3.65 \(\int (d x)^{5/2} \text{PolyLog}(3,a x) \, dx\)

Optimal. Leaf size=153 \[ -\frac{4 (d x)^{7/2} \text{PolyLog}(2,a x)}{49 d}+\frac{2 (d x)^{7/2} \text{PolyLog}(3,a x)}{7 d}+\frac{16 d^2 \sqrt{d x}}{343 a^3}-\frac{16 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{343 a^{7/2}}+\frac{16 d (d x)^{3/2}}{1029 a^2}+\frac{16 (d x)^{5/2}}{1715 a}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}+\frac{16 (d x)^{7/2}}{2401 d} \]

[Out]

(16*d^2*Sqrt[d*x])/(343*a^3) + (16*d*(d*x)^(3/2))/(1029*a^2) + (16*(d*x)^(5/2))/(1715*a) + (16*(d*x)^(7/2))/(2
401*d) - (16*d^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[d*x])/Sqrt[d]])/(343*a^(7/2)) - (8*(d*x)^(7/2)*Log[1 - a*x])/(343*d
) - (4*(d*x)^(7/2)*PolyLog[2, a*x])/(49*d) + (2*(d*x)^(7/2)*PolyLog[3, a*x])/(7*d)

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Rubi [A]  time = 0.0982584, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{4 (d x)^{7/2} \text{PolyLog}(2,a x)}{49 d}+\frac{2 (d x)^{7/2} \text{PolyLog}(3,a x)}{7 d}+\frac{16 d^2 \sqrt{d x}}{343 a^3}-\frac{16 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{343 a^{7/2}}+\frac{16 d (d x)^{3/2}}{1029 a^2}+\frac{16 (d x)^{5/2}}{1715 a}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}+\frac{16 (d x)^{7/2}}{2401 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(5/2)*PolyLog[3, a*x],x]

[Out]

(16*d^2*Sqrt[d*x])/(343*a^3) + (16*d*(d*x)^(3/2))/(1029*a^2) + (16*(d*x)^(5/2))/(1715*a) + (16*(d*x)^(7/2))/(2
401*d) - (16*d^(5/2)*ArcTanh[(Sqrt[a]*Sqrt[d*x])/Sqrt[d]])/(343*a^(7/2)) - (8*(d*x)^(7/2)*Log[1 - a*x])/(343*d
) - (4*(d*x)^(7/2)*PolyLog[2, a*x])/(49*d) + (2*(d*x)^(7/2)*PolyLog[3, a*x])/(7*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)^{5/2} \text{Li}_3(a x) \, dx &=\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{2}{7} \int (d x)^{5/2} \text{Li}_2(a x) \, dx\\ &=-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{4}{49} \int (d x)^{5/2} \log (1-a x) \, dx\\ &=-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{(8 a) \int \frac{(d x)^{7/2}}{1-a x} \, dx}{343 d}\\ &=\frac{16 (d x)^{7/2}}{2401 d}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{8}{343} \int \frac{(d x)^{5/2}}{1-a x} \, dx\\ &=\frac{16 (d x)^{5/2}}{1715 a}+\frac{16 (d x)^{7/2}}{2401 d}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{(8 d) \int \frac{(d x)^{3/2}}{1-a x} \, dx}{343 a}\\ &=\frac{16 d (d x)^{3/2}}{1029 a^2}+\frac{16 (d x)^{5/2}}{1715 a}+\frac{16 (d x)^{7/2}}{2401 d}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{\left (8 d^2\right ) \int \frac{\sqrt{d x}}{1-a x} \, dx}{343 a^2}\\ &=\frac{16 d^2 \sqrt{d x}}{343 a^3}+\frac{16 d (d x)^{3/2}}{1029 a^2}+\frac{16 (d x)^{5/2}}{1715 a}+\frac{16 (d x)^{7/2}}{2401 d}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{\left (8 d^3\right ) \int \frac{1}{\sqrt{d x} (1-a x)} \, dx}{343 a^3}\\ &=\frac{16 d^2 \sqrt{d x}}{343 a^3}+\frac{16 d (d x)^{3/2}}{1029 a^2}+\frac{16 (d x)^{5/2}}{1715 a}+\frac{16 (d x)^{7/2}}{2401 d}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}-\frac{\left (16 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{a x^2}{d}} \, dx,x,\sqrt{d x}\right )}{343 a^3}\\ &=\frac{16 d^2 \sqrt{d x}}{343 a^3}+\frac{16 d (d x)^{3/2}}{1029 a^2}+\frac{16 (d x)^{5/2}}{1715 a}+\frac{16 (d x)^{7/2}}{2401 d}-\frac{16 d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d x}}{\sqrt{d}}\right )}{343 a^{7/2}}-\frac{8 (d x)^{7/2} \log (1-a x)}{343 d}-\frac{4 (d x)^{7/2} \text{Li}_2(a x)}{49 d}+\frac{2 (d x)^{7/2} \text{Li}_3(a x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.260023, size = 98, normalized size = 0.64 \[ \frac{2 (d x)^{5/2} \left (-1470 x^3 \text{PolyLog}(2,a x)+5145 x^3 \text{PolyLog}(3,a x)+\frac{8 \left (15 a^3 x^3+21 a^2 x^2+35 a x+105\right )}{a^3}-\frac{840 \tanh ^{-1}\left (\sqrt{a} \sqrt{x}\right )}{a^{7/2} \sqrt{x}}-420 x^3 \log (1-a x)\right )}{36015 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(5/2)*PolyLog[3, a*x],x]

[Out]

