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

Optimal. Leaf size=129 \[ -\frac{16 a q^3 x^{q-1} \text{Hypergeometric2F1}\left (1,\frac{1}{2} \left (2-\frac{3}{q}\right ),\frac{1}{2} \left (4-\frac{3}{q}\right ),a x^q\right )}{27 d^2 (3-2 q) \sqrt{d x}}-\frac{4 q \text{PolyLog}\left (2,a x^q\right )}{9 d (d x)^{3/2}}-\frac{2 \text{PolyLog}\left (3,a x^q\right )}{3 d (d x)^{3/2}}+\frac{8 q^2 \log \left (1-a x^q\right )}{27 d (d x)^{3/2}} \]

[Out]

(-16*a*q^3*x^(-1 + q)*Hypergeometric2F1[1, (2 - 3/q)/2, (4 - 3/q)/2, a*x^q])/(27*d^2*(3 - 2*q)*Sqrt[d*x]) + (8
*q^2*Log[1 - a*x^q])/(27*d*(d*x)^(3/2)) - (4*q*PolyLog[2, a*x^q])/(9*d*(d*x)^(3/2)) - (2*PolyLog[3, a*x^q])/(3
*d*(d*x)^(3/2))

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Rubi [A]  time = 0.0749586, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {6591, 2455, 20, 364} \[ -\frac{4 q \text{PolyLog}\left (2,a x^q\right )}{9 d (d x)^{3/2}}-\frac{2 \text{PolyLog}\left (3,a x^q\right )}{3 d (d x)^{3/2}}-\frac{16 a q^3 x^{q-1} \, _2F_1\left (1,\frac{1}{2} \left (2-\frac{3}{q}\right );\frac{1}{2} \left (4-\frac{3}{q}\right );a x^q\right )}{27 d^2 (3-2 q) \sqrt{d x}}+\frac{8 q^2 \log \left (1-a x^q\right )}{27 d (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^q]/(d*x)^(5/2),x]

[Out]

(-16*a*q^3*x^(-1 + q)*Hypergeometric2F1[1, (2 - 3/q)/2, (4 - 3/q)/2, a*x^q])/(27*d^2*(3 - 2*q)*Sqrt[d*x]) + (8
*q^2*Log[1 - a*x^q])/(27*d*(d*x)^(3/2)) - (4*q*PolyLog[2, a*x^q])/(9*d*(d*x)^(3/2)) - (2*PolyLog[3, a*x^q])/(3
*d*(d*x)^(3/2))

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 2455

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\text{Li}_3\left (a x^q\right )}{(d x)^{5/2}} \, dx &=-\frac{2 \text{Li}_3\left (a x^q\right )}{3 d (d x)^{3/2}}+\frac{1}{3} (2 q) \int \frac{\text{Li}_2\left (a x^q\right )}{(d x)^{5/2}} \, dx\\ &=-\frac{4 q \text{Li}_2\left (a x^q\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^q\right )}{3 d (d x)^{3/2}}-\frac{1}{9} \left (4 q^2\right ) \int \frac{\log \left (1-a x^q\right )}{(d x)^{5/2}} \, dx\\ &=\frac{8 q^2 \log \left (1-a x^q\right )}{27 d (d x)^{3/2}}-\frac{4 q \text{Li}_2\left (a x^q\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^q\right )}{3 d (d x)^{3/2}}+\frac{\left (8 a q^3\right ) \int \frac{x^{-1+q}}{(d x)^{3/2} \left (1-a x^q\right )} \, dx}{27 d}\\ &=\frac{8 q^2 \log \left (1-a x^q\right )}{27 d (d x)^{3/2}}-\frac{4 q \text{Li}_2\left (a x^q\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^q\right )}{3 d (d x)^{3/2}}+\frac{\left (8 a q^3 \sqrt{x}\right ) \int \frac{x^{-\frac{5}{2}+q}}{1-a x^q} \, dx}{27 d^2 \sqrt{d x}}\\ &=-\frac{16 a q^3 x^{-1+q} \, _2F_1\left (1,\frac{1}{2} \left (2-\frac{3}{q}\right );\frac{1}{2} \left (4-\frac{3}{q}\right );a x^q\right )}{27 d^2 (3-2 q) \sqrt{d x}}+\frac{8 q^2 \log \left (1-a x^q\right )}{27 d (d x)^{3/2}}-\frac{4 q \text{Li}_2\left (a x^q\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^q\right )}{3 d (d x)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0240634, size = 50, normalized size = 0.39 \[ -\frac{x G_{5,5}^{1,5}\left (-a x^q|\begin{array}{c} 1,1,1,1,1+\frac{3}{2 q} \\ 1,0,0,0,\frac{3}{2 q} \\\end{array}\right )}{q (d x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((x*MeijerG[{{1, 1, 1, 1, 1 + 3/(2*q)}, {}}, {{1}, {0, 0, 0, 3/(2*q)}}, -(a*x^q)])/(q*(d*x)^(5/2)))

