3.1.87 \(\int \sqrt {d x} \text {PolyLog}(2,a x^q) \, dx\) [87]

Optimal. Leaf size=100 \[ \frac {8 a q^2 x^{1+q} \sqrt {d x} \, _2F_1\left (1,\frac {\frac {3}{2}+q}{q};\frac {1}{2} \left (4+\frac {3}{q}\right );a x^q\right )}{9 (3+2 q)}+\frac {4 q (d x)^{3/2} \log \left (1-a x^q\right )}{9 d}+\frac {2 (d x)^{3/2} \text {PolyLog}\left (2,a x^q\right )}{3 d} \]

[Out]

4/9*q*(d*x)^(3/2)*ln(1-a*x^q)/d+2/3*(d*x)^(3/2)*polylog(2,a*x^q)/d+8/9*a*q^2*x^(1+q)*hypergeom([1, (3/2+q)/q],
[2+3/2/q],a*x^q)*(d*x)^(1/2)/(3+2*q)

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Rubi [A]
time = 0.04, antiderivative size = 100, 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 = {6726, 2505, 20, 371} \begin {gather*} \frac {8 a q^2 \sqrt {d x} x^{q+1} \, _2F_1\left (1,\frac {q+\frac {3}{2}}{q};\frac {1}{2} \left (4+\frac {3}{q}\right );a x^q\right )}{9 (2 q+3)}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^q\right )}{3 d}+\frac {4 q (d x)^{3/2} \log \left (1-a x^q\right )}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]*PolyLog[2, a*x^q],x]

[Out]

(8*a*q^2*x^(1 + q)*Sqrt[d*x]*Hypergeometric2F1[1, (3/2 + q)/q, (4 + 3/q)/2, a*x^q])/(9*(3 + 2*q)) + (4*q*(d*x)
^(3/2)*Log[1 - a*x^q])/(9*d) + (2*(d*x)^(3/2)*PolyLog[2, a*x^q])/(3*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 \sqrt {d x} \text {Li}_2\left (a x^q\right ) \, dx &=\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^q\right )}{3 d}+\frac {1}{3} (2 q) \int \sqrt {d x} \log \left (1-a x^q\right ) \, dx\\ &=\frac {4 q (d x)^{3/2} \log \left (1-a x^q\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^q\right )}{3 d}+\frac {\left (4 a q^2\right ) \int \frac {x^{-1+q} (d x)^{3/2}}{1-a x^q} \, dx}{9 d}\\ &=\frac {4 q (d x)^{3/2} \log \left (1-a x^q\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^q\right )}{3 d}+\frac {\left (4 a q^2 \sqrt {d x}\right ) \int \frac {x^{\frac {1}{2}+q}}{1-a x^q} \, dx}{9 \sqrt {x}}\\ &=\frac {8 a q^2 x^{1+q} \sqrt {d x} \, _2F_1\left (1,\frac {\frac {3}{2}+q}{q};\frac {1}{2} \left (4+\frac {3}{q}\right );a x^q\right )}{9 (3+2 q)}+\frac {4 q (d x)^{3/2} \log \left (1-a x^q\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^q\right )}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 82, normalized size = 0.82 \begin {gather*} \frac {2 x \sqrt {d x} \left (4 a q^2 x^q \, _2F_1\left (1,\frac {\frac {3}{2}+q}{q};2+\frac {3}{2 q};a x^q\right )+(3+2 q) \left (2 q \log \left (1-a x^q\right )+3 \text {PolyLog}\left (2,a x^q\right )\right )\right )}{9 (3+2 q)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.52, size = 121, normalized size = 1.21

method	result	size
meijerg	\(-\frac {\sqrt {d x}\, \left (-a \right )^{-\frac {3}{2 q}} \left (-\frac {4 q^{2} x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2 q}} \ln \left (1-a \,x^{q}\right )}{9}-\frac {2 q \,x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{2 q}} \left (1+\frac {2 q}{3}\right ) \polylog \left (2, a \,x^{q}\right )}{3+2 q}-\frac {4 q^{2} x^{\frac {3}{2}+q} a \left (-a \right )^{\frac {3}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {3+2 q}{2 q}\right )}{9}\right )}{\sqrt {x}\, q}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x)^(1/2)/x^(1/2)*(-a)^(-3/2/q)/q*(-4/9*q^2*x^(3/2)*(-a)^(3/2/q)*ln(1-a*x^q)-2*q/(3+2*q)*x^(3/2)*(-a)^(3/2/
q)*(1+2/3*q)*polylog(2,a*x^q)-4/9*q^2*x^(3/2+q)*a*(-a)^(3/2/q)*LerchPhi(a*x^q,1,1/2*(3+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)^(1/2)*polylog(2,a*x^q),x, algorithm="maxima")

[Out]

8*sqrt(d)*q^3*integrate(1/9*sqrt(x)/((2*a^2*q - 3*a^2)*x^(2*q) - 2*(2*a*q - 3*a)*x^q + 2*q - 3), x) + 2/27*(9*
((2*a*sqrt(d)*q - 3*a*sqrt(d))*x*x^q - (2*sqrt(d)*q - 3*sqrt(d))*x)*sqrt(x)*dilog(a*x^q) + 6*((2*a*sqrt(d)*q^2
- 3*a*sqrt(d)*q)*x*x^q - (2*sqrt(d)*q^2 - 3*sqrt(d)*q)*x)*sqrt(x)*log(-a*x^q + 1) + 4*(2*sqrt(d)*q^3*x - (2*a
*sqrt(d)*q^3 - 3*a*sqrt(d)*q^2)*x*x^q)*sqrt(x))/((2*a*q - 3*a)*x^q - 2*q + 3)

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

[Out]

integral(sqrt(d*x)*dilog(a*x^q), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(sqrt(d*x)*dilog(a*x^q), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)*polylog(2, a*x^q),x)

[Out]

int((d*x)^(1/2)*polylog(2, a*x^q), x)

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