(2*(d*x)^(5/2)*((8*(105 + 35*a*x + 21*a^2*x^2 + 15*a^3*x^3))/a^3 - (840*ArcTanh[Sqrt[a]*Sqrt[x]])/(a^(7/2)*Sqr
t[x]) - 420*x^3*Log[1 - a*x] - 1470*x^3*PolyLog[2, a*x] + 5145*x^3*PolyLog[3, a*x]))/(36015*x^2)

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Maple [A]  time = 0.182, size = 149, normalized size = 1. \begin{align*}{\frac{1}{a} \left ( dx \right ) ^{{\frac{5}{2}}} \left ({\frac{720\,{x}^{3}{a}^{3}+1008\,{a}^{2}{x}^{2}+1680\,ax+5040}{108045\,{a}^{4}}\sqrt{x} \left ( -a \right ) ^{{\frac{9}{2}}}}+{\frac{8}{343\,{a}^{4}}\sqrt{x} \left ( -a \right ) ^{{\frac{9}{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 ) }{343\,a}{x}^{{\frac{7}{2}}} \left ( -a \right ) ^{{\frac{9}{2}}}}-{\frac{4\,{\it polylog} \left ( 2,ax \right ) }{49\,a}{x}^{{\frac{7}{2}}} \left ( -a \right ) ^{{\frac{9}{2}}}}+{\frac{2\,{\it polylog} \left ( 3,ax \right ) }{7\,a}{x}^{{\frac{7}{2}}} \left ( -a \right ) ^{{\frac{9}{2}}}} \right ){x}^{-{\frac{5}{2}}} \left ( -a \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(5/2)*polylog(3,a*x),x)

[Out]

(d*x)^(5/2)/x^(5/2)/(-a)^(5/2)/a*(2/108045*x^(1/2)*(-a)^(9/2)*(360*a^3*x^3+504*a^2*x^2+840*a*x+2520)/a^4+8/343
*x^(1/2)*(-a)^(9/2)/a^4/(a*x)^(1/2)*(ln(1-(a*x)^(1/2))-ln(1+(a*x)^(1/2)))-8/343*x^(7/2)*(-a)^(9/2)/a*ln(-a*x+1
)-4/49*x^(7/2)*(-a)^(9/2)*polylog(2,a*x)/a+2/7*x^(7/2)*(-a)^(9/2)/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)^(5/2)*polylog(3,a*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.97483, size = 822, normalized size = 5.37 \begin{align*} \left [-\frac{2 \,{\left (1470 \, \sqrt{d x} a^{3} d^{2} x^{3}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x + 1\right )}{a}, -\frac{\log \left (-a x + 1\right )}{x}\right ) - 5145 \, \sqrt{d x} a^{3} d^{2} x^{3}{\rm polylog}\left (3, a x\right ) - 420 \, d^{2} \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 (105 \, a^{3} d^{2} x^{3} \log \left (-a x + 1\right ) - 30 \, a^{3} d^{2} x^{3} - 42 \, a^{2} d^{2} x^{2} - 70 \, a d^{2} x - 210 \, d^{2}\right )} \sqrt{d x}\right )}}{36015 \, a^{3}}, -\frac{2 \,{\left (1470 \, \sqrt{d x} a^{3} d^{2} x^{3}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x + 1\right )}{a}, -\frac{\log \left (-a x + 1\right )}{x}\right ) - 5145 \, \sqrt{d x} a^{3} d^{2} x^{3}{\rm polylog}\left (3, a x\right ) - 840 \, d^{2} \sqrt{-\frac{d}{a}} \arctan \left (\frac{\sqrt{d x} a \sqrt{-\frac{d}{a}}}{d}\right ) + 4 \,{\left (105 \, a^{3} d^{2} x^{3} \log \left (-a x + 1\right ) - 30 \, a^{3} d^{2} x^{3} - 42 \, a^{2} d^{2} x^{2} - 70 \, a d^{2} x - 210 \, d^{2}\right )} \sqrt{d x}\right )}}{36015 \, a^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*polylog(3,a*x),x, algorithm="fricas")

[Out]

[-2/36015*(1470*sqrt(d*x)*a^3*d^2*x^3*\%iint(a, x, -log(-a*x + 1)/a, -log(-a*x + 1)/x) - 5145*sqrt(d*x)*a^3*d^2
*x^3*polylog(3, a*x) - 420*d^2*sqrt(d/a)*log((a*d*x - 2*sqrt(d*x)*a*sqrt(d/a) + d)/(a*x - 1)) + 4*(105*a^3*d^2
*x^3*log(-a*x + 1) - 30*a^3*d^2*x^3 - 42*a^2*d^2*x^2 - 70*a*d^2*x - 210*d^2)*sqrt(d*x))/a^3, -2/36015*(1470*sq
rt(d*x)*a^3*d^2*x^3*\%iint(a, x, -log(-a*x + 1)/a, -log(-a*x + 1)/x) - 5145*sqrt(d*x)*a^3*d^2*x^3*polylog(3, a*
x) - 840*d^2*sqrt(-d/a)*arctan(sqrt(d*x)*a*sqrt(-d/a)/d) + 4*(105*a^3*d^2*x^3*log(-a*x + 1) - 30*a^3*d^2*x^3 -
42*a^2*d^2*x^2 - 70*a*d^2*x - 210*d^2)*sqrt(d*x))/a^3]

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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)**(5/2)*polylog(3,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{5}{2}}{\rm Li}_{3}(a x)\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*polylog(3,a*x),x, algorithm="giac")

[Out]

integrate((d*x)^(5/2)*polylog(3, a*x), x)