3.88 \(\int \frac {\text {Li}_2(a x^q)}{\sqrt {d x}} \, dx\)

Optimal. Leaf size=93 \[ \frac {8 a q^2 \sqrt {d x} x^q \, _2F_1\left (1,\frac {q+\frac {1}{2}}{q};\frac {1}{2} \left (4+\frac {1}{q}\right );a x^q\right )}{d (2 q+1)}+\frac {2 \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {4 q \sqrt {d x} \log \left (1-a x^q\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6591, 2455, 20, 364} \[ \frac {2 \sqrt {d x} \text {PolyLog}\left (2,a x^q\right )}{d}+\frac {8 a q^2 \sqrt {d x} x^q \, _2F_1\left (1,\frac {q+\frac {1}{2}}{q};\frac {1}{2} \left (4+\frac {1}{q}\right );a x^q\right )}{d (2 q+1)}+\frac {4 q \sqrt {d x} \log \left (1-a x^q\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^q]/Sqrt[d*x],x]

[Out]

(8*a*q^2*x^q*Sqrt[d*x]*Hypergeometric2F1[1, (1/2 + q)/q, (4 + q^(-1))/2, a*x^q])/(d*(1 + 2*q)) + (4*q*Sqrt[d*x
]*Log[1 - a*x^q])/d + (2*Sqrt[d*x]*PolyLog[2, 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 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}_2\left (a x^q\right )}{\sqrt {d x}} \, dx &=\frac {2 \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+(2 q) \int \frac {\log \left (1-a x^q\right )}{\sqrt {d x}} \, dx\\ &=\frac {4 q \sqrt {d x} \log \left (1-a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {\left (4 a q^2\right ) \int \frac {x^{-1+q} \sqrt {d x}}{1-a x^q} \, dx}{d}\\ &=\frac {4 q \sqrt {d x} \log \left (1-a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}+\frac {\left (4 a q^2 \sqrt {d x}\right ) \int \frac {x^{-\frac {1}{2}+q}}{1-a x^q} \, dx}{d \sqrt {x}}\\ &=\frac {8 a q^2 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 {4 q \sqrt {d x} \log \left (1-a x^q\right )}{d}+\frac {2 \sqrt {d x} \text {Li}_2\left (a x^q\right )}{d}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 48, normalized size = 0.52 \[ -\frac {x G_{4,4}^{1,4}\left (-a x^q|\begin {array}{c} 1,1,1,1-\frac {1}{2 q} \\ 1,0,0,-\frac {1}{2 q} \\\end {array}\right )}{q \sqrt {d x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[2, a*x^q]/Sqrt[d*x],x]

[Out]

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

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x} {\rm Li}_2\left (a x^{q}\right )}{d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x^{q}\right )}{\sqrt {d x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^q)/sqrt(d*x), x)

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maple [C]  time = 0.13, size = 109, normalized size = 1.17 \[ -\frac {\sqrt {x}\, \left (-a \right )^{-\frac {1}{2 q}} \left (-4 q^{2} \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \ln \left (1-a \,x^{q}\right )-2 q \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \polylog \left (2, a \,x^{q}\right )-4 q^{2} x^{\frac {1}{2}+q} a \left (-a \right )^{\frac {1}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {1+2 q}{2 q}\right )\right )}{\sqrt {d x}\, q} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^q)/(d*x)^(1/2),x)

[Out]

-1/(d*x)^(1/2)*x^(1/2)*(-a)^(-1/2/q)/q*(-4*q^2*x^(1/2)*(-a)^(1/2/q)*ln(1-a*x^q)-2*q*x^(1/2)*(-a)^(1/2/q)*polyl
og(2,a*x^q)-4*q^2*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 \[ 8 \, q^{3} \int \frac {1}{{\left ({\left (2 \, a^{2} \sqrt {d} q - a^{2} \sqrt {d}\right )} x^{2 \, q} - 2 \, {\left (2 \, a \sqrt {d} q - a \sqrt {d}\right )} x^{q} + 2 \, \sqrt {d} q - \sqrt {d}\right )} \sqrt {x}}\,{d x} - \frac {2 \, {\left (\frac {{\left ({\left (2 \, a \sqrt {d} q - a \sqrt {d}\right )} x x^{q} - {\left (2 \, \sqrt {d} q - \sqrt {d}\right )} x\right )} {\rm Li}_2\left (a x^{q}\right )}{\sqrt {x}} + \frac {2 \, {\left ({\left (2 \, a \sqrt {d} q^{2} - a \sqrt {d} q\right )} x x^{q} - {\left (2 \, \sqrt {d} q^{2} - \sqrt {d} q\right )} x\right )} \log \left (-a x^{q} + 1\right )}{\sqrt {x}} + \frac {4 \, {\left (2 \, \sqrt {d} q^{3} x - {\left (2 \, a \sqrt {d} q^{3} - a \sqrt {d} q^{2}\right )} x x^{q}\right )}}{\sqrt {x}}\right )}}{2 \, d q - {\left (2 \, a d q - a d\right )} x^{q} - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^q)/(d*x)^(1/2),x)

[Out]

int(polylog(2, a*x^q)/(d*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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