3.59 \(\int (d x)^{3/2} \text{PolyLog}(2,a x) \, dx\)

Optimal. Leaf size=117 \[ \frac{2 (d x)^{5/2} \text{PolyLog}(2,a x)}{5 d}+\frac{8 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/2}}-\frac{8 d \sqrt{d x}}{25 a^2}-\frac{8 (d x)^{3/2}}{75 a}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}-\frac{8 (d x)^{5/2}}{125 d} \]

[Out]

(-8*d*Sqrt[d*x])/(25*a^2) - (8*(d*x)^(3/2))/(75*a) - (8*(d*x)^(5/2))/(125*d) + (8*d^(3/2)*ArcTanh[(Sqrt[a]*Sqr
t[d*x])/Sqrt[d]])/(25*a^(5/2)) + (4*(d*x)^(5/2)*Log[1 - a*x])/(25*d) + (2*(d*x)^(5/2)*PolyLog[2, a*x])/(5*d)

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Rubi [A]  time = 0.0721405, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {6591, 2395, 50, 63, 206} \[ \frac{2 (d x)^{5/2} \text{PolyLog}(2,a x)}{5 d}+\frac{8 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/2}}-\frac{8 d \sqrt{d x}}{25 a^2}-\frac{8 (d x)^{3/2}}{75 a}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}-\frac{8 (d x)^{5/2}}{125 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[2, a*x],x]

[Out]

(-8*d*Sqrt[d*x])/(25*a^2) - (8*(d*x)^(3/2))/(75*a) - (8*(d*x)^(5/2))/(125*d) + (8*d^(3/2)*ArcTanh[(Sqrt[a]*Sqr
t[d*x])/Sqrt[d]])/(25*a^(5/2)) + (4*(d*x)^(5/2)*Log[1 - a*x])/(25*d) + (2*(d*x)^(5/2)*PolyLog[2, a*x])/(5*d)

Rule 6591

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]

Rule 2395

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (d x)^{3/2} \text{Li}_2(a x) \, dx &=\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}+\frac{2}{5} \int (d x)^{3/2} \log (1-a x) \, dx\\ &=\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}+\frac{(4 a) \int \frac{(d x)^{5/2}}{1-a x} \, dx}{25 d}\\ &=-\frac{8 (d x)^{5/2}}{125 d}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}+\frac{4}{25} \int \frac{(d x)^{3/2}}{1-a x} \, dx\\ &=-\frac{8 (d x)^{3/2}}{75 a}-\frac{8 (d x)^{5/2}}{125 d}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}+\frac{(4 d) \int \frac{\sqrt{d x}}{1-a x} \, dx}{25 a}\\ &=-\frac{8 d \sqrt{d x}}{25 a^2}-\frac{8 (d x)^{3/2}}{75 a}-\frac{8 (d x)^{5/2}}{125 d}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}+\frac{\left (4 d^2\right ) \int \frac{1}{\sqrt{d x} (1-a x)} \, dx}{25 a^2}\\ &=-\frac{8 d \sqrt{d x}}{25 a^2}-\frac{8 (d x)^{3/2}}{75 a}-\frac{8 (d x)^{5/2}}{125 d}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}+\frac{(8 d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{a x^2}{d}} \, dx,x,\sqrt{d x}\right )}{25 a^2}\\ &=-\frac{8 d \sqrt{d x}}{25 a^2}-\frac{8 (d x)^{3/2}}{75 a}-\frac{8 (d x)^{5/2}}{125 d}+\frac{8 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/2}}+\frac{4 (d x)^{5/2} \log (1-a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_2(a x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.102173, size = 90, normalized size = 0.77 \[ \frac{2 (d x)^{3/2} \left (x^{5/2} \text{PolyLog}(2,a x)+\frac{2}{75} \sqrt{x} \left (15 x^2 \log (1-a x)-\frac{2 \left (3 a^2 x^2+5 a x+15\right )}{a^2}\right )+\frac{4 \tanh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{5 a^{5/2}}\right )}{5 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(3/2)*PolyLog[2, a*x],x]

