3.1.21 \(\int x \text {PolyLog}(2,a x^2) \, dx\) [21]

Optimal. Leaf size=46 \[ -\frac {x^2}{2}-\frac {\left (1-a x^2\right ) \log \left (1-a x^2\right )}{2 a}+\frac {1}{2} x^2 \text {PolyLog}\left (2,a x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6726, 2504, 2436, 2332} \begin {gather*} \frac {1}{2} x^2 \text {Li}_2\left (a x^2\right )-\frac {\left (1-a x^2\right ) \log \left (1-a x^2\right )}{2 a}-\frac {x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[2, a*x^2],x]

[Out]

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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 x \text {Li}_2\left (a x^2\right ) \, dx &=\frac {1}{2} x^2 \text {Li}_2\left (a x^2\right )+\int x \log \left (1-a x^2\right ) \, dx\\ &=\frac {1}{2} x^2 \text {Li}_2\left (a x^2\right )+\frac {1}{2} \text {Subst}\left (\int \log (1-a x) \, dx,x,x^2\right )\\ &=\frac {1}{2} x^2 \text {Li}_2\left (a x^2\right )-\frac {\text {Subst}\left (\int \log (x) \, dx,x,1-a x^2\right )}{2 a}\\ &=-\frac {x^2}{2}-\frac {\left (1-a x^2\right ) \log \left (1-a x^2\right )}{2 a}+\frac {1}{2} x^2 \text {Li}_2\left (a x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.93 \begin {gather*} \frac {-a x^2+\left (-1+a x^2\right ) \log \left (1-a x^2\right )+a x^2 \text {PolyLog}\left (2,a x^2\right )}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[2, a*x^2],x]

[Out]

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

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Maple [A]
time = 0.29, size = 45, normalized size = 0.98

method	result	size
meijerg	\(\frac {-a \,x^{2}-\frac {\left (-2 a \,x^{2}+2\right ) \ln \left (-a \,x^{2}+1\right )}{2}+a \,x^{2} \polylog \left (2, a \,x^{2}\right )}{2 a}\)	\(44\)
derivativedivides	\(\frac {a \,x^{2} \polylog \left (2, a \,x^{2}\right )-\ln \left (-a \,x^{2}+1\right ) \left (-a \,x^{2}+1\right )+1-a \,x^{2}}{2 a}\)	\(45\)
default	\(\frac {a \,x^{2} \polylog \left (2, a \,x^{2}\right )-\ln \left (-a \,x^{2}+1\right ) \left (-a \,x^{2}+1\right )+1-a \,x^{2}}{2 a}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 40, normalized size = 0.87 \begin {gather*} \frac {a x^{2} {\rm Li}_2\left (a x^{2}\right ) - a x^{2} + {\left (a x^{2} - 1\right )} \log \left (-a x^{2} + 1\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 40, normalized size = 0.87 \begin {gather*} \frac {a x^{2} {\rm Li}_2\left (a x^{2}\right ) - a x^{2} + {\left (a x^{2} - 1\right )} \log \left (-a x^{2} + 1\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.67, size = 39, normalized size = 0.85 \begin {gather*} \begin {cases} - \frac {x^{2} \operatorname {Li}_{1}\left (a x^{2}\right )}{2} + \frac {x^{2} \operatorname {Li}_{2}\left (a x^{2}\right )}{2} - \frac {x^{2}}{2} + \frac {\operatorname {Li}_{1}\left (a x^{2}\right )}{2 a} & \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**2),x)

[Out]

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

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

[Out]

integrate(x*dilog(a*x^2), x)

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Mupad [B]
time = 0.21, size = 45, normalized size = 0.98 \begin {gather*} \frac {x^2\,\mathrm {polylog}\left (2,a\,x^2\right )}{2}-\frac {\ln \left (a\,x^2-1\right )}{2\,a}+\frac {x^2\,\ln \left (1-a\,x^2\right )}{2}-\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(2, a*x^2),x)

[Out]

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

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