Integrand size = 13, antiderivative size = 94 \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\frac {4 a (d x)^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},a x^2\right )}{d^3 (1+m)^2 (3+m)}+\frac {2 (d x)^{1+m} \log \left (1-a x^2\right )}{d (1+m)^2}+\frac {(d x)^{1+m} \operatorname {PolyLog}\left (2,a x^2\right )}{d (1+m)} \]
4*a*(d*x)^(3+m)*hypergeom([1, 3/2+1/2*m],[5/2+1/2*m],a*x^2)/d^3/(1+m)^2/(3 +m)+2*(d*x)^(1+m)*ln(-a*x^2+1)/d/(1+m)^2+(d*x)^(1+m)*polylog(2,a*x^2)/d/(1 +m)
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\frac {x (d x)^m \left (4 a x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},a x^2\right )+(3+m) \left (2 \log \left (1-a x^2\right )+(1+m) \operatorname {PolyLog}\left (2,a x^2\right )\right )\right )}{(1+m)^2 (3+m)} \]
(x*(d*x)^m*(4*a*x^2*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, a*x^2] + (3 + m)*(2*Log[1 - a*x^2] + (1 + m)*PolyLog[2, a*x^2])))/((1 + m)^2*(3 + m))
Time = 0.28 (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, 278}
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.
\(\displaystyle \int \operatorname {PolyLog}\left (2,a x^2\right ) (d x)^m \, dx\) |
\(\Big \downarrow \) 7145 |
\(\displaystyle \frac {\operatorname {PolyLog}\left (2,a x^2\right ) (d x)^{m+1}}{d (m+1)}-\frac {2 \int -(d x)^m \log \left (1-a x^2\right )dx}{m+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \int (d x)^m \log \left (1-a x^2\right )dx}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^2\right ) (d x)^{m+1}}{d (m+1)}\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2 \left (\frac {2 a \int \frac {x (d x)^{m+1}}{1-a x^2}dx}{d (m+1)}+\frac {\log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)}\right )}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^2\right ) (d x)^{m+1}}{d (m+1)}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle \frac {2 \left (\frac {2 a \int \frac {(d x)^{m+2}}{1-a x^2}dx}{d^2 (m+1)}+\frac {\log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)}\right )}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^2\right ) (d x)^{m+1}}{d (m+1)}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {2 \left (\frac {2 a (d x)^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},a x^2\right )}{d^3 (m+1) (m+3)}+\frac {\log \left (1-a x^2\right ) (d x)^{m+1}}{d (m+1)}\right )}{m+1}+\frac {\operatorname {PolyLog}\left (2,a x^2\right ) (d x)^{m+1}}{d (m+1)}\) |
(2*((2*a*(d*x)^(3 + m)*Hypergeometric2F1[1, (3 + m)/2, (5 + m)/2, a*x^2])/ (d^3*(1 + m)*(3 + m)) + ((d*x)^(1 + m)*Log[1 - a*x^2])/(d*(1 + m))))/(1 + m) + ((d*x)^(1 + m)*PolyLog[2, a*x^2])/(d*(1 + m))
3.2.5.3.1 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_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[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 = 1.61 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.88
method | result | size |
meijerg | \(-\frac {\left (d x \right )^{m} x^{-m} \left (-a \right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-12-4 m \right )}{\left (3+m \right ) \left (1+m \right )^{3} a}-\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (-6-2 m \right ) \ln \left (-a \,x^{2}+1\right )}{\left (3+m \right ) \left (1+m \right )^{2} a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \operatorname {polylog}\left (2, a \,x^{2}\right )}{\left (1+m \right ) a}+\frac {2 x^{1+m} \left (-a \right )^{\frac {3}{2}+\frac {m}{2}} \left (6+2 m \right ) \operatorname {LerchPhi}\left (a \,x^{2}, 1, \frac {1}{2}+\frac {m}{2}\right )}{\left (3+m \right ) \left (1+m \right )^{2} a}\right )}{2}\) | \(177\) |
-1/2*(d*x)^m*x^(-m)*(-a)^(-1/2-1/2*m)*(2/(3+m)*x^(1+m)*(-a)^(3/2+1/2*m)*(- 12-4*m)/(1+m)^3/a-2/(3+m)*x^(1+m)*(-a)^(3/2+1/2*m)*(-6-2*m)/(1+m)^2*ln(-a* x^2+1)/a+2*x^(1+m)*(-a)^(3/2+1/2*m)/(1+m)/a*polylog(2,a*x^2)+2/(3+m)*x^(1+ m)*(-a)^(3/2+1/2*m)*(6+2*m)/(1+m)^2/a*LerchPhi(a*x^2,1,1/2+1/2*m))
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{2}\right ) \,d x } \]
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\int \left (d x\right )^{m} \operatorname {Li}_{2}\left (a x^{2}\right )\, dx \]
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{2}\right ) \,d x } \]
-4*a*d^m*integrate(x^2*x^m/((a*m^2 + 2*a*m + a)*x^2 - m^2 - 2*m - 1), x) + ((d^m*m + d^m)*x*x^m*dilog(a*x^2) + 2*d^m*x*x^m*log(-a*x^2 + 1))/(m^2 + 2 *m + 1)
\[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\int { \left (d x\right )^{m} {\rm Li}_2\left (a x^{2}\right ) \,d x } \]
Timed out. \[ \int (d x)^m \operatorname {PolyLog}\left (2,a x^2\right ) \, dx=\int \mathrm {polylog}\left (2,a\,x^2\right )\,{\left (d\,x\right )}^m \,d x \]