[Out]

(2*(d*x)^(3/2)*((4*ArcTanh[Sqrt[a]*Sqrt[x]])/(5*a^(5/2)) + (2*Sqrt[x]*((-2*(15 + 5*a*x + 3*a^2*x^2))/a^2 + 15*
x^2*Log[1 - a*x]))/75 + x^(5/2)*PolyLog[2, a*x]))/(5*x^(3/2))

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Maple [A]  time = 0.194, size = 96, normalized size = 0.8 \begin{align*}{\frac{2\,{\it polylog} \left ( 2,ax \right ) }{5\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}+{\frac{4}{25\,d} \left ( dx \right ) ^{{\frac{5}{2}}}\ln \left ({\frac{-adx+d}{d}} \right ) }-{\frac{8}{125\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}-{\frac{8}{75\,a} \left ( dx \right ) ^{{\frac{3}{2}}}}-{\frac{8\,d}{25\,{a}^{2}}\sqrt{dx}}+{\frac{8\,{d}^{2}}{25\,{a}^{2}}{\it Artanh} \left ({a\sqrt{dx}{\frac{1}{\sqrt{ad}}}} \right ){\frac{1}{\sqrt{ad}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*polylog(2,a*x),x)

[Out]

2/5*(d*x)^(5/2)*polylog(2,a*x)/d+4/25/d*(d*x)^(5/2)*ln((-a*d*x+d)/d)-8/125*(d*x)^(5/2)/d-8/75*(d*x)^(3/2)/a-8/
25*d*(d*x)^(1/2)/a^2+8/25*d^2/a^2/(a*d)^(1/2)*arctanh(a*(d*x)^(1/2)/(a*d)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.82002, size = 470, normalized size = 4.02 \begin{align*} \left [\frac{2 \,{\left (30 \, d \sqrt{\frac{d}{a}} \log \left (\frac{a d x + 2 \, \sqrt{d x} a \sqrt{\frac{d}{a}} + d}{a x - 1}\right ) +{\left (75 \, a^{2} d x^{2}{\rm Li}_2\left (a x\right ) + 30 \, a^{2} d x^{2} \log \left (-a x + 1\right ) - 12 \, a^{2} d x^{2} - 20 \, a d x - 60 \, d\right )} \sqrt{d x}\right )}}{375 \, a^{2}}, -\frac{2 \,{\left (60 \, d \sqrt{-\frac{d}{a}} \arctan \left (\frac{\sqrt{d x} a \sqrt{-\frac{d}{a}}}{d}\right ) -{\left (75 \, a^{2} d x^{2}{\rm Li}_2\left (a x\right ) + 30 \, a^{2} d x^{2} \log \left (-a x + 1\right ) - 12 \, a^{2} d x^{2} - 20 \, a d x - 60 \, d\right )} \sqrt{d x}\right )}}{375 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/375*(30*d*sqrt(d/a)*log((a*d*x + 2*sqrt(d*x)*a*sqrt(d/a) + d)/(a*x - 1)) + (75*a^2*d*x^2*dilog(a*x) + 30*a^
2*d*x^2*log(-a*x + 1) - 12*a^2*d*x^2 - 20*a*d*x - 60*d)*sqrt(d*x))/a^2, -2/375*(60*d*sqrt(-d/a)*arctan(sqrt(d*
x)*a*sqrt(-d/a)/d) - (75*a^2*d*x^2*dilog(a*x) + 30*a^2*d*x^2*log(-a*x + 1) - 12*a^2*d*x^2 - 20*a*d*x - 60*d)*s
qrt(d*x))/a^2]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{\frac{3}{2}}{\rm Li}_2\left (a x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x)^(3/2)*dilog(a*x), x)