Integrand size = 24, antiderivative size = 490 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=-\frac {4 a b e^4 n \sqrt [3]{x}}{3 d^4}+\frac {568 b^2 e^4 n^2 \sqrt [3]{x}}{315 d^4}-\frac {32 b^2 e^3 n^2 x}{105 d^3}+\frac {8 b^2 e^2 n^2 x^{5/3}}{105 d^2}-\frac {1408 b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{315 d^{9/2}}-\frac {4 i b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{3 d^{9/2}}+\frac {8 b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{3 d^{9/2}}-\frac {4 b^2 e^4 n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 d^4}+\frac {4 b e^3 n x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{9 d^3}-\frac {4 b e^2 n x^{5/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{15 d^2}+\frac {4 b e n x^{7/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{21 d}+\frac {4 b e^{9/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 d^{9/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2-\frac {4 i b^2 e^{9/2} n^2 \operatorname {PolyLog}\left (2,-1+\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{3 d^{9/2}} \]
-4/3*a*b*e^4*n*x^(1/3)/d^4+568/315*b^2*e^4*n^2*x^(1/3)/d^4-32/105*b^2*e^3* n^2*x/d^3+8/105*b^2*e^2*n^2*x^(5/3)/d^2-1408/315*b^2*e^(9/2)*n^2*arctan(x^ (1/3)*d^(1/2)/e^(1/2))/d^(9/2)-4/3*I*b^2*e^(9/2)*n^2*arctan(x^(1/3)*d^(1/2 )/e^(1/2))^2/d^(9/2)-4/3*b^2*e^4*n*x^(1/3)*ln(c*(d+e/x^(2/3))^n)/d^4+4/9*b *e^3*n*x*(a+b*ln(c*(d+e/x^(2/3))^n))/d^3-4/15*b*e^2*n*x^(5/3)*(a+b*ln(c*(d +e/x^(2/3))^n))/d^2+4/21*b*e*n*x^(7/3)*(a+b*ln(c*(d+e/x^(2/3))^n))/d+4/3*b *e^(9/2)*n*arctan(x^(1/3)*d^(1/2)/e^(1/2))*(a+b*ln(c*(d+e/x^(2/3))^n))/d^( 9/2)+1/3*x^3*(a+b*ln(c*(d+e/x^(2/3))^n))^2+8/3*b^2*e^(9/2)*n^2*arctan(x^(1 /3)*d^(1/2)/e^(1/2))*ln(2-2*e^(1/2)/(-I*x^(1/3)*d^(1/2)+e^(1/2)))/d^(9/2)- 4/3*I*b^2*e^(9/2)*n^2*polylog(2,-1+2*e^(1/2)/(-I*x^(1/3)*d^(1/2)+e^(1/2))) /d^(9/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.43 (sec) , antiderivative size = 735, normalized size of antiderivative = 1.50 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=\frac {1}{3} \left (x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2-4 b e n \left (\frac {a e^3 \sqrt [3]{x}}{d^4}-\frac {2 b e^{7/2} n \arctan \left (\frac {\sqrt {e}}{\sqrt {d} \sqrt [3]{x}}\right )}{d^{9/2}}-\frac {2 b e n x^{5/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\frac {e}{d x^{2/3}}\right )}{35 d^2}+\frac {2 b e^2 n x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {e}{d x^{2/3}}\right )}{15 d^3}-\frac {2 b e^3 n \sqrt [3]{x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {e}{d x^{2/3}}\right )}{3 d^4}+\frac {b e^3 \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d^4}-\frac {e^2 x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 d^3}+\frac {e x^{5/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{5 d^2}-\frac {x^{7/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{7 d}+\frac {e^{7/2} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \log \left (\sqrt {e}-\sqrt {-d} \sqrt [3]{x}\right )}{2 (-d)^{9/2}}-\frac {e^{7/2} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \log \left (\sqrt {e}+\sqrt {-d} \sqrt [3]{x}\right )}{2 (-d)^{9/2}}-\frac {b e^{7/2} n \left (\log \left (\sqrt {e}-\sqrt {-d} \sqrt [3]{x}\right ) \left (\log \left (\sqrt {e}-\sqrt {-d} \sqrt [3]{x}\right )+2 \log \left (\frac {1}{2} \left (1+\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )-4 \log \left (\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )-4 \operatorname {PolyLog}\left (2,1-\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )+2 \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {\sqrt {-d} \sqrt [3]{x}}{2 \sqrt {e}}\right )\right )}{4 (-d)^{9/2}}+\frac {b e^{7/2} n \left (\log \left (\sqrt {e}+\sqrt {-d} \sqrt [3]{x}\right ) \left (\log \left (\sqrt {e}+\sqrt {-d} \sqrt [3]{x}\right )+2 \log \left (\frac {1}{2}-\frac {\sqrt {-d} \sqrt [3]{x}}{2 \sqrt {e}}\right )-4 \log \left (-\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1+\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )-4 \operatorname {PolyLog}\left (2,1+\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )}{4 (-d)^{9/2}}\right )\right ) \]
(x^3*(a + b*Log[c*(d + e/x^(2/3))^n])^2 - 4*b*e*n*((a*e^3*x^(1/3))/d^4 - ( 2*b*e^(7/2)*n*ArcTan[Sqrt[e]/(Sqrt[d]*x^(1/3))])/d^(9/2) - (2*b*e*n*x^(5/3 )*Hypergeometric2F1[-5/2, 1, -3/2, -(e/(d*x^(2/3)))])/(35*d^2) + (2*b*e^2* n*x*Hypergeometric2F1[-3/2, 1, -1/2, -(e/(d*x^(2/3)))])/(15*d^3) - (2*b*e^ 3*n*x^(1/3)*Hypergeometric2F1[-1/2, 1, 1/2, -(e/(d*x^(2/3)))])/(3*d^4) + ( b*e^3*x^(1/3)*Log[c*(d + e/x^(2/3))^n])/d^4 - (e^2*x*(a + b*Log[c*(d + e/x ^(2/3))^n]))/(3*d^3) + (e*x^(5/3)*(a + b*Log[c*(d + e/x^(2/3))^n]))/(5*d^2 ) - (x^(7/3)*(a + b*Log[c*(d + e/x^(2/3))^n]))/(7*d) + (e^(7/2)*(a + b*Log [c*(d + e/x^(2/3))^n])*Log[Sqrt[e] - Sqrt[-d]*x^(1/3)])/(2*(-d)^(9/2)) - ( e^(7/2)*(a + b*Log[c*(d + e/x^(2/3))^n])*Log[Sqrt[e] + Sqrt[-d]*x^(1/3)])/ (2*(-d)^(9/2)) - (b*e^(7/2)*n*(Log[Sqrt[e] - Sqrt[-d]*x^(1/3)]*(Log[Sqrt[e ] - Sqrt[-d]*x^(1/3)] + 2*Log[(1 + (Sqrt[-d]*x^(1/3))/Sqrt[e])/2] - 4*Log[ (Sqrt[-d]*x^(1/3))/Sqrt[e]]) - 4*PolyLog[2, 1 - (Sqrt[-d]*x^(1/3))/Sqrt[e] ] + 2*PolyLog[2, 1/2 - (Sqrt[-d]*x^(1/3))/(2*Sqrt[e])]))/(4*(-d)^(9/2)) + (b*e^(7/2)*n*(Log[Sqrt[e] + Sqrt[-d]*x^(1/3)]*(Log[Sqrt[e] + Sqrt[-d]*x^(1 /3)] + 2*Log[1/2 - (Sqrt[-d]*x^(1/3))/(2*Sqrt[e])] - 4*Log[-((Sqrt[-d]*x^( 1/3))/Sqrt[e])]) + 2*PolyLog[2, (1 + (Sqrt[-d]*x^(1/3))/Sqrt[e])/2] - 4*Po lyLog[2, 1 + (Sqrt[-d]*x^(1/3))/Sqrt[e]]))/(4*(-d)^(9/2))))/3
Time = 0.84 (sec) , antiderivative size = 441, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2908, 2907, 2005, 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.
