Integrand size = 22, antiderivative size = 275 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=-\frac {3 b^2 d n^2 \left (d+e x^{2/3}\right )^2}{4 e^3}+\frac {b^2 n^2 \left (d+e x^{2/3}\right )^3}{9 e^3}+\frac {3 b^2 d^2 n^2 x^{2/3}}{e^2}-\frac {b^2 d^3 n^2 \log ^2\left (d+e x^{2/3}\right )}{2 e^3}-\frac {3 b d^2 n \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{e^3}+\frac {3 b d n \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{2 e^3}-\frac {b n \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 e^3}+\frac {b d^3 n \log \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{e^3}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \]
-3/4*b^2*d*n^2*(d+e*x^(2/3))^2/e^3+1/9*b^2*n^2*(d+e*x^(2/3))^3/e^3+3*b^2*d ^2*n^2*x^(2/3)/e^2-1/2*b^2*d^3*n^2*ln(d+e*x^(2/3))^2/e^3-3*b*d^2*n*(d+e*x^ (2/3))*(a+b*ln(c*(d+e*x^(2/3))^n))/e^3+3/2*b*d*n*(d+e*x^(2/3))^2*(a+b*ln(c *(d+e*x^(2/3))^n))/e^3-1/3*b*n*(d+e*x^(2/3))^3*(a+b*ln(c*(d+e*x^(2/3))^n)) /e^3+b*d^3*n*ln(d+e*x^(2/3))*(a+b*ln(c*(d+e*x^(2/3))^n))/e^3+1/2*x^2*(a+b* ln(c*(d+e*x^(2/3))^n))^2
Time = 0.10 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.87 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {18 a^2 d^3-36 a b d^2 e n x^{2/3}+66 b^2 d^2 e n^2 x^{2/3}+18 a b d e^2 n x^{4/3}-15 b^2 d e^2 n^2 x^{4/3}+18 a^2 e^3 x^2-12 a b e^3 n x^2+4 b^2 e^3 n^2 x^2-30 b^2 d^3 n^2 \log \left (d+e x^{2/3}\right )+6 b \left (6 a \left (d^3+e^3 x^2\right )-b n \left (6 d^3+6 d^2 e x^{2/3}-3 d e^2 x^{4/3}+2 e^3 x^2\right )\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )+18 b^2 \left (d^3+e^3 x^2\right ) \log ^2\left (c \left (d+e x^{2/3}\right )^n\right )}{36 e^3} \]
(18*a^2*d^3 - 36*a*b*d^2*e*n*x^(2/3) + 66*b^2*d^2*e*n^2*x^(2/3) + 18*a*b*d *e^2*n*x^(4/3) - 15*b^2*d*e^2*n^2*x^(4/3) + 18*a^2*e^3*x^2 - 12*a*b*e^3*n* x^2 + 4*b^2*e^3*n^2*x^2 - 30*b^2*d^3*n^2*Log[d + e*x^(2/3)] + 6*b*(6*a*(d^ 3 + e^3*x^2) - b*n*(6*d^3 + 6*d^2*e*x^(2/3) - 3*d*e^2*x^(4/3) + 2*e^3*x^2) )*Log[c*(d + e*x^(2/3))^n] + 18*b^2*(d^3 + e^3*x^2)*Log[c*(d + e*x^(2/3))^ n]^2)/(36*e^3)
Time = 0.47 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.72, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, 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.
