Integrand size = 13, antiderivative size = 94 \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\frac {9 a (d x)^{4+m} \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{3},\frac {7+m}{3},a x^3\right )}{d^4 (1+m)^2 (4+m)}+\frac {3 (d x)^{1+m} \log \left (1-a x^3\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \operatorname {PolyLog}\left (2,a x^3\right )}{d (1+m)} \]
9*a*(d*x)^(4+m)*hypergeom([1, 4/3+1/3*m],[7/3+1/3*m],a*x^3)/d^4/(1+m)^2/(4 +m)+3*(d*x)^(1+m)*ln(-a*x^3+1)/d/(1+m)^2+(d*x)^(1+m)*polylog(2,a*x^3)/d/(1 +m)
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\frac {x (d x)^m \left (9 a x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {4+m}{3},\frac {7+m}{3},a x^3\right )+(4+m) \left (3 \log \left (1-a x^3\right )+(1+m) \operatorname {PolyLog}\left (2,a x^3\right )\right )\right )}{(1+m)^2 (4+m)} \]
(x*(d*x)^m*(9*a*x^3*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, a*x^3] + (4 + m)*(3*Log[1 - a*x^3] + (1 + m)*PolyLog[2, a*x^3])))/((1 + m)^2*(4 + m))
Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {7145, 25, 2905, 8, 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.
\(\displaystyle \int \operatorname {PolyLog}\left (2,a x^3\right ) (d x)^m \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {\operatorname {PolyLog}\left (2,a x^3\right ) (d x)^{m+1}}{d (m+1)}-\frac {3 \int -(d x)^m \log \left (1-a x^3\right )dx}{m+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {3 \int (d x)^m \log \left (1-a x^3\right )dx}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^3\right ) (d x)^{m+1}}{d (m+1)}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {3 \left (\frac {3 a \int \frac {x^2 (d x)^{m+1}}{1-a x^3}dx}{d (m+1)}+\frac {\log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)}\right )}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^3\right ) (d x)^{m+1}}{d (m+1)}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle \frac {3 \left (\frac {3 a \int \frac {(d x)^{m+3}}{1-a x^3}dx}{d^3 (m+1)}+\frac {\log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)}\right )}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^3\right ) (d x)^{m+1}}{d (m+1)}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {3 \left (\frac {3 a (d x)^{m+4} \operatorname {Hypergeometric2F1}\left (1,\frac {m+4}{3},\frac {m+7}{3},a x^3\right )}{d^4 (m+1) (m+4)}+\frac {\log \left (1-a x^3\right ) (d x)^{m+1}}{d (m+1)}\right )}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^3\right ) (d x)^{m+1}}{d (m+1)}\) |
(3*((3*a*(d*x)^(4 + m)*Hypergeometric2F1[1, (4 + m)/3, (7 + m)/3, a*x^3])/ (d^4*(1 + m)*(4 + m)) + ((d*x)^(1 + m)*Log[1 - a*x^3])/(d*(1 + m))))/(1 + m) + ((d*x)^(1 + m)*PolyLog[2, a*x^3])/(d*(1 + m))
3.2.8.3.1 Defintions of rubi rules used
Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m Int[u*(a* x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
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 = 3.65 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.88
method | result | size |
meijerg | \(-\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {1}{3}-\frac {m}{3}} \left (\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (-36-9 m \right )}{\left (4+m \right ) \left (1+m \right )^{3} a}-\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (-12-3 m \right ) \ln \left (-a \,x^{3}+1\right )}{\left (4+m \right ) \left (1+m \right )^{2} a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \operatorname {polylog}\left (2, a \,x^{3}\right )}{\left (1+m \right ) a}+\frac {3 x^{1+m} \left (-a \right )^{\frac {4}{3}+\frac {m}{3}} \left (12+3 m \right ) \operatorname {LerchPhi}\left (a \,x^{3}, 1, \frac {m}{3}+\frac {1}{3}\right )}{\left (4+m \right ) \left (1+m \right )^{2} a}\right )}{3}\) | \(177\) |
-1/3*(d*x)^m*x^(-m)*(-a)^(-1/3-1/3*m)*(3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(- 36-9*m)/(1+m)^3/a-3/(4+m)*x^(1+m)*(-a)^(4/3+1/3*m)*(-12-3*m)/(1+m)^2*ln(-a *x^3+1)/a+3*x^(1+m)*(-a)^(4/3+1/3*m)/(1+m)/a*polylog(2,a*x^3)+3/(4+m)*x^(1 +m)*(-a)^(4/3+1/3*m)*(12+3*m)/(1+m)^2/a*LerchPhi(a*x^3,1,1/3*m+1/3))
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{3}\right ) \,d x } \]
Timed out. \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\text {Timed out} \]
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{3}\right ) \,d x } \]
-9*a*d^m*integrate(x^3*x^m/((a*m^2 + 2*a*m + a)*x^3 - m^2 - 2*m - 1), x) + ((d^m*m + d^m)*x*x^m*dilog(a*x^3) + 3*d^m*x*x^m*log(-a*x^3 + 1))/(m^2 + 2 *m + 1)
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{3}\right ) \,d x } \]
Timed out. \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^3\right ) \, dx=\int \mathrm {polylog}\left (2,a\,x^3\right )\,{\left (d\,x\right )}^m \,d x \]