Integrand size = 24, antiderivative size = 269 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=\frac {3 b^2 d n^2 \left (d+\frac {e}{\sqrt [3]{x}}\right )^2}{2 e^3}-\frac {2 b^2 n^2 \left (d+\frac {e}{\sqrt [3]{x}}\right )^3}{9 e^3}-\frac {6 b^2 d^2 n^2}{e^2 \sqrt [3]{x}}+\frac {b^2 d^3 n^2 \log ^2\left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^3}+\frac {6 b d^2 n \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )}{e^3}-\frac {3 b d n \left (d+\frac {e}{\sqrt [3]{x}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )}{e^3}+\frac {2 b n \left (d+\frac {e}{\sqrt [3]{x}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )}{3 e^3}-\frac {2 b d^3 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )}{e^3}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x} \]
3/2*b^2*d*n^2*(d+e/x^(1/3))^2/e^3-2/9*b^2*n^2*(d+e/x^(1/3))^3/e^3-6*b^2*d^ 2*n^2/e^2/x^(1/3)+b^2*d^3*n^2*ln(d+e/x^(1/3))^2/e^3+6*b*d^2*n*(d+e/x^(1/3) )*(a+b*ln(c*(d+e/x^(1/3))^n))/e^3-3*b*d*n*(d+e/x^(1/3))^2*(a+b*ln(c*(d+e/x ^(1/3))^n))/e^3+2/3*b*n*(d+e/x^(1/3))^3*(a+b*ln(c*(d+e/x^(1/3))^n))/e^3-2* b*d^3*n*ln(d+e/x^(1/3))*(a+b*ln(c*(d+e/x^(1/3))^n))/e^3-(a+b*ln(c*(d+e/x^( 1/3))^n))^2/x
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.26 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=\frac {-18 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2+\frac {b n \left (-2 b e n \left (2 e^2-3 d e \sqrt [3]{x}+6 d^2 x^{2/3}\right )+9 b d e n \left (e-2 d \sqrt [3]{x}\right ) \sqrt [3]{x}+36 a d^2 e x^{2/3}-36 b d^2 e n x^{2/3}+30 b d^3 n x \log \left (d+\frac {e}{\sqrt [3]{x}}\right )+36 b d^2 \left (e+d \sqrt [3]{x}\right ) x^{2/3} \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+12 e^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-18 d e^2 \sqrt [3]{x} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-36 d^3 x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \log \left (e+d \sqrt [3]{x}\right )-36 d^3 x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \log \left (-\frac {e}{d \sqrt [3]{x}}\right )+18 b d^3 n x \log \left (e+d \sqrt [3]{x}\right ) \left (\log \left (e+d \sqrt [3]{x}\right )-2 \log \left (-\frac {d \sqrt [3]{x}}{e}\right )\right )-36 b d^3 n x \operatorname {PolyLog}\left (2,1+\frac {e}{d \sqrt [3]{x}}\right )-36 b d^3 n x \operatorname {PolyLog}\left (2,1+\frac {d \sqrt [3]{x}}{e}\right )\right )}{e^3}}{18 x} \]
(-18*(a + b*Log[c*(d + e/x^(1/3))^n])^2 + (b*n*(-2*b*e*n*(2*e^2 - 3*d*e*x^ (1/3) + 6*d^2*x^(2/3)) + 9*b*d*e*n*(e - 2*d*x^(1/3))*x^(1/3) + 36*a*d^2*e* x^(2/3) - 36*b*d^2*e*n*x^(2/3) + 30*b*d^3*n*x*Log[d + e/x^(1/3)] + 36*b*d^ 2*(e + d*x^(1/3))*x^(2/3)*Log[c*(d + e/x^(1/3))^n] + 12*e^3*(a + b*Log[c*( d + e/x^(1/3))^n]) - 18*d*e^2*x^(1/3)*(a + b*Log[c*(d + e/x^(1/3))^n]) - 3 6*d^3*x*(a + b*Log[c*(d + e/x^(1/3))^n])*Log[e + d*x^(1/3)] - 36*d^3*x*(a + b*Log[c*(d + e/x^(1/3))^n])*Log[-(e/(d*x^(1/3)))] + 18*b*d^3*n*x*Log[e + d*x^(1/3)]*(Log[e + d*x^(1/3)] - 2*Log[-((d*x^(1/3))/e)]) - 36*b*d^3*n*x* PolyLog[2, 1 + e/(d*x^(1/3))] - 36*b*d^3*n*x*PolyLog[2, 1 + (d*x^(1/3))/e] ))/e^3)/(18*x)
