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

Optimal. Leaf size=136 \[ -\frac{4 (d x)^{5/2} \text{PolyLog}(2,a x)}{25 d}+\frac{2 (d x)^{5/2} \text{PolyLog}(3,a x)}{5 d}-\frac{16 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 a^{5/2}}+\frac{16 d \sqrt{d x}}{125 a^2}+\frac{16 (d x)^{3/2}}{375 a}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}+\frac{16 (d x)^{5/2}}{625 d} \]

[Out]

(16*d*Sqrt[d*x])/(125*a^2) + (16*(d*x)^(3/2))/(375*a) + (16*(d*x)^(5/2))/(625*d) - (16*d^(3/2)*ArcTanh[(Sqrt[a
]*Sqrt[d*x])/Sqrt[d]])/(125*a^(5/2)) - (8*(d*x)^(5/2)*Log[1 - a*x])/(125*d) - (4*(d*x)^(5/2)*PolyLog[2, a*x])/
(25*d) + (2*(d*x)^(5/2)*PolyLog[3, a*x])/(5*d)

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Rubi [A]  time = 0.0834981, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{4 (d x)^{5/2} \text{PolyLog}(2,a x)}{25 d}+\frac{2 (d x)^{5/2} \text{PolyLog}(3,a x)}{5 d}-\frac{16 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 a^{5/2}}+\frac{16 d \sqrt{d x}}{125 a^2}+\frac{16 (d x)^{3/2}}{375 a}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}+\frac{16 (d x)^{5/2}}{625 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[3, a*x],x]

[Out]

(16*d*Sqrt[d*x])/(125*a^2) + (16*(d*x)^(3/2))/(375*a) + (16*(d*x)^(5/2))/(625*d) - (16*d^(3/2)*ArcTanh[(Sqrt[a
]*Sqrt[d*x])/Sqrt[d]])/(125*a^(5/2)) - (8*(d*x)^(5/2)*Log[1 - a*x])/(125*d) - (4*(d*x)^(5/2)*PolyLog[2, a*x])/
(25*d) + (2*(d*x)^(5/2)*PolyLog[3, 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}_3(a x) \, dx &=\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{2}{5} \int (d x)^{3/2} \text{Li}_2(a x) \, dx\\ &=-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{4}{25} \int (d x)^{3/2} \log (1-a x) \, dx\\ &=-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{(8 a) \int \frac{(d x)^{5/2}}{1-a x} \, dx}{125 d}\\ &=\frac{16 (d x)^{5/2}}{625 d}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{8}{125} \int \frac{(d x)^{3/2}}{1-a x} \, dx\\ &=\frac{16 (d x)^{3/2}}{375 a}+\frac{16 (d x)^{5/2}}{625 d}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{(8 d) \int \frac{\sqrt{d x}}{1-a x} \, dx}{125 a}\\ &=\frac{16 d \sqrt{d x}}{125 a^2}+\frac{16 (d x)^{3/2}}{375 a}+\frac{16 (d x)^{5/2}}{625 d}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{\left (8 d^2\right ) \int \frac{1}{\sqrt{d x} (1-a x)} \, dx}{125 a^2}\\ &=\frac{16 d \sqrt{d x}}{125 a^2}+\frac{16 (d x)^{3/2}}{375 a}+\frac{16 (d x)^{5/2}}{625 d}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}-\frac{(16 d) \operatorname{Subst}\left (\int \frac{1}{1-\frac{a x^2}{d}} \, dx,x,\sqrt{d x}\right )}{125 a^2}\\ &=\frac{16 d \sqrt{d x}}{125 a^2}+\frac{16 (d x)^{3/2}}{375 a}+\frac{16 (d x)^{5/2}}{625 d}-\frac{16 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 a^{5/2}}-\frac{8 (d x)^{5/2} \log (1-a x)}{125 d}-\frac{4 (d x)^{5/2} \text{Li}_2(a x)}{25 d}+\frac{2 (d x)^{5/2} \text{Li}_3(a x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.217824, size = 88, normalized size = 0.65 \[ \frac{2 d \sqrt{d x} \left (-150 x^2 \text{PolyLog}(2,a x)+375 x^2 \text{PolyLog}(3,a x)+4 \left (-\frac{30 \tanh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{a^{5/2} \sqrt{x}}+\frac{30}{a^2}-15 x^2 \log (1-a x)+\frac{10 x}{a}+6 x^2\right )\right )}{1875} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*Sqrt[d*x]*(4*(30/a^2 + (10*x)/a + 6*x^2 - (30*ArcTanh[Sqrt[a]*Sqrt[x]])/(a^(5/2)*Sqrt[x]) - 15*x^2*Log[1
- a*x]) - 150*x^2*PolyLog[2, a*x] + 375*x^2*PolyLog[3, a*x]))/1875

