Integrand size = 15, antiderivative size = 115 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=-\frac {16 a q^3 x^q \sqrt {d x} \operatorname {Hypergeometric2F1}\left (1,\frac {\frac {1}{2}+q}{q},\frac {1}{2} \left (4+\frac {1}{q}\right ),a x^q\right )}{d (1+2 q)}-\frac {8 q^2 \sqrt {d x} \log \left (1-a x^q\right )}{d}-\frac {4 q \sqrt {d x} \operatorname {PolyLog}\left (2,a x^q\right )}{d}+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d} \]
-16*a*q^3*x^q*hypergeom([1, (1/2+q)/q],[2+1/2/q],a*x^q)*(d*x)^(1/2)/d/(1+2 *q)-8*q^2*ln(1-a*x^q)*(d*x)^(1/2)/d-4*q*polylog(2,a*x^q)*(d*x)^(1/2)/d+2*p olylog(3,a*x^q)*(d*x)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.43 \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=-\frac {x G_{5,5}^{1,5}\left (-a x^q|\begin {array}{c} 1,1,1,1,1-\frac {1}{2 q} \\ 1,0,0,0,-\frac {1}{2 q} \\\end {array}\right )}{q \sqrt {d x}} \]
-((x*MeijerG[{{1, 1, 1, 1, 1 - 1/(2*q)}, {}}, {{1}, {0, 0, 0, -1/2*1/q}}, -(a*x^q)])/(q*Sqrt[d*x]))
Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d}-2 q \int \frac {\operatorname {PolyLog}\left (2,a x^q\right )}{\sqrt {d x}}dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d}-2 q \left (\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^q\right )}{d}-2 q \int -\frac {\log \left (1-a x^q\right )}{\sqrt {d x}}dx\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d}-2 q \left (2 q \int \frac {\log \left (1-a x^q\right )}{\sqrt {d x}}dx+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^q\right )}{d}\right )\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d}-2 q \left (2 q \left (\frac {2 a q \int \frac {x^{q-1} \sqrt {d x}}{1-a x^q}dx}{d}+\frac {2 \sqrt {d x} \log \left (1-a x^q\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^q\right )}{d}\right )\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d}-2 q \left (2 q \left (\frac {2 a q \sqrt {d x} \int \frac {x^{q-\frac {1}{2}}}{1-a x^q}dx}{d \sqrt {x}}+\frac {2 \sqrt {d x} \log \left (1-a x^q\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^q\right )}{d}\right )\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 \sqrt {d x} \operatorname {PolyLog}\left (3,a x^q\right )}{d}-2 q \left (2 q \left (\frac {4 a q \sqrt {d x} x^q \operatorname {Hypergeometric2F1}\left (1,\frac {q+\frac {1}{2}}{q},\frac {1}{2} \left (4+\frac {1}{q}\right ),a x^q\right )}{d (2 q+1)}+\frac {2 \sqrt {d x} \log \left (1-a x^q\right )}{d}\right )+\frac {2 \sqrt {d x} \operatorname {PolyLog}\left (2,a x^q\right )}{d}\right )\) |
-2*q*(2*q*((4*a*q*x^q*Sqrt[d*x]*Hypergeometric2F1[1, (1/2 + q)/q, (4 + q^( -1))/2, a*x^q])/(d*(1 + 2*q)) + (2*Sqrt[d*x]*Log[1 - a*x^q])/d) + (2*Sqrt[ d*x]*PolyLog[2, a*x^q])/d) + (2*Sqrt[d*x]*PolyLog[3, a*x^q])/d
3.1.93.3.1 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 = 0.37 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.16
method | result | size |
meijerg | \(-\frac {\sqrt {x}\, \left (-a \right )^{-\frac {1}{2 q}} \left (8 q^{3} \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \ln \left (1-a \,x^{q}\right )+4 q^{2} \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \operatorname {polylog}\left (2, a \,x^{q}\right )-2 q \sqrt {x}\, \left (-a \right )^{\frac {1}{2 q}} \operatorname {polylog}\left (3, a \,x^{q}\right )+8 q^{3} x^{\frac {1}{2}+q} a \left (-a \right )^{\frac {1}{2 q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {1+2 q}{2 q}\right )\right )}{\sqrt {d x}\, q}\) | \(133\) |
-1/(d*x)^(1/2)*x^(1/2)*(-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*x^(1/2)*(-a)^(1/2/q)* polylog(3,a*x^q)+8*q^3*x^(1/2+q)*a*(-a)^(1/2/q)*LerchPhi(a*x^q,1,1/2*(1+2* q)/q))
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{\sqrt {d x}} \,d x } \]
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=\int \frac {\operatorname {Li}_{3}\left (a x^{q}\right )}{\sqrt {d x}}\, dx \]
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{\sqrt {d x}} \,d x } \]
-16*q^4*integrate(1/((a^2*sqrt(d)*(2*q - 1)*x^(2*q) - 2*a*sqrt(d)*(2*q - 1 )*x^q + sqrt(d)*(2*q - 1))*sqrt(x)), x) - 2*(2*((2*q^2 - q)*a*x*x^q - (2*q ^2 - q)*x)*dilog(a*x^q)/sqrt(x) + 4*((2*q^3 - q^2)*a*x*x^q - (2*q^3 - q^2) *x)*log(-a*x^q + 1)/sqrt(x) - (a*(2*q - 1)*x*x^q - (2*q - 1)*x)*polylog(3, a*x^q)/sqrt(x) + 8*(2*q^4*x - (2*q^4 - q^3)*a*x*x^q)/sqrt(x))/(a*sqrt(d)* (2*q - 1)*x^q - sqrt(d)*(2*q - 1))
\[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=\int { \frac {{\rm Li}_{3}(a x^{q})}{\sqrt {d x}} \,d x } \]
Timed out. \[ \int \frac {\operatorname {PolyLog}\left (3,a x^q\right )}{\sqrt {d x}} \, dx=\int \frac {\mathrm {polylog}\left (3,a\,x^q\right )}{\sqrt {d\,x}} \,d x \]