Time = 0.47 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.73, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2904, 2845, 2858, 25, 27, 2772, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -3 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^{2/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 2845 |
\(\displaystyle -3 \left (\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{3 x}-\frac {2}{3} b e n \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{\left (d+\frac {e}{\sqrt [3]{x}}\right ) x}d\frac {1}{\sqrt [3]{x}}\right )\) |
\(\Big \downarrow \) 2858 |
\(\displaystyle -3 \left (\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{3 x}-\frac {2}{3} b n \int \frac {a+b \log \left (c x^{-n/3}\right )}{x^{2/3}}d\left (d+\frac {e}{\sqrt [3]{x}}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -3 \left (\frac {2}{3} b n \int -\frac {a+b \log \left (c x^{-n/3}\right )}{x^{2/3}}d\left (d+\frac {e}{\sqrt [3]{x}}\right )+\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{3 x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -3 \left (\frac {2 b n \int -\frac {e^3 \left (a+b \log \left (c x^{-n/3}\right )\right )}{x^{2/3}}d\left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 e^3}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{3 x}\right )\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -3 \left (\frac {2 b n \left (-b n \int \left (\sqrt [3]{x} \log \left (d+\frac {e}{\sqrt [3]{x}}\right ) d^3-3 d^2+\frac {3}{2} \left (d+\frac {e}{\sqrt [3]{x}}\right ) d-\frac {1}{3 x^{2/3}}\right )d\left (d+\frac {e}{\sqrt [3]{x}}\right )+d^3 \log \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c x^{-n/3}\right )\right )-3 d^2 \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c x^{-n/3}\right )\right )+\frac {3 d \left (a+b \log \left (c x^{-n/3}\right )\right )}{2 x^{2/3}}-\frac {a+b \log \left (c x^{-n/3}\right )}{3 x}\right )}{3 e^3}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{3 x}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\frac {2 b n \left (d^3 \log \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c x^{-n/3}\right )\right )-3 d^2 \left (d+\frac {e}{\sqrt [3]{x}}\right ) \left (a+b \log \left (c x^{-n/3}\right )\right )+\frac {3 d \left (a+b \log \left (c x^{-n/3}\right )\right )}{2 x^{2/3}}-\frac {a+b \log \left (c x^{-n/3}\right )}{3 x}-b n \left (\frac {1}{2} d^3 \log ^2\left (d+\frac {e}{\sqrt [3]{x}}\right )-3 d^2 \left (d+\frac {e}{\sqrt [3]{x}}\right )+\frac {3 d}{4 x^{2/3}}-\frac {1}{9 x}\right )\right )}{3 e^3}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{3 x}\right )\) |
