Integrand size = 15, antiderivative size = 125 \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=-\frac {16 a d q^3 x^{2+q} \sqrt {d x} \operatorname {Hypergeometric2F1}\left (1,\frac {\frac {5}{2}+q}{q},\frac {1}{2} \left (4+\frac {5}{q}\right ),a x^q\right )}{125 (5+2 q)}-\frac {8 q^2 (d x)^{5/2} \log \left (1-a x^q\right )}{125 d}-\frac {4 q (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{25 d}+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d} \]
-8/125*q^2*(d*x)^(5/2)*ln(1-a*x^q)/d-4/25*q*(d*x)^(5/2)*polylog(2,a*x^q)/d +2/5*(d*x)^(5/2)*polylog(3,a*x^q)/d-16/125*a*d*q^3*x^(2+q)*hypergeom([1, ( 5/2+q)/q],[2+5/2/q],a*x^q)*(d*x)^(1/2)/(5+2*q)
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.40 \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=-\frac {x (d x)^{3/2} G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1-\frac {5}{2 q} \\ 1,0,0,0,-\frac {5}{2 q} \\\end {array}\right )}{q} \]
-((x*(d*x)^(3/2)*MeijerG[{{1, 1, 1, 1, 1 - 5/(2*q)}, {}}, {{1}, {0, 0, 0, -5/(2*q)}}, -(a*x^q)])/q)
Time = 0.37 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {7145, 7145, 25, 2905, 30, 888}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d}-\frac {2}{5} q \int (d x)^{3/2} \operatorname {PolyLog}\left (2,a x^q\right )dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d}-\frac {2}{5} q \left (\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}-\frac {2}{5} q \int -(d x)^{3/2} \log \left (1-a x^q\right )dx\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d}-\frac {2}{5} q \left (\frac {2}{5} q \int (d x)^{3/2} \log \left (1-a x^q\right )dx+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\right )\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d}-\frac {2}{5} q \left (\frac {2}{5} q \left (\frac {2 a q \int \frac {x^{q-1} (d x)^{5/2}}{1-a x^q}dx}{5 d}+\frac {2 (d x)^{5/2} \log \left (1-a x^q\right )}{5 d}\right )+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\right )\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d}-\frac {2}{5} q \left (\frac {2}{5} q \left (\frac {2 a d q \sqrt {d x} \int \frac {x^{q+\frac {3}{2}}}{1-a x^q}dx}{5 \sqrt {x}}+\frac {2 (d x)^{5/2} \log \left (1-a x^q\right )}{5 d}\right )+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\right )\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (3,a x^q\right )}{5 d}-\frac {2}{5} q \left (\frac {2}{5} q \left (\frac {4 a d q \sqrt {d x} x^{q+2} \operatorname {Hypergeometric2F1}\left (1,\frac {q+\frac {5}{2}}{q},\frac {1}{2} \left (4+\frac {5}{q}\right ),a x^q\right )}{5 (2 q+5)}+\frac {2 (d x)^{5/2} \log \left (1-a x^q\right )}{5 d}\right )+\frac {2 (d x)^{5/2} \operatorname {PolyLog}\left (2,a x^q\right )}{5 d}\right )\) |
(-2*q*((2*q*((4*a*d*q*x^(2 + q)*Sqrt[d*x]*Hypergeometric2F1[1, (5/2 + q)/q , (4 + 5/q)/2, a*x^q])/(5*(5 + 2*q)) + (2*(d*x)^(5/2)*Log[1 - a*x^q])/(5*d )))/5 + (2*(d*x)^(5/2)*PolyLog[2, a*x^q])/(5*d)))/5 + (2*(d*x)^(5/2)*PolyL og[3, a*x^q])/(5*d)
3.1.91.3.1 Defintions of rubi rules used
Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & & !IntegerQ[p]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
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] - Simp[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]
Int[((d_.)*(x_))^(m_.)*PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbo l] :> Simp[(d*x)^(m + 1)*(PolyLog[n, a*(b*x^p)^q]/(d*(m + 1))), x] - Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 5.
Time = 0.36 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.16
method | result | size |
meijerg | \(-\frac {\left (d x \right )^{\frac {3}{2}} \left (-a \right )^{-\frac {5}{2 q}} \left (\frac {8 q^{3} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \ln \left (1-a \,x^{q}\right )}{125}+\frac {4 q^{2} x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \operatorname {polylog}\left (2, a \,x^{q}\right )}{25}-\frac {2 q \,x^{\frac {5}{2}} \left (-a \right )^{\frac {5}{2 q}} \left (1+\frac {2 q}{5}\right ) \operatorname {polylog}\left (3, a \,x^{q}\right )}{5+2 q}+\frac {8 q^{3} x^{\frac {5}{2}+q} a \left (-a \right )^{\frac {5}{2 q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {5+2 q}{2 q}\right )}{125}\right )}{x^{\frac {3}{2}} q}\) | \(145\) |
-(d*x)^(3/2)/x^(3/2)*(-a)^(-5/2/q)/q*(8/125*q^3*x^(5/2)*(-a)^(5/2/q)*ln(1- a*x^q)+4/25*q^2*x^(5/2)*(-a)^(5/2/q)*polylog(2,a*x^q)-2*q/(5+2*q)*x^(5/2)* (-a)^(5/2/q)*(1+2/5*q)*polylog(3,a*x^q)+8/125*q^3*x^(5/2+q)*a*(-a)^(5/2/q) *LerchPhi(a*x^q,1,1/2*(5+2*q)/q))
\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\rm Li}_{3}(a x^{q}) \,d x } \]
\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=\int \left (d x\right )^{\frac {3}{2}} \operatorname {Li}_{3}\left (a x^{q}\right )\, dx \]
\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\rm Li}_{3}(a x^{q}) \,d x } \]
-16*d^(3/2)*q^4*integrate(1/125*x^(3/2)/(a^2*(2*q - 5)*x^(2*q) - 2*a*(2*q - 5)*x^q + 2*q - 5), x) - 2/625*(50*((2*q^2 - 5*q)*a*d^(3/2)*x*x^q - (2*q^ 2 - 5*q)*d^(3/2)*x)*x^(3/2)*dilog(a*x^q) + 20*((2*q^3 - 5*q^2)*a*d^(3/2)*x *x^q - (2*q^3 - 5*q^2)*d^(3/2)*x)*x^(3/2)*log(-a*x^q + 1) - 125*(a*d^(3/2) *(2*q - 5)*x*x^q - d^(3/2)*(2*q - 5)*x)*x^(3/2)*polylog(3, a*x^q) + 8*(2*d ^(3/2)*q^4*x - (2*q^4 - 5*q^3)*a*d^(3/2)*x*x^q)*x^(3/2))/(a*(2*q - 5)*x^q - 2*q + 5)
\[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=\int { \left (d x\right )^{\frac {3}{2}} {\rm Li}_{3}(a x^{q}) \,d x } \]
Timed out. \[ \int (d x)^{3/2} \operatorname {PolyLog}\left (3,a x^q\right ) \, dx=\int {\left (d\,x\right )}^{3/2}\,\mathrm {polylog}\left (3,a\,x^q\right ) \,d x \]