3.86 \(\int (d x)^{3/2} \text {Li}_2(a x^q) \, dx\)

Optimal. Leaf size=101 \[ \frac {8 a d q^2 \sqrt {d x} x^{q+2} \, _2F_1\left (1,\frac {q+\frac {5}{2}}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{25 (2 q+5)}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d} \]

[Out]

4/25*q*(d*x)^(5/2)*ln(1-a*x^q)/d+2/5*(d*x)^(5/2)*polylog(2,a*x^q)/d+8/25*a*d*q^2*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 = 0.06, antiderivative size = 101, 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} \[ \frac {2 (d x)^{5/2} \text {PolyLog}\left (2,a x^q\right )}{5 d}+\frac {8 a d q^2 \sqrt {d x} x^{q+2} \, _2F_1\left (1,\frac {q+\frac {5}{2}}{q};\frac {1}{2} \left (4+\frac {5}{q}\right );a x^q\right )}{25 (2 q+5)}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[2, a*x^q],x]

[Out]

(8*a*d*q^2*x^(2 + q)*Sqrt[d*x]*Hypergeometric2F1[1, (5/2 + q)/q, (4 + 5/q)/2, a*x^q])/(25*(5 + 2*q)) + (4*q*(d
*x)^(5/2)*Log[1 - a*x^q])/(25*d) + (2*(d*x)^(5/2)*PolyLog[2, 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 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 (d x)^{3/2} \text {Li}_2\left (a x^q\right ) \, dx &=\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {1}{5} (2 q) \int (d x)^{3/2} \log \left (1-a x^q\right ) \, dx\\ &=\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {\left (4 a q^2\right ) \int \frac {x^{-1+q} (d x)^{5/2}}{1-a x^q} \, dx}{25 d}\\ &=\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}+\frac {\left (4 a d q^2 \sqrt {d x}\right ) \int \frac {x^{\frac {3}{2}+q}}{1-a x^q} \, dx}{25 \sqrt {x}}\\ &=\frac {8 a d q^2 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 )}{25 (5+2 q)}+\frac {4 q (d x)^{5/2} \log \left (1-a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^q\right )}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 82, normalized size = 0.81 \[ \frac {2 x (d x)^{3/2} \left (4 a q^2 x^q \, _2F_1\left (1,\frac {q+\frac {5}{2}}{q};2+\frac {5}{2 q};a x^q\right )+(2 q+5) \left (5 \text {Li}_2\left (a x^q\right )+2 q \log \left (1-a x^q\right )\right )\right )}{25 (2 q+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {d x} d x {\rm Li}_2\left (a x^{q}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {3}{2}} {\rm Li}_2\left (a x^{q}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.16, size = 121, normalized size = 1.20 \[ -\frac {\left (d x \right )^{\frac {3}{2}} \left (-a \right )^{-\frac {5}{2 q}} \left (-\frac {4 q^{2} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \ln \left (1-a \,x^{q}\right )}{25}-\frac {2 q \,x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \left (1+\frac {2 q}{5}\right ) \polylog \left (2, a \,x^{q}\right )}{5+2 q}-\frac {4 q^{2} x^{\frac {5}{2}+q} a \left (-a \right )^{\frac {5}{2 q}} \Phi \left (a \,x^{q}, 1, \frac {5+2 q}{2 q}\right )}{25}\right )}{x^{\frac {3}{2}} q} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*x)^(3/2)/x^(3/2)*(-a)^(-5/2/q)/q*(-4/25*q^2*x^(5/2)*(-a)^(5/2/q)*ln(1-a*x^q)-2*q/(5+2*q)*x^(5/2)*(-a)^(5/2
/q)*(1+2/5*q)*polylog(2,a*x^q)-4/25*q^2*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 \[ 8 \, d^{\frac {3}{2}} q^{3} \int \frac {x^{\frac {3}{2}}}{25 \, {\left ({\left (2 \, a^{2} q - 5 \, a^{2}\right )} x^{2 \, q} - 2 \, {\left (2 \, a q - 5 \, a\right )} x^{q} + 2 \, q - 5\right )}}\,{d x} + \frac {2 \, {\left (25 \, {\left ({\left (2 \, a d^{\frac {3}{2}} q - 5 \, a d^{\frac {3}{2}}\right )} x x^{q} - {\left (2 \, d^{\frac {3}{2}} q - 5 \, d^{\frac {3}{2}}\right )} x\right )} x^{\frac {3}{2}} {\rm Li}_2\left (a x^{q}\right ) + 10 \, {\left ({\left (2 \, a d^{\frac {3}{2}} q^{2} - 5 \, a d^{\frac {3}{2}} q\right )} x x^{q} - {\left (2 \, d^{\frac {3}{2}} q^{2} - 5 \, d^{\frac {3}{2}} q\right )} x\right )} x^{\frac {3}{2}} \log \left (-a x^{q} + 1\right ) + 4 \, {\left (2 \, d^{\frac {3}{2}} q^{3} x - {\left (2 \, a d^{\frac {3}{2}} q^{3} - 5 \, a d^{\frac {3}{2}} q^{2}\right )} x x^{q}\right )} x^{\frac {3}{2}}\right )}}{125 \, {\left ({\left (2 \, a q - 5 \, a\right )} x^{q} - 2 \, q + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*d^(3/2)*q^3*integrate(1/25*x^(3/2)/((2*a^2*q - 5*a^2)*x^(2*q) - 2*(2*a*q - 5*a)*x^q + 2*q - 5), x) + 2/125*(
25*((2*a*d^(3/2)*q - 5*a*d^(3/2))*x*x^q - (2*d^(3/2)*q - 5*d^(3/2))*x)*x^(3/2)*dilog(a*x^q) + 10*((2*a*d^(3/2)
*q^2 - 5*a*d^(3/2)*q)*x*x^q - (2*d^(3/2)*q^2 - 5*d^(3/2)*q)*x)*x^(3/2)*log(-a*x^q + 1) + 4*(2*d^(3/2)*q^3*x -
(2*a*d^(3/2)*q^3 - 5*a*d^(3/2)*q^2)*x*x^q)*x^(3/2))/((2*a*q - 5*a)*x^q - 2*q + 5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*polylog(2, a*x^q),x)

[Out]

int((d*x)^(3/2)*polylog(2, a*x^q), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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