-3*((a + b*Log[c*(d + e/x^(1/3))^n])^2/(3*x) + (2*b*n*(-(b*n*(-3*d^2*(d + e/x^(1/3)) - 1/(9*x) + (3*d)/(4*x^(2/3)) + (d^3*Log[d + e/x^(1/3)]^2)/2)) - 3*d^2*(d + e/x^(1/3))*(a + b*Log[c/x^(n/3)]) - (a + b*Log[c/x^(n/3)])/(3 *x) + (3*d*(a + b*Log[c/x^(n/3)]))/(2*x^(2/3)) + d^3*Log[d + e/x^(1/3)]*(a + b*Log[c/x^(n/3)])))/(3*e^3))
3.6.1.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ .))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^r)^q, x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] && !(EqQ[q , 1] && EqQ[m, -1])
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^ n])^p/(g*(q + 1))), x] - Simp[b*e*n*(p/(g*(q + 1))) Int[(f + g*x)^(q + 1) *((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && In tegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_ .)*(x_))^(q_.)*((h_.) + (i_.)*(x_))^(r_.), x_Symbol] :> Simp[1/e Subst[In t[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[r, 0]) && IntegerQ[2*r]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )\right )}^{2}}{x^{2}}d x\]
Time = 0.33 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=-\frac {4 \, b^{2} e^{3} n^{2} - 12 \, a b e^{3} n + 18 \, a^{2} e^{3} - 18 \, {\left (b^{2} e^{3} x - b^{2} e^{3}\right )} \log \left (c\right )^{2} + 18 \, {\left (b^{2} d^{3} n^{2} x + b^{2} e^{3} n^{2}\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right )^{2} - 2 \, {\left (2 \, b^{2} e^{3} n^{2} - 6 \, a b e^{3} n + 9 \, a^{2} e^{3}\right )} x - 12 \, {\left (b^{2} e^{3} n - 3 \, a b e^{3} - {\left (b^{2} e^{3} n - 3 \, a b e^{3}\right )} x\right )} \log \left (c\right ) - 6 \, {\left (6 \, b^{2} d^{2} e n^{2} x^{\frac {2}{3}} - 3 \, b^{2} d e^{2} n^{2} x^{\frac {1}{3}} + 2 \, b^{2} e^{3} n^{2} - 6 \, a b e^{3} n + {\left (11 \, b^{2} d^{3} n^{2} - 6 \, a b d^{3} n\right )} x - 6 \, {\left (b^{2} d^{3} n x + b^{2} e^{3} n\right )} \log \left (c\right )\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 6 \, {\left (11 \, b^{2} d^{2} e n^{2} - 6 \, b^{2} d^{2} e n \log \left (c\right ) - 6 \, a b d^{2} e n\right )} x^{\frac {2}{3}} - 3 \, {\left (5 \, b^{2} d e^{2} n^{2} - 6 \, b^{2} d e^{2} n \log \left (c\right ) - 6 \, a b d e^{2} n\right )} x^{\frac {1}{3}}}{18 \, e^{3} x} \]
-1/18*(4*b^2*e^3*n^2 - 12*a*b*e^3*n + 18*a^2*e^3 - 18*(b^2*e^3*x - b^2*e^3 )*log(c)^2 + 18*(b^2*d^3*n^2*x + b^2*e^3*n^2)*log((d*x + e*x^(2/3))/x)^2 - 2*(2*b^2*e^3*n^2 - 6*a*b*e^3*n + 9*a^2*e^3)*x - 12*(b^2*e^3*n - 3*a*b*e^3 - (b^2*e^3*n - 3*a*b*e^3)*x)*log(c) - 6*(6*b^2*d^2*e*n^2*x^(2/3) - 3*b^2* d*e^2*n^2*x^(1/3) + 2*b^2*e^3*n^2 - 6*a*b*e^3*n + (11*b^2*d^3*n^2 - 6*a*b* d^3*n)*x - 6*(b^2*d^3*n*x + b^2*e^3*n)*log(c))*log((d*x + e*x^(2/3))/x) + 6*(11*b^2*d^2*e*n^2 - 6*b^2*d^2*e*n*log(c) - 6*a*b*d^2*e*n)*x^(2/3) - 3*(5 *b^2*d*e^2*n^2 - 6*b^2*d*e^2*n*log(c) - 6*a*b*d*e^2*n)*x^(1/3))/(e^3*x)
