3.1.93 \(\int \frac {\text {PolyLog}(3,a x^q)}{\sqrt {d x}} \, dx\) [93]

Optimal. Leaf size=115 \[ -\frac {16 a q^3 x^q \sqrt {d x} \, _2F_1\left (1,\frac {\frac {1}{2}+q}{q};\frac {1}{2} \left (4+\frac {1}{q}\right );a x^q\right )}{d (1+2 q)}-\frac {8 q^2 \sqrt {d x} \log \left (1-a x^q\right )}{d}-\frac {4 q \sqrt {d x} \text {PolyLog}\left (2,a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {PolyLog}\left (3,a x^q\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 115, 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 q^3 \sqrt {d x} x^q \, _2F_1\left (1,\frac {q+\frac {1}{2}}{q};\frac {1}{2} \left (4+\frac {1}{q}\right );a x^q\right )}{d (2 q+1)}-\frac {4 q \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_3\left (a x^q\right )}{d}-\frac {8 q^2 \sqrt {d x} \log \left (1-a x^q\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^q]/Sqrt[d*x],x]

[Out]

(-16*a*q^3*x^q*Sqrt[d*x]*Hypergeometric2F1[1, (1/2 + q)/q, (4 + q^(-1))/2, a*x^q])/(d*(1 + 2*q)) - (8*q^2*Sqrt
[d*x]*Log[1 - a*x^q])/d - (4*q*Sqrt[d*x]*PolyLog[2, a*x^q])/d + (2*Sqrt[d*x]*PolyLog[3, a*x^q])/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 \frac {\text {Li}_3\left (a x^q\right )}{\sqrt {d x}} \, dx &=\frac {2 \sqrt {d x} \text {Li}_3\left (a x^q\right )}{d}-(2 q) \int \frac {\text {Li}_2\left (a x^q\right )}{\sqrt {d x}} \, dx\\ &=-\frac {4 q \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_3\left (a x^q\right )}{d}-\left (4 q^2\right ) \int \frac {\log \left (1-a x^q\right )}{\sqrt {d x}} \, dx\\ &=-\frac {8 q^2 \sqrt {d x} \log \left (1-a x^q\right )}{d}-\frac {4 q \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_3\left (a x^q\right )}{d}-\frac {\left (8 a q^3\right ) \int \frac {x^{-1+q} \sqrt {d x}}{1-a x^q} \, dx}{d}\\ &=-\frac {8 q^2 \sqrt {d x} \log \left (1-a x^q\right )}{d}-\frac {4 q \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_3\left (a x^q\right )}{d}-\frac {\left (8 a q^3 \sqrt {d x}\right ) \int \frac {x^{-\frac {1}{2}+q}}{1-a x^q} \, dx}{d \sqrt {x}}\\ &=-\frac {16 a q^3 x^q \sqrt {d x} \, _2F_1\left (1,\frac {\frac {1}{2}+q}{q};\frac {1}{2} \left (4+\frac {1}{q}\right );a x^q\right )}{d (1+2 q)}-\frac {8 q^2 \sqrt {d x} \log \left (1-a x^q\right )}{d}-\frac {4 q \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_3\left (a x^q\right )}{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 = 0.43 \begin {gather*} -\frac {x G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1-\frac {1}{2 q} \\ 1,0,0,0,-\frac {1}{2 q} \\\end {array}\right )}{q \sqrt {d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x^q]/Sqrt[d*x],x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.26, size = 133, normalized size = 1.16

method	result	size
meijerg	\(-\frac {\sqrt {x}\, \left (-a \right )^{-\frac {1}{2 q}} \left (8 q^{3} \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \ln \left (1-a \,x^{q}\right )+4 q^{2} \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \polylog \left (2, a \,x^{q}\right )-2 q \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \polylog \left (3, a \,x^{q}\right )+8 q^{3} x^{\frac {1}{2}+q} a \left (-a \right )^{\frac {1}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {1+2 q}{2 q}\right )\right )}{\sqrt {d x}\, q}\)	\(133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^q)/(d*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/(d*x)^(1/2)*x^(1/2)*(-a)^(-1/2/q)/q*(8*q^3*x^(1/2)*(-a)^(1/2/q)*ln(1-a*x^q)+4*q^2*x^(1/2)*(-a)^(1/2/q)*poly
log(2,a*x^q)-2*q*x^(1/2)*(-a)^(1/2/q)*polylog(3,a*x^q)+8*q^3*x^(1/2+q)*a*(-a)^(1/2/q)*LerchPhi(a*x^q,1,1/2*(1+
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(polylog(3,a*x^q)/(d*x)^(1/2),x, algorithm="maxima")

[Out]

-16*q^4*integrate(1/((a^2*sqrt(d)*(2*q - 1)*x^(2*q) - 2*a*sqrt(d)*(2*q - 1)*x^q + sqrt(d)*(2*q - 1))*sqrt(x)),
x) - 2*(2*((2*q^2 - q)*a*x*x^q - (2*q^2 - q)*x)*dilog(a*x^q)/sqrt(x) + 4*((2*q^3 - q^2)*a*x*x^q - (2*q^3 - q^
2)*x)*log(-a*x^q + 1)/sqrt(x) - (a*(2*q - 1)*x*x^q - (2*q - 1)*x)*polylog(3, a*x^q)/sqrt(x) + 8*(2*q^4*x - (2*
q^4 - q^3)*a*x*x^q)/sqrt(x))/(a*sqrt(d)*(2*q - 1)*x^q - sqrt(d)*(2*q - 1))

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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(polylog(3,a*x^q)/(d*x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**q)/(d*x)**(1/2),x)

[Out]

Integral(polylog(3, a*x**q)/sqrt(d*x), 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(polylog(3,a*x^q)/(d*x)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^q)/(d*x)^(1/2),x)

[Out]

int(polylog(3, a*x^q)/(d*x)^(1/2), x)

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