Integrand size = 15, antiderivative size = 119 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=-\frac {16 a q^3 x^q \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (2-\frac {1}{q}\right ),\frac {1}{2} \left (4-\frac {1}{q}\right ),a x^q\right )}{d (1-2 q) \sqrt {d x}}+\frac {8 q^2 \log \left (1-a x^q\right )}{d \sqrt {d x}}-\frac {4 q \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}} \]
-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*pol ylog(3,a*x^q)/d/(d*x)^(1/2)
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 = 0.42 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=-\frac {x G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1+\frac {1}{2 q} \\ 1,0,0,0,\frac {1}{2 q} \\\end {array}\right )}{q (d x)^{3/2}} \]
-((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)))
Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle 2 q \int \frac {\operatorname {PolyLog}\left (2,a x^q\right )}{(d x)^{3/2}}dx-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle 2 q \left (2 q \int -\frac {\log \left (1-a x^q\right )}{(d x)^{3/2}}dx-\frac {2 \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 q \left (-2 q \int \frac {\log \left (1-a x^q\right )}{(d x)^{3/2}}dx-\frac {2 \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle 2 q \left (-2 q \left (-\frac {2 a q \int \frac {x^{q-1}}{\sqrt {d x} \left (1-a x^q\right )}dx}{d}-\frac {2 \log \left (1-a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle 2 q \left (-2 q \left (-\frac {2 a q \sqrt {x} \int \frac {x^{q-\frac {3}{2}}}{1-a x^q}dx}{d \sqrt {d x}}-\frac {2 \log \left (1-a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle 2 q \left (-2 q \left (\frac {4 a q x^q \operatorname {Hypergeometric2F1}\left (1,-\frac {\frac {1}{2}-q}{q},\frac {1}{2} \left (4-\frac {1}{q}\right ),a x^q\right )}{d (1-2 q) \sqrt {d x}}-\frac {2 \log \left (1-a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (2,a x^q\right )}{d \sqrt {d x}}\right )-\frac {2 \operatorname {PolyLog}\left (3,a x^q\right )}{d \sqrt {d x}}\) |
2*q*(-2*q*((4*a*q*x^q*Hypergeometric2F1[1, -((1/2 - q)/q), (4 - q^(-1))/2, a*x^q])/(d*(1 - 2*q)*Sqrt[d*x]) - (2*Log[1 - a*x^q])/(d*Sqrt[d*x])) - (2* PolyLog[2, a*x^q])/(d*Sqrt[d*x])) - (2*PolyLog[3, a*x^q])/(d*Sqrt[d*x])
3.1.94.3.1 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 = 0.34 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.22
method | result | size |
meijerg | \(-\frac {x^{\frac {3}{2}} \left (-a \right )^{\frac {1}{2 q}} \left (-\frac {8 q^{3} \left (-a \right )^{-\frac {1}{2 q}} \ln \left (1-a \,x^{q}\right )}{\sqrt {x}}+\frac {4 q^{2} \left (-a \right )^{-\frac {1}{2 q}} \operatorname {polylog}\left (2, a \,x^{q}\right )}{\sqrt {x}}-\frac {2 q \left (-a \right )^{-\frac {1}{2 q}} \left (1-2 q \right ) \operatorname {polylog}\left (3, a \,x^{q}\right )}{\left (2 q -1\right ) \sqrt {x}}-8 q^{3} x^{q -\frac {1}{2}} a \left (-a \right )^{-\frac {1}{2 q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {2 q -1}{2 q}\right )\right )}{\left (d x \right )^{\frac {3}{2}} q}\) | \(145\) |
-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)*polylog(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)*LerchP hi(a*x^q,1,1/2*(2*q-1)/q))
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{q}\right )}{\left (d x\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]
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))
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{(d x)^{3/2}} \, dx=\int \frac {\mathrm {polylog}\left (3,a\,x^q\right )}{{\left (d\,x\right )}^{3/2}} \,d x \]