^3-4*(-a*x+1)^2+2-6*a*x)/a^4/x^4)

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Maxima [A]
time = 0.26, size = 58, normalized size = 0.74 \begin {gather*} \frac {1}{16} \, a^{4} \log \left (x\right ) - \frac {6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x + 6 \, {\left (a^{4} x^{4} - 1\right )} \log \left (-a x + 1\right ) + 24 \, {\rm Li}_2\left (a x\right )}{96 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a^4*log(x) - 1/96*(6*a^3*x^3 + 3*a^2*x^2 + 2*a*x + 6*(a^4*x^4 - 1)*log(-a*x + 1) + 24*dilog(a*x))/x^4

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Fricas [A]
time = 0.37, size = 65, normalized size = 0.83 \begin {gather*} -\frac {6 \, a^{4} x^{4} \log \left (a x - 1\right ) - 6 \, a^{4} x^{4} \log \left (x\right ) + 6 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + 2 \, a x + 24 \, {\rm Li}_2\left (a x\right ) - 6 \, \log \left (-a x + 1\right )}{96 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*a^4*x^4*log(a*x - 1) - 6*a^4*x^4*log(x) + 6*a^3*x^3 + 3*a^2*x^2 + 2*a*x + 24*dilog(a*x) - 6*log(-a*x
+ 1))/x^4

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Sympy [A]
time = 2.29, size = 60, normalized size = 0.77 \begin {gather*} \frac {a^{4} \log {\left (x \right )}}{16} + \frac {a^{4} \operatorname {Li}_{1}\left (a x\right )}{16} - \frac {a^{3}}{16 x} - \frac {a^{2}}{32 x^{2}} - \frac {a}{48 x^{3}} - \frac {\operatorname {Li}_{1}\left (a x\right )}{16 x^{4}} - \frac {\operatorname {Li}_{2}\left (a x\right )}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*log(x)/16 + a**4*polylog(1, a*x)/16 - a**3/(16*x) - a**2/(32*x**2) - a/(48*x**3) - polylog(1, a*x)/(16*x*
*4) - polylog(2, a*x)/(4*x**4)

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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(2,a*x)/x^5,x, algorithm="giac")

[Out]

integrate(dilog(a*x)/x^5, x)

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Mupad [B]
time = 0.44, size = 60, normalized size = 0.77 \begin {gather*} \frac {\ln \left (1-a\,x\right )}{16\,x^4}-\frac {\mathrm {polylog}\left (2,a\,x\right )}{4\,x^4}-\frac {a^3\,x^2+\frac {a^2\,x}{2}+\frac {a}{3}}{16\,x^3}-\frac {a^4\,\mathrm {atan}\left (a\,x\,2{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x)/x^5,x)

[Out]

log(1 - a*x)/(16*x^4) - (a^4*atan(a*x*2i - 1i)*1i)/8 - polylog(2, a*x)/(4*x^4) - (a/3 + (a^2*x)/2 + a^3*x^2)/(
16*x^3)

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