3.1.4 \(\int x \text {PolyLog}(2,a x) \, dx\) [4]

Optimal. Leaf size=56 \[ -\frac {x}{4 a}-\frac {x^2}{8}-\frac {\log (1-a x)}{4 a^2}+\frac {1}{4} x^2 \log (1-a x)+\frac {1}{2} x^2 \text {PolyLog}(2,a x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6726, 2442, 45} \begin {gather*} -\frac {\log (1-a x)}{4 a^2}+\frac {1}{2} x^2 \text {Li}_2(a x)+\frac {1}{4} x^2 \log (1-a x)-\frac {x}{4 a}-\frac {x^2}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[2, a*x],x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 x \text {Li}_2(a x) \, dx &=\frac {1}{2} x^2 \text {Li}_2(a x)+\frac {1}{2} \int x \log (1-a x) \, dx\\ &=\frac {1}{4} x^2 \log (1-a x)+\frac {1}{2} x^2 \text {Li}_2(a x)+\frac {1}{4} a \int \frac {x^2}{1-a x} \, dx\\ &=\frac {1}{4} x^2 \log (1-a x)+\frac {1}{2} x^2 \text {Li}_2(a x)+\frac {1}{4} a \int \left (-\frac {1}{a^2}-\frac {x}{a}-\frac {1}{a^2 (-1+a x)}\right ) \, dx\\ &=-\frac {x}{4 a}-\frac {x^2}{8}-\frac {\log (1-a x)}{4 a^2}+\frac {1}{4} x^2 \log (1-a x)+\frac {1}{2} x^2 \text {Li}_2(a x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 48, normalized size = 0.86 \begin {gather*} \frac {-a x (2+a x)+2 \left (-1+a^2 x^2\right ) \log (1-a x)+4 a^2 x^2 \text {PolyLog}(2,a x)}{8 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[2, a*x],x]

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.35, size = 66, normalized size = 1.18

method	result	size
meijerg	\(-\frac {\frac {a x \left (3 a x +6\right )}{24}+\frac {\left (-3 a^{2} x^{2}+3\right ) \ln \left (-a x +1\right )}{12}-\frac {a^{2} x^{2} \polylog \left (2, a x \right )}{2}}{a^{2}}\)	\(49\)
derivativedivides	\(\frac {\frac {a^{2} x^{2} \polylog \left (2, a x \right )}{2}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{4}-\frac {\left (-a x +1\right )^{2}}{8}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{2}+\frac {1}{2}-\frac {a x}{2}}{a^{2}}\)	\(66\)
default	\(\frac {\frac {a^{2} x^{2} \polylog \left (2, a x \right )}{2}+\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )^{2}}{4}-\frac {\left (-a x +1\right )^{2}}{8}-\frac {\ln \left (-a x +1\right ) \left (-a x +1\right )}{2}+\frac {1}{2}-\frac {a x}{2}}{a^{2}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 48, normalized size = 0.86 \begin {gather*} \frac {4 \, a^{2} x^{2} {\rm Li}_2\left (a x\right ) - a^{2} x^{2} - 2 \, a x + 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}{8 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 48, normalized size = 0.86 \begin {gather*} \frac {4 \, a^{2} x^{2} {\rm Li}_2\left (a x\right ) - a^{2} x^{2} - 2 \, a x + 2 \, {\left (a^{2} x^{2} - 1\right )} \log \left (-a x + 1\right )}{8 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.48, size = 41, normalized size = 0.73 \begin {gather*} \begin {cases} - \frac {x^{2} \operatorname {Li}_{1}\left (a x\right )}{4} + \frac {x^{2} \operatorname {Li}_{2}\left (a x\right )}{2} - \frac {x^{2}}{8} - \frac {x}{4 a} + \frac {\operatorname {Li}_{1}\left (a x\right )}{4 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-x**2*polylog(1, a*x)/4 + x**2*polylog(2, a*x)/2 - x**2/8 - x/(4*a) + polylog(1, a*x)/(4*a**2), Ne(
a, 0)), (0, True))

________________________________________________________________________________________

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(x*polylog(2,a*x),x, algorithm="giac")

[Out]

integrate(x*dilog(a*x), x)

________________________________________________________________________________________

Mupad [B]
time = 0.34, size = 46, normalized size = 0.82 \begin {gather*} \frac {x^2\,\ln \left (1-a\,x\right )}{4}-\frac {\ln \left (1-a\,x\right )}{4\,a^2}-\frac {x}{4\,a}-\frac {x^2}{8}+\frac {x^2\,\mathrm {polylog}\left (2,a\,x\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(2, a*x),x)

[Out]

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

________________________________________________________________________________________