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Maple [A]  time = 0.061, size = 141, normalized size = 1. \begin{align*}{\frac{1}{a} \left ( dx \right ) ^{{\frac{3}{2}}} \left ({\frac{336\,{a}^{2}{x}^{2}+560\,ax+1680}{13125\,{a}^{3}}\sqrt{x} \left ( -a \right ) ^{{\frac{7}{2}}}}+{\frac{8}{125\,{a}^{3}}\sqrt{x} \left ( -a \right ) ^{{\frac{7}{2}}} \left ( \ln \left ( 1-\sqrt{ax} \right ) -\ln \left ( 1+\sqrt{ax} \right ) \right ){\frac{1}{\sqrt{ax}}}}-{\frac{8\,\ln \left ( -ax+1 \right ) }{125\,a}{x}^{{\frac{5}{2}}} \left ( -a \right ) ^{{\frac{7}{2}}}}-{\frac{4\,{\it polylog} \left ( 2,ax \right ) }{25\,a}{x}^{{\frac{5}{2}}} \left ( -a \right ) ^{{\frac{7}{2}}}}+{\frac{2\,{\it polylog} \left ( 3,ax \right ) }{5\,a}{x}^{{\frac{5}{2}}} \left ( -a \right ) ^{{\frac{7}{2}}}} \right ){x}^{-{\frac{3}{2}}} \left ( -a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x)^(3/2)/x^(3/2)/(-a)^(3/2)/a*(2/13125*x^(1/2)*(-a)^(7/2)*(168*a^2*x^2+280*a*x+840)/a^3+8/125*x^(1/2)*(-a)^
(7/2)/a^3/(a*x)^(1/2)*(ln(1-(a*x)^(1/2))-ln(1+(a*x)^(1/2)))-8/125*x^(5/2)*(-a)^(7/2)/a*ln(-a*x+1)-4/25*x^(5/2)
*(-a)^(7/2)*polylog(2,a*x)/a+2/5*x^(5/2)*(-a)^(7/2)/a*polylog(3,a*x))

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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(3,a*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.87358, size = 721, normalized size = 5.3 \begin{align*} \left [-\frac{2 \,{\left (150 \, \sqrt{d x} a^{2} d x^{2}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x + 1\right )}{a}, -\frac{\log \left (-a x + 1\right )}{x}\right ) - 375 \, \sqrt{d x} a^{2} d x^{2}{\rm polylog}\left (3, a x\right ) - 60 \, 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 ) + 4 \,{\left (15 \, a^{2} d x^{2} \log \left (-a x + 1\right ) - 6 \, a^{2} d x^{2} - 10 \, a d x - 30 \, d\right )} \sqrt{d x}\right )}}{1875 \, a^{2}}, -\frac{2 \,{\left (150 \, \sqrt{d x} a^{2} d x^{2}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x + 1\right )}{a}, -\frac{\log \left (-a x + 1\right )}{x}\right ) - 375 \, \sqrt{d x} a^{2} d x^{2}{\rm polylog}\left (3, a x\right ) - 120 \, d \sqrt{-\frac{d}{a}} \arctan \left (\frac{\sqrt{d x} a \sqrt{-\frac{d}{a}}}{d}\right ) + 4 \,{\left (15 \, a^{2} d x^{2} \log \left (-a x + 1\right ) - 6 \, a^{2} d x^{2} - 10 \, a d x - 30 \, d\right )} \sqrt{d x}\right )}}{1875 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-2/1875*(150*sqrt(d*x)*a^2*d*x^2*\%iint(a, x, -log(-a*x + 1)/a, -log(-a*x + 1)/x) - 375*sqrt(d*x)*a^2*d*x^2*po
lylog(3, a*x) - 60*d*sqrt(d/a)*log((a*d*x - 2*sqrt(d*x)*a*sqrt(d/a) + d)/(a*x - 1)) + 4*(15*a^2*d*x^2*log(-a*x
+ 1) - 6*a^2*d*x^2 - 10*a*d*x - 30*d)*sqrt(d*x))/a^2, -2/1875*(150*sqrt(d*x)*a^2*d*x^2*\%iint(a, x, -log(-a*x
+ 1)/a, -log(-a*x + 1)/x) - 375*sqrt(d*x)*a^2*d*x^2*polylog(3, a*x) - 120*d*sqrt(-d/a)*arctan(sqrt(d*x)*a*sqrt
(-d/a)/d) + 4*(15*a^2*d*x^2*log(-a*x + 1) - 6*a^2*d*x^2 - 10*a*d*x - 30*d)*sqrt(d*x))/a^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}_{3}(a x)\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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