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Maple [C]  time = 0.381, size = 145, normalized size = 1.1 \begin{align*} -{\frac{1}{q}{x}^{{\frac{5}{2}}} \left ( -a \right ) ^{{\frac{3}{2\,q}}} \left ( -{\frac{8\,{q}^{3}\ln \left ( 1-a{x}^{q} \right ) }{27} \left ( -a \right ) ^{-{\frac{3}{2\,q}}}{x}^{-{\frac{3}{2}}}}+{\frac{4\,{q}^{2}{\it polylog} \left ( 2,a{x}^{q} \right ) }{9} \left ( -a \right ) ^{-{\frac{3}{2\,q}}}{x}^{-{\frac{3}{2}}}}-2\,{\frac{q \left ( 1-2/3\,q \right ){\it polylog} \left ( 3,a{x}^{q} \right ) }{ \left ( -3+2\,q \right ){x}^{3/2}} \left ( -a \right ) ^{-3/2\,{q}^{-1}}}-{\frac{8\,{q}^{3}a}{27}{x}^{q-{\frac{3}{2}}} \left ( -a \right ) ^{-{\frac{3}{2\,q}}}{\it LerchPhi} \left ( a{x}^{q},1,{\frac{-3+2\,q}{2\,q}} \right ) } \right ) \left ( dx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*x)^(5/2)*x^(5/2)*(-a)^(3/2/q)/q*(-8/27*q^3/x^(3/2)*(-a)^(-3/2/q)*ln(1-a*x^q)+4/9*q^2/x^(3/2)*(-a)^(-3/2/
q)*polylog(2,a*x^q)-2*q/(-3+2*q)/x^(3/2)*(-a)^(-3/2/q)*(1-2/3*q)*polylog(3,a*x^q)-8/27*q^3*x^(q-3/2)*a*(-a)^(-
3/2/q)*LerchPhi(a*x^q,1,1/2*(-3+2*q)/q))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 16 \, q^{4} \int \frac{1}{27 \,{\left (a^{2} d^{\frac{5}{2}}{\left (2 \, q + 3\right )} x^{2 \, q} - 2 \, a d^{\frac{5}{2}}{\left (2 \, q + 3\right )} x^{q} + d^{\frac{5}{2}}{\left (2 \, q + 3\right )}\right )} x^{\frac{5}{2}}}\,{d x} - \frac{2 \,{\left (\frac{18 \,{\left ({\left (2 \, q^{2} + 3 \, q\right )} a x x^{q} -{\left (2 \, q^{2} + 3 \, q\right )} x\right )}{\rm Li}_2\left (a x^{q}\right )}{x^{\frac{5}{2}}} - \frac{12 \,{\left ({\left (2 \, q^{3} + 3 \, q^{2}\right )} a x x^{q} -{\left (2 \, q^{3} + 3 \, q^{2}\right )} x\right )} \log \left (-a x^{q} + 1\right )}{x^{\frac{5}{2}}} + \frac{27 \,{\left (a{\left (2 \, q + 3\right )} x x^{q} -{\left (2 \, q + 3\right )} x\right )}{\rm Li}_{3}(a x^{q})}{x^{\frac{5}{2}}} + \frac{8 \,{\left (2 \, q^{4} x -{\left (2 \, q^{4} + 3 \, q^{3}\right )} a x x^{q}\right )}}{x^{\frac{5}{2}}}\right )}}{81 \,{\left (a d^{\frac{5}{2}}{\left (2 \, q + 3\right )} x^{q} - d^{\frac{5}{2}}{\left (2 \, q + 3\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*q^4*integrate(1/27/((a^2*d^(5/2)*(2*q + 3)*x^(2*q) - 2*a*d^(5/2)*(2*q + 3)*x^q + d^(5/2)*(2*q + 3))*x^(5/2)
), x) - 2/81*(18*((2*q^2 + 3*q)*a*x*x^q - (2*q^2 + 3*q)*x)*dilog(a*x^q)/x^(5/2) - 12*((2*q^3 + 3*q^2)*a*x*x^q
- (2*q^3 + 3*q^2)*x)*log(-a*x^q + 1)/x^(5/2) + 27*(a*(2*q + 3)*x*x^q - (2*q + 3)*x)*polylog(3, a*x^q)/x^(5/2)
+ 8*(2*q^4*x - (2*q^4 + 3*q^3)*a*x*x^q)/x^(5/2))/(a*d^(5/2)*(2*q + 3)*x^q - d^(5/2)*(2*q + 3))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d x}{\rm polylog}\left (3, a x^{q}\right )}{d^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*x)*polylog(3, a*x^q)/(d^3*x^3), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_{3}(a x^{q})}{\left (d x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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