3.94 \(\int \frac {\text {Li}_3(a x^q)}{(d x)^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {16 a q^3 x^q \, _2F_1\left (1,\frac {1}{2} \left (2-\frac {1}{q}\right );\frac {1}{2} \left (4-\frac {1}{q}\right );a x^q\right )}{d (1-2 q) \sqrt {d x}}-\frac {4 q \text {Li}_2\left (a x^q\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^q\right )}{d \sqrt {d x}}+\frac {8 q^2 \log \left (1-a x^q\right )}{d \sqrt {d x}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 119, 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 = {6591, 2455, 20, 364} \[ -\frac {4 q \text {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}-\frac {2 \text {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}-\frac {16 a q^3 x^q \, _2F_1\left (1,\frac {1}{2} \left (2-\frac {1}{q}\right );\frac {1}{2} \left (4-\frac {1}{q}\right );a x^q\right )}{d (1-2 q) \sqrt {d x}}+\frac {8 q^2 \log \left (1-a x^q\right )}{d \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^q]/(d*x)^(3/2),x]

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 50, normalized size = 0.42 \[ -\frac {x G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1+\frac {1}{2 q} \\ 1,0,0,0,\frac {1}{2 q} \\\end {array}\right )}{q (d x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x} {\rm polylog}\left (3, a x^{q}\right )}{d^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_{3}(a x^{q})}{\left (d x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.28, size = 145, normalized size = 1.22 \[ -\frac {x^{\frac {3}{2}} \left (-a \right )^{\frac {1}{2 q}} \left (-\frac {8 q^{3} \left (-a \right )^{-\frac {1}{2 q}} \ln \left (1-a \,x^{q}\right )}{\sqrt {x}}+\frac {4 q^{2} \left (-a \right )^{-\frac {1}{2 q}} \polylog \left (2, a \,x^{q}\right )}{\sqrt {x}}-\frac {2 q \left (-a \right )^{-\frac {1}{2 q}} \left (1-2 q \right ) \polylog \left (3, a \,x^{q}\right )}{\left (2 q -1\right ) \sqrt {x}}-8 q^{3} x^{q -\frac {1}{2}} a \left (-a \right )^{-\frac {1}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {2 q -1}{2 q}\right )\right )}{\left (d x \right )^{\frac {3}{2}} q} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(d*x)^(3/2)*x^(3/2)*(-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)*po
lylog(2,a*x^q)-2*q/(2*q-1)/x^(1/2)*(-a)^(-1/2/q)*(1-2*q)*polylog(3,a*x^q)-8*q^3*x^(q-1/2)*a*(-a)^(-1/2/q)*Lerc
hPhi(a*x^q,1,1/2*(2*q-1)/q))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 16 \, q^{4} \int \frac {1}{{\left (a^{2} d^{\frac {3}{2}} {\left (2 \, q + 1\right )} x^{2 \, q} - 2 \, a d^{\frac {3}{2}} {\left (2 \, q + 1\right )} x^{q} + d^{\frac {3}{2}} {\left (2 \, q + 1\right )}\right )} x^{\frac {3}{2}}}\,{d x} - \frac {2 \, {\left (\frac {2 \, {\left ({\left (2 \, q^{2} + q\right )} a x x^{q} - {\left (2 \, q^{2} + q\right )} x\right )} {\rm Li}_2\left (a x^{q}\right )}{x^{\frac {3}{2}}} - \frac {4 \, {\left ({\left (2 \, q^{3} + q^{2}\right )} a x x^{q} - {\left (2 \, q^{3} + q^{2}\right )} x\right )} \log \left (-a x^{q} + 1\right )}{x^{\frac {3}{2}}} + \frac {{\left (a {\left (2 \, q + 1\right )} x x^{q} - {\left (2 \, q + 1\right )} x\right )} {\rm Li}_{3}(a x^{q})}{x^{\frac {3}{2}}} + \frac {8 \, {\left (2 \, q^{4} x - {\left (2 \, q^{4} + q^{3}\right )} a x x^{q}\right )}}{x^{\frac {3}{2}}}\right )}}{a d^{\frac {3}{2}} {\left (2 \, q + 1\right )} x^{q} - d^{\frac {3}{2}} {\left (2 \, q + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^q)/(d*x)^(3/2),x)

[Out]

int(polylog(3, a*x^q)/(d*x)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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