3.1.91 \(\int (d x)^{3/2} \text {PolyLog}(3,a x^q) \, dx\) [91]

Optimal. Leaf size=125 \[ -\frac {16 a d q^3 x^{2+q} \sqrt {d x} \, _2F_1\left (1,\frac {\frac {5}{2}+q}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{125 (5+2 q)}-\frac {8 q^2 (d x)^{5/2} \log \left (1-a x^q\right )}{125 d}-\frac {4 q (d x)^{5/2} \text {PolyLog}\left (2,a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {PolyLog}\left (3,a x^q\right )}{5 d} \]

[Out]

-8/125*q^2*(d*x)^(5/2)*ln(1-a*x^q)/d-4/25*q*(d*x)^(5/2)*polylog(2,a*x^q)/d+2/5*(d*x)^(5/2)*polylog(3,a*x^q)/d-
16/125*a*d*q^3*x^(2+q)*hypergeom([1, (5/2+q)/q],[2+5/2/q],a*x^q)*(d*x)^(1/2)/(5+2*q)

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Rubi [A]
time = 0.05, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 = {6726, 2505, 20, 371} \begin {gather*} -\frac {16 a d q^3 \sqrt {d x} x^{q+2} \, _2F_1\left (1,\frac {q+\frac {5}{2}}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{125 (2 q+5)}-\frac {4 q (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^q\right )}{5 d}-\frac {8 q^2 (d x)^{5/2} \log \left (1-a x^q\right )}{125 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[3, a*x^q],x]

[Out]

(-16*a*d*q^3*x^(2 + q)*Sqrt[d*x]*Hypergeometric2F1[1, (5/2 + q)/q, (4 + 5/q)/2, a*x^q])/(125*(5 + 2*q)) - (8*q
^2*(d*x)^(5/2)*Log[1 - a*x^q])/(125*d) - (4*q*(d*x)^(5/2)*PolyLog[2, a*x^q])/(25*d) + (2*(d*x)^(5/2)*PolyLog[3
, a*x^q])/(5*d)

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2505

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 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 (d x)^{3/2} \text {Li}_3\left (a x^q\right ) \, dx &=\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^q\right )}{5 d}-\frac {1}{5} (2 q) \int (d x)^{3/2} \text {Li}_2\left (a x^q\right ) \, dx\\ &=-\frac {4 q (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^q\right )}{5 d}-\frac {1}{25} \left (4 q^2\right ) \int (d x)^{3/2} \log \left (1-a x^q\right ) \, dx\\ &=-\frac {8 q^2 (d x)^{5/2} \log \left (1-a x^q\right )}{125 d}-\frac {4 q (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^q\right )}{5 d}-\frac {\left (8 a q^3\right ) \int \frac {x^{-1+q} (d x)^{5/2}}{1-a x^q} \, dx}{125 d}\\ &=-\frac {8 q^2 (d x)^{5/2} \log \left (1-a x^q\right )}{125 d}-\frac {4 q (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^q\right )}{5 d}-\frac {\left (8 a d q^3 \sqrt {d x}\right ) \int \frac {x^{\frac {3}{2}+q}}{1-a x^q} \, dx}{125 \sqrt {x}}\\ &=-\frac {16 a d q^3 x^{2+q} \sqrt {d x} \, _2F_1\left (1,\frac {\frac {5}{2}+q}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{125 (5+2 q)}-\frac {8 q^2 (d x)^{5/2} \log \left (1-a x^q\right )}{125 d}-\frac {4 q (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^q\right )}{5 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
time = 0.02, size = 50, normalized size = 0.40 \begin {gather*} -\frac {x (d x)^{3/2} G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1-\frac {5}{2 q} \\ 1,0,0,0,-\frac {5}{2 q} \\\end {array}\right )}{q} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.25, size = 145, normalized size = 1.16

method	result	size
meijerg	\(-\frac {\left (d x \right )^{\frac {3}{2}} \left (-a \right )^{-\frac {5}{2 q}} \left (\frac {8 q^{3} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \ln \left (1-a \,x^{q}\right )}{125}+\frac {4 q^{2} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \polylog \left (2, a \,x^{q}\right )}{25}-\frac {2 q \,x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \left (1+\frac {2 q}{5}\right ) \polylog \left (3, a \,x^{q}\right )}{5+2 q}+\frac {8 q^{3} x^{\frac {5}{2}+q} a \left (-a \right )^{\frac {5}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {5+2 q}{2 q}\right )}{125}\right )}{x^{\frac {3}{2}} q}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*polylog(3,a*x^q),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*d^(3/2)*q^4*integrate(1/125*x^(3/2)/(a^2*(2*q - 5)*x^(2*q) - 2*a*(2*q - 5)*x^q + 2*q - 5), x) - 2/625*(50*
((2*q^2 - 5*q)*a*d^(3/2)*x*x^q - (2*q^2 - 5*q)*d^(3/2)*x)*x^(3/2)*dilog(a*x^q) + 20*((2*q^3 - 5*q^2)*a*d^(3/2)
*x*x^q - (2*q^3 - 5*q^2)*d^(3/2)*x)*x^(3/2)*log(-a*x^q + 1) - 125*(a*d^(3/2)*(2*q - 5)*x*x^q - d^(3/2)*(2*q -
5)*x)*x^(3/2)*polylog(3, a*x^q) + 8*(2*d^(3/2)*q^4*x - (2*q^4 - 5*q^3)*a*d^(3/2)*x*x^q)*x^(3/2))/(a*(2*q - 5)*
x^q - 2*q + 5)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{\frac {3}{2}} \operatorname {Li}_{3}\left (a x^{q}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^{3/2}\,\mathrm {polylog}\left (3,a\,x^q\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*polylog(3, a*x^q),x)

[Out]

int((d*x)^(3/2)*polylog(3, a*x^q), x)

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