| \(\displaystyle \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 2908 |
| \(\displaystyle 3 \int x^{8/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 2907 |
| \(\displaystyle 3 \left (\frac {4}{9} b e n \int \frac {x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d+\frac {e}{x^{2/3}}}d\sqrt [3]{x}+\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2\right )\) |
| \(\Big \downarrow \) 2005 |
| \(\displaystyle 3 \left (\frac {4}{9} b e n \int \frac {x^{8/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{x^{2/3} d+e}d\sqrt [3]{x}+\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2\right )\) |
| \(\Big \downarrow \) 2926 |
| \(\displaystyle 3 \left (\frac {4}{9} b e n \int \left (\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) e^4}{d^4 \left (x^{2/3} d+e\right )}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) e^3}{d^4}+\frac {x^{2/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) e^2}{d^3}-\frac {x^{4/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) e}{d^2}+\frac {x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d}\right )d\sqrt [3]{x}+\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {4}{9} b e n \left (\frac {e^{7/2} \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{9/2}}+\frac {e^2 x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 d^3}-\frac {e x^{5/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{5 d^2}+\frac {x^{7/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{7 d}-\frac {a e^3 \sqrt [3]{x}}{d^4}-\frac {i b e^{7/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{9/2}}-\frac {352 b e^{7/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{105 d^{9/2}}+\frac {2 b e^{7/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{d^{9/2}}-\frac {b e^3 \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d^4}-\frac {i b e^{7/2} n \operatorname {PolyLog}\left (2,\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}-1\right )}{d^{9/2}}+\frac {142 b e^3 n \sqrt [3]{x}}{105 d^4}-\frac {8 b e^2 n x}{35 d^3}+\frac {2 b e n x^{5/3}}{35 d^2}\right )\right )\) |
3*((x^3*(a + b*Log[c*(d + e/x^(2/3))^n])^2)/9 + (4*b*e*n*(-((a*e^3*x^(1/3) )/d^4) + (142*b*e^3*n*x^(1/3))/(105*d^4) - (8*b*e^2*n*x)/(35*d^3) + (2*b*e *n*x^(5/3))/(35*d^2) - (352*b*e^(7/2)*n*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]) /(105*d^(9/2)) - (I*b*e^(7/2)*n*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]^2)/d^(9/ 2) + (2*b*e^(7/2)*n*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]*Log[2 - (2*Sqrt[e])/ (Sqrt[e] - I*Sqrt[d]*x^(1/3))])/d^(9/2) - (b*e^3*x^(1/3)*Log[c*(d + e/x^(2 /3))^n])/d^4 + (e^2*x*(a + b*Log[c*(d + e/x^(2/3))^n]))/(3*d^3) - (e*x^(5/ 3)*(a + b*Log[c*(d + e/x^(2/3))^n]))/(5*d^2) + (x^(7/3)*(a + b*Log[c*(d + e/x^(2/3))^n]))/(7*d) + (e^(7/2)*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]*(a + b* Log[c*(d + e/x^(2/3))^n]))/d^(9/2) - (I*b*e^(7/2)*n*PolyLog[2, -1 + (2*Sqr t[e])/(Sqrt[e] - I*Sqrt[d]*x^(1/3))])/d^(9/2)))/9)
3.6.21.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
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]
\[\int x^{2} {\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )}^{2}d x\]
\[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=\int { {\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{2} x^{2} \,d x } \]
integral(b^2*x^2*log(c*((d*x + e*x^(1/3))/x)^n)^2 + 2*a*b*x^2*log(c*((d*x + e*x^(1/3))/x)^n) + a^2*x^2, x)
Timed out. \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=\text {Timed out} \]
Exception generated. \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=\int { {\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{2} x^{2} \,d x } \]
Timed out. \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx=\int x^2\,{\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )\right )}^2 \,d x \]