Integrand size = 13, antiderivative size = 101 \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\frac {a q^2 x^{1+q} (d x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+q}{q},\frac {1+m+2 q}{q},a x^q\right )}{(1+m)^2 (1+m+q)}+\frac {q (d x)^{1+m} \log \left (1-a x^q\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \operatorname {PolyLog}\left (2,a x^q\right )}{d (1+m)} \]
a*q^2*x^(1+q)*(d*x)^m*hypergeom([1, (1+m+q)/q],[(1+m+2*q)/q],a*x^q)/(1+m)^ 2/(1+m+q)+q*(d*x)^(1+m)*ln(1-a*x^q)/d/(1+m)^2+(d*x)^(1+m)*polylog(2,a*x^q) /d/(1+m)
Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.79 \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\frac {x (d x)^m \left (a q^2 x^q \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+q}{q},\frac {1+m+2 q}{q},a x^q\right )+(1+m+q) \left (q \log \left (1-a x^q\right )+(1+m) \operatorname {PolyLog}\left (2,a x^q\right )\right )\right )}{(1+m)^2 (1+m+q)} \]
(x*(d*x)^m*(a*q^2*x^q*Hypergeometric2F1[1, (1 + m + q)/q, (1 + m + 2*q)/q, a*x^q] + (1 + m + q)*(q*Log[1 - a*x^q] + (1 + m)*PolyLog[2, a*x^q])))/((1 + m)^2*(1 + m + q))
Time = 0.30 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05, 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, 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.
\(\displaystyle \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}-\frac {q \int -(d x)^m \log \left (1-a x^q\right )dx}{m+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {q \int (d x)^m \log \left (1-a x^q\right )dx}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {q \left (\frac {a q \int \frac {x^{q-1} (d x)^{m+1}}{1-a x^q}dx}{d (m+1)}+\frac {(d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)}\right )}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\) |
\(\Big \downarrow \) 30 |
\(\displaystyle \frac {q \left (\frac {a q x^{-m} (d x)^m \int \frac {x^{m+q}}{1-a x^q}dx}{m+1}+\frac {(d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)}\right )}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {q \left (\frac {a q x^{q+1} (d x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {m+q+1}{q},\frac {m+2 q+1}{q},a x^q\right )}{(m+1) (m+q+1)}+\frac {(d x)^{m+1} \log \left (1-a x^q\right )}{d (m+1)}\right )}{m+1}+\frac {(d x)^{m+1} \operatorname {PolyLog}\left (2,a x^q\right )}{d (m+1)}\) |
(q*((a*q*x^(1 + q)*(d*x)^m*Hypergeometric2F1[1, (1 + m + q)/q, (1 + m + 2* q)/q, a*x^q])/((1 + m)*(1 + m + q)) + ((d*x)^(1 + m)*Log[1 - a*x^q])/(d*(1 + m))))/(1 + m) + ((d*x)^(1 + m)*PolyLog[2, a*x^q])/(d*(1 + m))
3.2.11.3.1 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 = 2.92 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.47
method | result | size |
meijerg | \(-\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {m}{q}-\frac {1}{q}} \left (-\frac {q^{2} x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \ln \left (1-a \,x^{q}\right )}{\left (1+m \right )^{2}}-\frac {q \,x^{1+m} \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \operatorname {polylog}\left (2, a \,x^{q}\right )}{1+m}-\frac {q^{2} x^{1+m +q} a \left (-a \right )^{\frac {m}{q}+\frac {1}{q}} \operatorname {LerchPhi}\left (a \,x^{q}, 1, \frac {1+m +q}{q}\right )}{\left (1+m \right )^{2}}\right )}{q}\) | \(148\) |
-(d*x)^m*x^(-m)*(-a)^(-m/q-1/q)/q*(-q^2*x^(1+m)*(-a)^(m/q+1/q)/(1+m)^2*ln( 1-a*x^q)-q*x^(1+m)*(-a)^(m/q+1/q)/(1+m)*polylog(2,a*x^q)-q^2*x^(1+m+q)*a*( -a)^(m/q+1/q)/(1+m)^2*LerchPhi(a*x^q,1,(1+m+q)/q))
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{q}\right ) \,d x } \]
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int \left (d x\right )^{m} \operatorname {Li}_{2}\left (a x^{q}\right )\, dx \]
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{q}\right ) \,d x } \]
-d^m*q^2*integrate(-x^m/(m^2 - (a*m^2 + 2*a*m + a)*x^q + 2*m + 1), x) - (d ^m*q^2*x*x^m - (d^m*m + d^m)*q*x*x^m*log(-a*x^q + 1) - (d^m*m^2 + 2*d^m*m + d^m)*x*x^m*dilog(a*x^q))/(m^3 + 3*m^2 + 3*m + 1)
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{q}\right ) \,d x } \]
Timed out. \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^q\right ) \, dx=\int {\left (d\,x\right )}^m\,\mathrm {polylog}\left (2,a\,x^q\right ) \,d x \]