Integrand size = 24, antiderivative size = 640 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=-\frac {8 b^2 e^2 n^2}{715 d^2 x^{11/3}}+\frac {64 b^2 e^3 n^2}{2145 d^3 x^3}-\frac {2872 b^2 e^4 n^2}{45045 d^4 x^{7/3}}+\frac {1216 b^2 e^5 n^2}{9009 d^5 x^{5/3}}-\frac {224072 b^2 e^6 n^2}{675675 d^6 x}+\frac {344192 b^2 e^7 n^2}{225225 d^7 \sqrt [3]{x}}+\frac {704552 b^2 e^{15/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{225225 d^{15/2}}-\frac {4 i b^2 e^{15/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{5 d^{15/2}}-\frac {8 b^2 e^{15/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{5 d^{15/2}}-\frac {4 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{65 d x^{13/3}}+\frac {4 b e^2 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{55 d^2 x^{11/3}}-\frac {4 b e^3 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{45 d^3 x^3}+\frac {4 b e^4 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{35 d^4 x^{7/3}}-\frac {4 b e^5 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{25 d^5 x^{5/3}}+\frac {4 b e^6 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{15 d^6 x}-\frac {4 b e^7 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 d^7 \sqrt [3]{x}}-\frac {4 b e^{15/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 d^{15/2}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{5 x^5}-\frac {4 i b^2 e^{15/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{5 d^{15/2}} \]
-8/715*b^2*e^2*n^2/d^2/x^(11/3)+64/2145*b^2*e^3*n^2/d^3/x^3-2872/45045*b^2 *e^4*n^2/d^4/x^(7/3)+1216/9009*b^2*e^5*n^2/d^5/x^(5/3)-224072/675675*b^2*e ^6*n^2/d^6/x+344192/225225*b^2*e^7*n^2/d^7/x^(1/3)+704552/225225*b^2*e^(15 /2)*n^2*arctan(x^(1/3)*e^(1/2)/d^(1/2))/d^(15/2)-4/5*I*b^2*e^(15/2)*n^2*ar ctan(x^(1/3)*e^(1/2)/d^(1/2))^2/d^(15/2)-4/65*b*e*n*(a+b*ln(c*(d+e*x^(2/3) )^n))/d/x^(13/3)+4/55*b*e^2*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d^2/x^(11/3)-4/4 5*b*e^3*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d^3/x^3+4/35*b*e^4*n*(a+b*ln(c*(d+e* x^(2/3))^n))/d^4/x^(7/3)-4/25*b*e^5*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d^5/x^(5 /3)+4/15*b*e^6*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d^6/x-4/5*b*e^7*n*(a+b*ln(c*( d+e*x^(2/3))^n))/d^7/x^(1/3)-4/5*b*e^(15/2)*n*arctan(x^(1/3)*e^(1/2)/d^(1/ 2))*(a+b*ln(c*(d+e*x^(2/3))^n))/d^(15/2)-1/5*(a+b*ln(c*(d+e*x^(2/3))^n))^2 /x^5-8/5*b^2*e^(15/2)*n^2*arctan(x^(1/3)*e^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^ (1/2)+I*x^(1/3)*e^(1/2)))/d^(15/2)-4/5*I*b^2*e^(15/2)*n^2*polylog(2,1-2*d^ (1/2)/(d^(1/2)+I*x^(1/3)*e^(1/2)))/d^(15/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.76 (sec) , antiderivative size = 678, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{5 x^5}+\frac {4}{5} b e n \left (\frac {2 b e^{13/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{15/2}}-\frac {2 b e n \operatorname {Hypergeometric2F1}\left (-\frac {11}{2},1,-\frac {9}{2},-\frac {e x^{2/3}}{d}\right )}{143 d^2 x^{11/3}}+\frac {2 b e^2 n \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},1,-\frac {7}{2},-\frac {e x^{2/3}}{d}\right )}{99 d^3 x^3}-\frac {2 b e^3 n \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},-\frac {e x^{2/3}}{d}\right )}{63 d^4 x^{7/3}}+\frac {2 b e^4 n \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\frac {e x^{2/3}}{d}\right )}{35 d^5 x^{5/3}}-\frac {2 b e^5 n \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {e x^{2/3}}{d}\right )}{15 d^6 x}+\frac {2 b e^6 n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {e x^{2/3}}{d}\right )}{3 d^7 \sqrt [3]{x}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{13 d x^{13/3}}+\frac {e \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{11 d^2 x^{11/3}}-\frac {e^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{9 d^3 x^3}+\frac {e^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{7 d^4 x^{7/3}}-\frac {e^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 d^5 x^{5/3}}+\frac {e^5 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 d^6 x}-\frac {e^6 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^7 \sqrt [3]{x}}-\frac {e^{13/2} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{15/2}}-\frac {i b e^{13/2} n \left (\arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (\arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )-2 i \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}{-i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}\right )\right )}{d^{15/2}}\right ) \]
-1/5*(a + b*Log[c*(d + e*x^(2/3))^n])^2/x^5 + (4*b*e*n*((2*b*e^(13/2)*n*Ar cTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/d^(15/2) - (2*b*e*n*Hypergeometric2F1[-11 /2, 1, -9/2, -((e*x^(2/3))/d)])/(143*d^2*x^(11/3)) + (2*b*e^2*n*Hypergeome tric2F1[-9/2, 1, -7/2, -((e*x^(2/3))/d)])/(99*d^3*x^3) - (2*b*e^3*n*Hyperg eometric2F1[-7/2, 1, -5/2, -((e*x^(2/3))/d)])/(63*d^4*x^(7/3)) + (2*b*e^4* n*Hypergeometric2F1[-5/2, 1, -3/2, -((e*x^(2/3))/d)])/(35*d^5*x^(5/3)) - ( 2*b*e^5*n*Hypergeometric2F1[-3/2, 1, -1/2, -((e*x^(2/3))/d)])/(15*d^6*x) + (2*b*e^6*n*Hypergeometric2F1[-1/2, 1, 1/2, -((e*x^(2/3))/d)])/(3*d^7*x^(1 /3)) - (a + b*Log[c*(d + e*x^(2/3))^n])/(13*d*x^(13/3)) + (e*(a + b*Log[c* (d + e*x^(2/3))^n]))/(11*d^2*x^(11/3)) - (e^2*(a + b*Log[c*(d + e*x^(2/3)) ^n]))/(9*d^3*x^3) + (e^3*(a + b*Log[c*(d + e*x^(2/3))^n]))/(7*d^4*x^(7/3)) - (e^4*(a + b*Log[c*(d + e*x^(2/3))^n]))/(5*d^5*x^(5/3)) + (e^5*(a + b*Lo g[c*(d + e*x^(2/3))^n]))/(3*d^6*x) - (e^6*(a + b*Log[c*(d + e*x^(2/3))^n]) )/(d^7*x^(1/3)) - (e^(13/2)*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(a + b*Log[c *(d + e*x^(2/3))^n]))/d^(15/2) - (I*b*e^(13/2)*n*(ArcTan[(Sqrt[e]*x^(1/3)) /Sqrt[d]]*(ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]] - (2*I)*Log[(2*Sqrt[d])/(Sqrt [d] + I*Sqrt[e]*x^(1/3))]) + PolyLog[2, (I*Sqrt[d] + Sqrt[e]*x^(1/3))/((-I )*Sqrt[d] + Sqrt[e]*x^(1/3))]))/d^(15/2)))/5
Time = 1.01 (sec) , antiderivative size = 579, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2908, 2907, 2926, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx\) |
| \(\Big \downarrow \) 2908 |
| \(\displaystyle 3 \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^{16/3}}d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 2907 |
| \(\displaystyle 3 \left (\frac {4}{15} b e n \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{\left (d+e x^{2/3}\right ) x^{14/3}}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{15 x^5}\right )\) |
| \(\Big \downarrow \) 2926 |
| \(\displaystyle 3 \left (\frac {4}{15} b e n \int \left (-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^7}{d^7 \left (d+e x^{2/3}\right )}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^6}{d^7 x^{2/3}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^5}{d^6 x^{4/3}}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^4}{d^5 x^2}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^3}{d^4 x^{8/3}}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^2}{d^3 x^{10/3}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e}{d^2 x^4}+\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{d x^{14/3}}\right )d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{15 