| \(\displaystyle \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle \frac {3}{2} \int x^{4/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2dx^{2/3}\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {2}{3} b e n \int \frac {x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d+e x^{2/3}}dx^{2/3}\right )\) |
| \(\Big \downarrow \) 2858 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {2}{3} b n \int x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )d\left (d+e x^{2/3}\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3}{2} \left (\frac {2}{3} b n \int -x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )d\left (d+e x^{2/3}\right )+\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{2} \left (\frac {2 b n \int -e^3 x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )d\left (d+e x^{2/3}\right )}{3 e^3}+\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2\right )\) |
| \(\Big \downarrow \) 2772 |
| \(\displaystyle \frac {3}{2} \left (\frac {2 b n \left (-b n \int \left (\frac {\log \left (d+e x^{2/3}\right ) d^3}{x^{2/3}}-3 d^2+\frac {3}{2} \left (d+e x^{2/3}\right ) d-\frac {x^{4/3}}{3}\right )d\left (d+e x^{2/3}\right )+d^3 \log \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )-3 d^2 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {3}{2} d x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )-\frac {1}{3} x^2 \left (a+b \log \left (c x^{2 n/3}\right )\right )\right )}{3 e^3}+\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{2} \left (\frac {2 b n \left (d^3 \log \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )-3 d^2 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {3}{2} d x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )-\frac {1}{3} x^2 \left (a+b \log \left (c x^{2 n/3}\right )\right )-b n \left (\frac {1}{2} d^3 \log ^2\left (d+e x^{2/3}\right )-3 d^2 \left (d+e x^{2/3}\right )+\frac {3}{4} d x^{4/3}-\frac {x^2}{9}\right )\right )}{3 e^3}+\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2\right )\) |
(3*((x^2*(a + b*Log[c*(d + e*x^(2/3))^n])^2)/3 + (2*b*n*(-(b*n*(-3*d^2*(d + e*x^(2/3)) + (3*d*x^(4/3))/4 - x^2/9 + (d^3*Log[d + e*x^(2/3)]^2)/2)) - 3*d^2*(d + e*x^(2/3))*(a + b*Log[c*x^((2*n)/3)]) + (3*d*x^(4/3)*(a + b*Log [c*x^((2*n)/3)]))/2 - (x^2*(a + b*Log[c*x^((2*n)/3)]))/3 + d^3*Log[d + e*x ^(2/3)]*(a + b*Log[c*x^((2*n)/3)])))/(3*e^3)))/2
3.5.72.3.1 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])
\[\int x {\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{2}d x\]
Time = 0.37 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.11 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {18 \, b^{2} e^{3} x^{2} \log \left (c\right )^{2} - 12 \, {\left (b^{2} e^{3} n - 3 \, a b e^{3}\right )} x^{2} \log \left (c\right ) + 2 \, {\left (2 \, b^{2} e^{3} n^{2} - 6 \, a b e^{3} n + 9 \, a^{2} e^{3}\right )} x^{2} + 18 \, {\left (b^{2} e^{3} n^{2} x^{2} + b^{2} d^{3} n^{2}\right )} \log \left (e x^{\frac {2}{3}} + d\right )^{2} + 6 \, {\left (3 \, b^{2} d e^{2} n^{2} x^{\frac {4}{3}} - 6 \, b^{2} d^{2} e n^{2} x^{\frac {2}{3}} - 11 \, b^{2} d^{3} n^{2} + 6 \, a b d^{3} n - 2 \, {\left (b^{2} e^{3} n^{2} - 3 \, a b e^{3} n\right )} x^{2} + 6 \, {\left (b^{2} e^{3} n x^{2} + b^{2} d^{3} n\right )} \log \left (c\right )\right )} \log \left (e x^{\frac {2}{3}} + d\right ) + 6 \, {\left (11 \, b^{2} d^{2} e n^{2} - 6 \, b^{2} d^{2} e n \log \left (c\right ) - 6 \, a b d^{2} e n\right )} x^{\frac {2}{3}} + 3 \, {\left (6 \, b^{2} d e^{2} n x \log \left (c\right ) - {\left (5 \, b^{2} d e^{2} n^{2} - 6 \, a b d e^{2} n\right )} x\right )} x^{\frac {1}{3}}}{36 \, e^{3}} \]
1/36*(18*b^2*e^3*x^2*log(c)^2 - 12*(b^2*e^3*n - 3*a*b*e^3)*x^2*log(c) + 2* (2*b^2*e^3*n^2 - 6*a*b*e^3*n + 9*a^2*e^3)*x^2 + 18*(b^2*e^3*n^2*x^2 + b^2* d^3*n^2)*log(e*x^(2/3) + d)^2 + 6*(3*b^2*d*e^2*n^2*x^(4/3) - 6*b^2*d^2*e*n ^2*x^(2/3) - 11*b^2*d^3*n^2 + 6*a*b*d^3*n - 2*(b^2*e^3*n^2 - 3*a*b*e^3*n)* x^2 + 6*(b^2*e^3*n*x^2 + b^2*d^3*n)*log(c))*log(e*x^(2/3) + d) + 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*(6*b^2*d*e^ 2*n*x*log(c) - (5*b^2*d*e^2*n^2 - 6*a*b*d*e^2*n)*x)*x^(1/3))/e^3
Timed out. \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.84 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {1}{2} \, b^{2} x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )^{2} + \frac {1}{6} \, a b e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )} + a b x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {1}{2} \, a^{2} x^{2} + \frac {1}{36} \, {\left (6 \, e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {{\left (4 \, e^{3} x^{2} - 18 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )^{2} - 15 \, d e^{2} x^{\frac {4}{3}} - 66 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right ) + 66 \, d^{2} e x^{\frac {2}{3}}\right )} n^{2}}{e^{3}}\right )} b^{2} \]
1/2*b^2*x^2*log((e*x^(2/3) + d)^n*c)^2 + 1/6*a*b*e*n*(6*d^3*log(e*x^(2/3) + d)/e^4 - (2*e^2*x^2 - 3*d*e*x^(4/3) + 6*d^2*x^(2/3))/e^3) + a*b*x^2*log( (e*x^(2/3) + d)^n*c) + 1/2*a^2*x^2 + 1/36*(6*e*n*(6*d^3*log(e*x^(2/3) + d) /e^4 - (2*e^2*x^2 - 3*d*e*x^(4/3) + 6*d^2*x^(2/3))/e^3)*log((e*x^(2/3) + d )^n*c) + (4*e^3*x^2 - 18*d^3*log(e*x^(2/3) + d)^2 - 15*d*e^2*x^(4/3) - 66* d^3*log(e*x^(2/3) + d) + 66*d^2*e*x^(2/3))*n^2/e^3)*b^2
Time = 0.50 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.14 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {1}{2} \, b^{2} x^{2} \log \left (c\right )^{2} + \frac {1}{36} \, {\left (18 \, x^{2} \log \left (e x^{\frac {2}{3}} + d\right )^{2} - {\left (6 \, {\left (\frac {2 \, {\left (e x^{\frac {2}{3}} + d\right )}^{3}}{e^{4}} - \frac {9 \, {\left (e x^{\frac {2}{3}} + d\right )}^{2} d}{e^{4}} + \frac {18 \, {\left (e x^{\frac {2}{3}} + d\right )} d^{2}}{e^{4}}\right )} \log \left (e x^{\frac {2}{3}} + d\right ) - \frac {18 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )^{2}}{e^{4}} - \frac {4 \, {\left (e x^{\frac {2}{3}} + d\right )}^{3}}{e^{4}} + \frac {27 \, {\left (e x^{\frac {2}{3}} + d\right )}^{2} d}{e^{4}} - \frac {108 \, {\left (e x^{\frac {2}{3}} + d\right )} d^{2}}{e^{4}}\right )} e\right )} b^{2} n^{2} + \frac {1}{6} \, {\left (6 \, x^{2} \log \left (e x^{\frac {2}{3}} + d\right ) + e {\left (\frac {6 \, d^{3} \log \left ({\left | e x^{\frac {2}{3}} + d \right |}\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )}\right )} b^{2} n \log \left (c\right ) + a b x^{2} \log \left (c\right ) + \frac {1}{6} \, {\left (6 \, x^{2} \log \left (e x^{\frac {2}{3}} + d\right ) + e {\left (\frac {6 \, d^{3} \log \left ({\left | e x^{\frac {2}{3}} + d \right |}\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )}\right )} a b n + \frac {1}{2} \, a^{2} x^{2} \]
1/2*b^2*x^2*log(c)^2 + 1/36*(18*x^2*log(e*x^(2/3) + d)^2 - (6*(2*(e*x^(2/3 ) + d)^3/e^4 - 9*(e*x^(2/3) + d)^2*d/e^4 + 18*(e*x^(2/3) + d)*d^2/e^4)*log (e*x^(2/3) + d) - 18*d^3*log(e*x^(2/3) + d)^2/e^4 - 4*(e*x^(2/3) + d)^3/e^ 4 + 27*(e*x^(2/3) + d)^2*d/e^4 - 108*(e*x^(2/3) + d)*d^2/e^4)*e)*b^2*n^2 + 1/6*(6*x^2*log(e*x^(2/3) + d) + e*(6*d^3*log(abs(e*x^(2/3) + d))/e^4 - (2 *e^2*x^2 - 3*d*e*x^(4/3) + 6*d^2*x^(2/3))/e^3))*b^2*n*log(c) + a*b*x^2*log (c) + 1/6*(6*x^2*log(e*x^(2/3) + d) + e*(6*d^3*log(abs(e*x^(2/3) + d))/e^4 - (2*e^2*x^2 - 3*d*e*x^(4/3) + 6*d^2*x^(2/3))/e^3))*a*b*n + 1/2*a^2*x^2
Time = 1.73 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.09 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx={\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^2\,\left (\frac {b^2\,x^2}{2}+\frac {b^2\,d^3}{2\,e^3}\right )-x^{4/3}\,\left (\frac {d\,\left (\frac {3\,a^2}{2}-a\,b\,n+\frac {b^2\,n^2}{3}\right )}{2\,e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{4\,e}\right )+x^2\,\left (\frac {a^2}{2}-\frac {a\,b\,n}{3}+\frac {b^2\,n^2}{9}\right )+\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\,\left (\frac {b\,x^2\,\left (3\,a-b\,n\right )}{3}-x^{4/3}\,\left (\frac {b\,d\,\left (3\,a-b\,n\right )}{2\,e}-\frac {3\,a\,b\,d}{2\,e}\right )+\frac {d\,x^{2/3}\,\left (\frac {b\,d\,\left (3\,a-b\,n\right )}{e}-\frac {3\,a\,b\,d}{e}\right )}{e}\right )+x^{2/3}\,\left (\frac {d\,\left (\frac {d\,\left (\frac {3\,a^2}{2}-a\,b\,n+\frac {b^2\,n^2}{3}\right )}{e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{2\,e}\right )}{e}+\frac {b^2\,d^2\,n^2}{e^2}\right )-\frac {\ln \left (d+e\,x^{2/3}\right )\,\left (11\,b^2\,d^3\,n^2-6\,a\,b\,d^3\,n\right )}{6\,e^3} \]
log(c*(d + e*x^(2/3))^n)^2*((b^2*x^2)/2 + (b^2*d^3)/(2*e^3)) - x^(4/3)*((d *((3*a^2)/2 + (b^2*n^2)/3 - a*b*n))/(2*e) - (d*(3*a^2 - b^2*n^2))/(4*e)) + x^2*(a^2/2 + (b^2*n^2)/9 - (a*b*n)/3) + log(c*(d + e*x^(2/3))^n)*((b*x^2* (3*a - b*n))/3 - x^(4/3)*((b*d*(3*a - b*n))/(2*e) - (3*a*b*d)/(2*e)) + (d* x^(2/3)*((b*d*(3*a - b*n))/e - (3*a*b*d)/e))/e) + x^(2/3)*((d*((d*((3*a^2) /2 + (b^2*n^2)/3 - a*b*n))/e - (d*(3*a^2 - b^2*n^2))/(2*e)))/e + (b^2*d^2* n^2)/e^2) - (log(d + e*x^(2/3))*(11*b^2*d^3*n^2 - 6*a*b*d^3*n))/(6*e^3)