3.1.19 \(\int x^5 \text {PolyLog}(2,a x^2) \, dx\) [19]

Optimal. Leaf size=74 \[ -\frac {x^2}{18 a^2}-\frac {x^4}{36 a}-\frac {x^6}{54}-\frac {\log \left (1-a x^2\right )}{18 a^3}+\frac {1}{18} x^6 \log \left (1-a x^2\right )+\frac {1}{6} x^6 \text {PolyLog}\left (2,a x^2\right ) \]

[Out]

-1/18*x^2/a^2-1/36*x^4/a-1/54*x^6-1/18*ln(-a*x^2+1)/a^3+1/18*x^6*ln(-a*x^2+1)+1/6*x^6*polylog(2,a*x^2)

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Rubi [A]
time = 0.04, 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 = {6726, 2504, 2442, 45} \begin {gather*} -\frac {\log \left (1-a x^2\right )}{18 a^3}-\frac {x^2}{18 a^2}+\frac {1}{6} x^6 \text {Li}_2\left (a x^2\right )-\frac {x^4}{36 a}+\frac {1}{18} x^6 \log \left (1-a x^2\right )-\frac {x^6}{54} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*PolyLog[2, a*x^2],x]

[Out]

-1/18*x^2/a^2 - x^4/(36*a) - x^6/54 - Log[1 - a*x^2]/(18*a^3) + (x^6*Log[1 - a*x^2])/18 + (x^6*PolyLog[2, a*x^
2])/6

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 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^5 \text {Li}_2\left (a x^2\right ) \, dx &=\frac {1}{6} x^6 \text {Li}_2\left (a x^2\right )+\frac {1}{3} \int x^5 \log \left (1-a x^2\right ) \, dx\\ &=\frac {1}{6} x^6 \text {Li}_2\left (a x^2\right )+\frac {1}{6} \text {Subst}\left (\int x^2 \log (1-a x) \, dx,x,x^2\right )\\ &=\frac {1}{18} x^6 \log \left (1-a x^2\right )+\frac {1}{6} x^6 \text {Li}_2\left (a x^2\right )+\frac {1}{18} a \text {Subst}\left (\int \frac {x^3}{1-a x} \, dx,x,x^2\right )\\ &=\frac {1}{18} x^6 \log \left (1-a x^2\right )+\frac {1}{6} x^6 \text {Li}_2\left (a x^2\right )+\frac {1}{18} a \text {Subst}\left (\int \left (-\frac {1}{a^3}-\frac {x}{a^2}-\frac {x^2}{a}-\frac {1}{a^3 (-1+a x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {x^2}{18 a^2}-\frac {x^4}{36 a}-\frac {x^6}{54}-\frac {\log \left (1-a x^2\right )}{18 a^3}+\frac {1}{18} x^6 \log \left (1-a x^2\right )+\frac {1}{6} x^6 \text {Li}_2\left (a x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 65, normalized size = 0.88 \begin {gather*} \frac {-a x^2 \left (6+3 a x^2+2 a^2 x^4\right )+6 \left (-1+a^3 x^6\right ) \log \left (1-a x^2\right )+18 a^3 x^6 \text {PolyLog}\left (2,a x^2\right )}{108 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*PolyLog[2, a*x^2],x]

[Out]

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

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Maple [A]
time = 0.06, size = 68, normalized size = 0.92

method	result	size
meijerg	\(\frac {-\frac {x^{2} a \left (4 a^{2} x^{4}+6 a \,x^{2}+12\right )}{108}-\frac {\left (-4 a^{3} x^{6}+4\right ) \ln \left (-a \,x^{2}+1\right )}{36}+\frac {x^{6} a^{3} \polylog \left (2, a \,x^{2}\right )}{3}}{2 a^{3}}\)	\(65\)
default	\(\frac {x^{6} \polylog \left (2, a \,x^{2}\right )}{6}+\frac {x^{6} \ln \left (-a \,x^{2}+1\right )}{18}+\frac {a \left (-\frac {\frac {1}{3} a^{2} x^{6}+\frac {1}{2} a \,x^{4}+x^{2}}{2 a^{3}}-\frac {\ln \left (a \,x^{2}-1\right )}{2 a^{4}}\right )}{9}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 62, normalized size = 0.84 \begin {gather*} \frac {18 \, a^{3} x^{6} {\rm Li}_2\left (a x^{2}\right ) - 2 \, a^{3} x^{6} - 3 \, a^{2} x^{4} - 6 \, a x^{2} + 6 \, {\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right )}{108 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 62, normalized size = 0.84 \begin {gather*} \frac {18 \, a^{3} x^{6} {\rm Li}_2\left (a x^{2}\right ) - 2 \, a^{3} x^{6} - 3 \, a^{2} x^{4} - 6 \, a x^{2} + 6 \, {\left (a^{3} x^{6} - 1\right )} \log \left (-a x^{2} + 1\right )}{108 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 4.65, size = 56, normalized size = 0.76 \begin {gather*} \begin {cases} - \frac {x^{6} \operatorname {Li}_{1}\left (a x^{2}\right )}{18} + \frac {x^{6} \operatorname {Li}_{2}\left (a x^{2}\right )}{6} - \frac {x^{6}}{54} - \frac {x^{4}}{36 a} - \frac {x^{2}}{18 a^{2}} + \frac {\operatorname {Li}_{1}\left (a x^{2}\right )}{18 a^{3}} & \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**5*polylog(2,a*x**2),x)

[Out]

Piecewise((-x**6*polylog(1, a*x**2)/18 + x**6*polylog(2, a*x**2)/6 - x**6/54 - x**4/(36*a) - x**2/(18*a**2) +
polylog(1, a*x**2)/(18*a**3), 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^5*polylog(2,a*x^2),x, algorithm="giac")

[Out]

integrate(x^5*dilog(a*x^2), x)

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Mupad [B]
time = 0.19, size = 61, normalized size = 0.82 \begin {gather*} \frac {x^6\,\mathrm {polylog}\left (2,a\,x^2\right )}{6}-\frac {\ln \left (a\,x^2-1\right )}{18\,a^3}+\frac {x^6\,\ln \left (1-a\,x^2\right )}{18}-\frac {x^6}{54}-\frac {x^2}{18\,a^2}-\frac {x^4}{36\,a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*polylog(2, a*x^2),x)

[Out]

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

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