x^5}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{15 x^5}+\frac {4}{15} b e n \left (-\frac {e^{13/2} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{15/2}}-\frac {e^6 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^7 \sqrt [3]{x}}+\frac {e^5 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 d^6 x}-\frac {e^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 d^5 x^{5/3}}+\frac {e^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{7 d^4 x^{7/3}}-\frac {e^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{9 d^3 x^3}+\frac {e \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{11 d^2 x^{11/3}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{13 d x^{13/3}}-\frac {i b e^{13/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{d^{15/2}}+\frac {176138 b e^{13/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{45045 d^{15/2}}-\frac {2 b e^{13/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{15/2}}-\frac {i b e^{13/2} n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{15/2}}+\frac {86048 b e^6 n}{45045 d^7 \sqrt [3]{x}}-\frac {56018 b e^5 n}{135135 d^6 x}+\frac {1520 b e^4 n}{9009 d^5 x^{5/3}}-\frac {718 b e^3 n}{9009 d^4 x^{7/3}}+\frac {16 b e^2 n}{429 d^3 x^3}-\frac {2 b e n}{143 d^2 x^{11/3}}\right )\right )\) |
3*(-1/15*(a + b*Log[c*(d + e*x^(2/3))^n])^2/x^5 + (4*b*e*n*((-2*b*e*n)/(14 3*d^2*x^(11/3)) + (16*b*e^2*n)/(429*d^3*x^3) - (718*b*e^3*n)/(9009*d^4*x^( 7/3)) + (1520*b*e^4*n)/(9009*d^5*x^(5/3)) - (56018*b*e^5*n)/(135135*d^6*x) + (86048*b*e^6*n)/(45045*d^7*x^(1/3)) + (176138*b*e^(13/2)*n*ArcTan[(Sqrt [e]*x^(1/3))/Sqrt[d]])/(45045*d^(15/2)) - (I*b*e^(13/2)*n*ArcTan[(Sqrt[e]* x^(1/3))/Sqrt[d]]^2)/d^(15/2) - (2*b*e^(13/2)*n*ArcTan[(Sqrt[e]*x^(1/3))/S qrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x^(1/3))])/d^(15/2) - (a + b* Log[c*(d + e*x^(2/3))^n])/(13*d*x^(13/3)) + (e*(a + b*Log[c*(d + e*x^(2/3) )^n]))/(11*d^2*x^(11/3)) - (e^2*(a + b*Log[c*(d + e*x^(2/3))^n]))/(9*d^3*x ^3) + (e^3*(a + b*Log[c*(d + e*x^(2/3))^n]))/(7*d^4*x^(7/3)) - (e^4*(a + b *Log[c*(d + e*x^(2/3))^n]))/(5*d^5*x^(5/3)) + (e^5*(a + b*Log[c*(d + e*x^( 2/3))^n]))/(3*d^6*x) - (e^6*(a + b*Log[c*(d + e*x^(2/3))^n]))/(d^7*x^(1/3) ) - (e^(13/2)*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(a + b*Log[c*(d + e*x^(2/3 ))^n]))/d^(15/2) - (I*b*e^(13/2)*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] + I *Sqrt[e]*x^(1/3))])/d^(15/2)))/15)
3.5.80.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*((f_.)*( x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])^q /(f*(m + 1))), x] - Simp[b*e*n*p*(q/(f^n*(m + 1))) Int[(f*x)^(m + n)*((a + b*Log[c*(d + e*x^n)^p])^(q - 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d , e, f, m, p}, x] && IGtQ[q, 1] && IntegerQ[n] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*(x_)^(m_ .), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k*(m + 1) - 1)*(a + b*Log[c*(d + e*x^(k*n))^p])^q, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, e, m, p, q}, x] && FractionQ[n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b *Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e , f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & & IntegerQ[s]
\[\int \frac {{\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{2}}{x^{6}}d x\]
\[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{2}}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{2}}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^6} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\right )}^2}{x^6} \,d x \]