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=\int \frac {\left (a + b \log {\left (c \left (d + \frac {e}{\sqrt [3]{x}}\right )^{n} \right )}\right )^{2}}{x^{2}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=-\frac {1}{3} \, a b e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{4}} - \frac {2 \, d^{3} \log \left (x\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {2}{3}} - 3 \, d e x^{\frac {1}{3}} + 2 \, e^{2}}{e^{3} x}\right )} - \frac {1}{18} \, {\left (6 \, e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{4}} - \frac {2 \, d^{3} \log \left (x\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {2}{3}} - 3 \, d e x^{\frac {1}{3}} + 2 \, e^{2}}{e^{3} x}\right )} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right ) - \frac {{\left (18 \, d^{3} x \log \left (d x^{\frac {1}{3}} + e\right )^{2} + 2 \, d^{3} x \log \left (x\right )^{2} - 22 \, d^{3} x \log \left (x\right ) - 66 \, d^{2} e x^{\frac {2}{3}} + 15 \, d e^{2} x^{\frac {1}{3}} - 4 \, e^{3} - 6 \, {\left (2 \, d^{3} x \log \left (x\right ) - 11 \, d^{3} x\right )} \log \left (d x^{\frac {1}{3}} + e\right )\right )} n^{2}}{e^{3} x}\right )} b^{2} - \frac {b^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )^{2}}{x} - \frac {2 \, a b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{x} - \frac {a^{2}}{x} \]
-1/3*a*b*e*n*(6*d^3*log(d*x^(1/3) + e)/e^4 - 2*d^3*log(x)/e^4 - (6*d^2*x^( 2/3) - 3*d*e*x^(1/3) + 2*e^2)/(e^3*x)) - 1/18*(6*e*n*(6*d^3*log(d*x^(1/3) + e)/e^4 - 2*d^3*log(x)/e^4 - (6*d^2*x^(2/3) - 3*d*e*x^(1/3) + 2*e^2)/(e^3 *x))*log(c*(d + e/x^(1/3))^n) - (18*d^3*x*log(d*x^(1/3) + e)^2 + 2*d^3*x*l og(x)^2 - 22*d^3*x*log(x) - 66*d^2*e*x^(2/3) + 15*d*e^2*x^(1/3) - 4*e^3 - 6*(2*d^3*x*log(x) - 11*d^3*x)*log(d*x^(1/3) + e))*n^2/(e^3*x))*b^2 - b^2*l og(c*(d + e/x^(1/3))^n)^2/x - 2*a*b*log(c*(d + e/x^(1/3))^n)/x - a^2/x
Time = 0.34 (sec) , antiderivative size = 428, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=-\frac {18 \, {\left (\frac {3 \, {\left (d x^{\frac {1}{3}} + e\right )} b^{2} d^{2} n^{2}}{e^{2} x^{\frac {1}{3}}} - \frac {3 \, {\left (d x^{\frac {1}{3}} + e\right )}^{2} b^{2} d n^{2}}{e^{2} x^{\frac {2}{3}}} + \frac {{\left (d x^{\frac {1}{3}} + e\right )}^{3} b^{2} n^{2}}{e^{2} x}\right )} \log \left (\frac {d x^{\frac {1}{3}} + e}{x^{\frac {1}{3}}}\right )^{2} - 6 \, {\left (\frac {2 \, {\left (b^{2} n^{2} - 3 \, b^{2} n \log \left (c\right ) - 3 \, a b n\right )} {\left (d x^{\frac {1}{3}} + e\right )}^{3}}{e^{2} x} - \frac {9 \, {\left (b^{2} d n^{2} - 2 \, b^{2} d n \log \left (c\right ) - 2 \, a b d n\right )} {\left (d x^{\frac {1}{3}} + e\right )}^{2}}{e^{2} x^{\frac {2}{3}}} + \frac {18 \, {\left (b^{2} d^{2} n^{2} - b^{2} d^{2} n \log \left (c\right ) - a b d^{2} n\right )} {\left (d x^{\frac {1}{3}} + e\right )}}{e^{2} x^{\frac {1}{3}}}\right )} \log \left (\frac {d x^{\frac {1}{3}} + e}{x^{\frac {1}{3}}}\right ) + \frac {2 \, {\left (2 \, b^{2} n^{2} - 6 \, b^{2} n \log \left (c\right ) + 9 \, b^{2} \log \left (c\right )^{2} - 6 \, a b n + 18 \, a b \log \left (c\right ) + 9 \, a^{2}\right )} {\left (d x^{\frac {1}{3}} + e\right )}^{3}}{e^{2} x} - \frac {27 \, {\left (b^{2} d n^{2} - 2 \, b^{2} d n \log \left (c\right ) + 2 \, b^{2} d \log \left (c\right )^{2} - 2 \, a b d n + 4 \, a b d \log \left (c\right ) + 2 \, a^{2} d\right )} {\left (d x^{\frac {1}{3}} + e\right )}^{2}}{e^{2} x^{\frac {2}{3}}} + \frac {54 \, {\left (2 \, b^{2} d^{2} n^{2} - 2 \, b^{2} d^{2} n \log \left (c\right ) + b^{2} d^{2} \log \left (c\right )^{2} - 2 \, a b d^{2} n + 2 \, a b d^{2} \log \left (c\right ) + a^{2} d^{2}\right )} {\left (d x^{\frac {1}{3}} + e\right )}}{e^{2} x^{\frac {1}{3}}}}{18 \, e} \]
-1/18*(18*(3*(d*x^(1/3) + e)*b^2*d^2*n^2/(e^2*x^(1/3)) - 3*(d*x^(1/3) + e) ^2*b^2*d*n^2/(e^2*x^(2/3)) + (d*x^(1/3) + e)^3*b^2*n^2/(e^2*x))*log((d*x^( 1/3) + e)/x^(1/3))^2 - 6*(2*(b^2*n^2 - 3*b^2*n*log(c) - 3*a*b*n)*(d*x^(1/3 ) + e)^3/(e^2*x) - 9*(b^2*d*n^2 - 2*b^2*d*n*log(c) - 2*a*b*d*n)*(d*x^(1/3) + e)^2/(e^2*x^(2/3)) + 18*(b^2*d^2*n^2 - b^2*d^2*n*log(c) - a*b*d^2*n)*(d *x^(1/3) + e)/(e^2*x^(1/3)))*log((d*x^(1/3) + e)/x^(1/3)) + 2*(2*b^2*n^2 - 6*b^2*n*log(c) + 9*b^2*log(c)^2 - 6*a*b*n + 18*a*b*log(c) + 9*a^2)*(d*x^( 1/3) + e)^3/(e^2*x) - 27*(b^2*d*n^2 - 2*b^2*d*n*log(c) + 2*b^2*d*log(c)^2 - 2*a*b*d*n + 4*a*b*d*log(c) + 2*a^2*d)*(d*x^(1/3) + e)^2/(e^2*x^(2/3)) + 54*(2*b^2*d^2*n^2 - 2*b^2*d^2*n*log(c) + b^2*d^2*log(c)^2 - 2*a*b*d^2*n + 2*a*b*d^2*log(c) + a^2*d^2)*(d*x^(1/3) + e)/(e^2*x^(1/3)))/e
Time = 1.81 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )^2}{x^2} \, dx=\frac {\frac {d\,\left (3\,a^2-2\,a\,b\,n+\frac {2\,b^2\,n^2}{3}\right )}{2\,e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{2\,e}}{x^{2/3}}-{\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}^2\,\left (\frac {b^2}{x}+\frac {b^2\,d^3}{e^3}\right )-\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )\,\left (\frac {2\,b\,\left (3\,a-b\,n\right )}{3\,x}-\frac {\frac {b\,d\,\left (3\,a-b\,n\right )}{e}-\frac {3\,a\,b\,d}{e}}{x^{2/3}}+\frac {d\,\left (\frac {2\,b\,d\,\left (3\,a-b\,n\right )}{e}-\frac {6\,a\,b\,d}{e}\right )}{e\,x^{1/3}}\right )-\frac {\frac {d\,\left (\frac {d\,\left (3\,a^2-2\,a\,b\,n+\frac {2\,b^2\,n^2}{3}\right )}{e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{e}\right )}{e}+\frac {2\,b^2\,d^2\,n^2}{e^2}}{x^{1/3}}-\frac {a^2-\frac {2\,a\,b\,n}{3}+\frac {2\,b^2\,n^2}{9}}{x}+\frac {\ln \left (d+\frac {e}{x^{1/3}}\right )\,\left (11\,b^2\,d^3\,n^2-6\,a\,b\,d^3\,n\right )}{3\,e^3} \]
((d*(3*a^2 + (2*b^2*n^2)/3 - 2*a*b*n))/(2*e) - (d*(3*a^2 - b^2*n^2))/(2*e) )/x^(2/3) - log(c*(d + e/x^(1/3))^n)^2*(b^2/x + (b^2*d^3)/e^3) - log(c*(d + e/x^(1/3))^n)*((2*b*(3*a - b*n))/(3*x) - ((b*d*(3*a - b*n))/e - (3*a*b*d )/e)/x^(2/3) + (d*((2*b*d*(3*a - b*n))/e - (6*a*b*d)/e))/(e*x^(1/3))) - (( d*((d*(3*a^2 + (2*b^2*n^2)/3 - 2*a*b*n))/e - (d*(3*a^2 - b^2*n^2))/e))/e + (2*b^2*d^2*n^2)/e^2)/x^(1/3) - (a^2 + (2*b^2*n^2)/9 - (2*a*b*n)/3)/x + (l og(d + e/x^(1/3))*(11*b^2*d^3*n^2 - 6*a*b*d^3*n))/(3*e^3)