Integrand size = 24, antiderivative size = 276 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=\frac {3 b^2 d n^2 \left (d+\frac {e}{x^{2/3}}\right )^2}{4 e^3}-\frac {b^2 n^2 \left (d+\frac {e}{x^{2/3}}\right )^3}{9 e^3}-\frac {3 b^2 d^2 n^2}{e^2 x^{2/3}}+\frac {b^2 d^3 n^2 \log ^2\left (d+\frac {e}{x^{2/3}}\right )}{2 e^3}+\frac {3 b d^2 n \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{e^3}-\frac {3 b d n \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{2 e^3}+\frac {b n \left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 e^3}-\frac {b d^3 n \log \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{e^3}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{2 x^2} \] Output:
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/e^2/x^(2/3)+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*(a+b*ln(c* (d+e/x^(2/3))^n))^2/x^2
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.63 (sec) , antiderivative size = 691, normalized size of antiderivative = 2.50 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(2/3))^n])^2/x^3,x]
Output:
(-18*e^3*(a + b*Log[c*(d + e/x^(2/3))^n])^2 + b*n*(9*b*d*n*x^(2/3)*(e*(e - 2*d*x^(2/3)) + 2*d^2*x^(4/3)*Log[d + e/x^(2/3)]) - 2*b*n*(e*(2*e^2 - 3*d* e*x^(2/3) + 6*d^2*x^(4/3)) - 6*d^3*x^2*Log[d + e/x^(2/3)]) + 12*e^3*(a + b *Log[c*(d + e/x^(2/3))^n]) - 18*d*e^2*x^(2/3)*(a + b*Log[c*(d + e/x^(2/3)) ^n]) + 36*d^2*x^(4/3)*(e*(a - b*n) + b*(e + d*x^(2/3))*Log[c*(d + e/x^(2/3 ))^n]) - 36*d^3*x^2*(a + b*Log[c*(d + e/x^(2/3))^n])*Log[Sqrt[e] - Sqrt[-d ]*x^(1/3)] - 36*d^3*x^2*(a + b*Log[c*(d + e/x^(2/3))^n])*Log[Sqrt[e] + Sqr t[-d]*x^(1/3)] - 36*d^3*x^2*((a + b*Log[c*(d + e/x^(2/3))^n])*Log[-(e/(d*x ^(2/3)))] + b*n*PolyLog[2, 1 + e/(d*x^(2/3))]) + 18*b*d^3*n*x^2*(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])]) + 18*b*d^3*n*x^2*(Log[Sqrt[e] + Sqrt[-d]*x^(1/3)]*(Log[Sqr t[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*PolyLog[2, 1 + (Sqrt[-d]*x^(1/3))/Sqrt[e]])))/(36*e^3*x^ 2)
Time = 0.86 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.72, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle -\frac {3}{2} \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^{4/3}}d\frac {1}{x^{2/3}}\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle -\frac {3}{2} \left (\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 x^2}-\frac {2}{3} b e n \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{\left (d+\frac {e}{x^{2/3}}\right ) x^2}d\frac {1}{x^{2/3}}\right )\) |
| \(\Big \downarrow \) 2858 |
| \(\displaystyle -\frac {3}{2} \left (\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 x^2}-\frac {2}{3} b n \int \frac {a+b \log \left (c x^{-2 n/3}\right )}{x^{4/3}}d\left (d+\frac {e}{x^{2/3}}\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {3}{2} \left (\frac {2}{3} b n \int -\frac {a+b \log \left (c x^{-2 n/3}\right )}{x^{4/3}}d\left (d+\frac {e}{x^{2/3}}\right )+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 x^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3}{2} \left (\frac {2 b n \int -\frac {e^3 \left (a+b \log \left (c x^{-2 n/3}\right )\right )}{x^{4/3}}d\left (d+\frac {e}{x^{2/3}}\right )}{3 e^3}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 x^2}\right )\) |
| \(\Big \downarrow \) 2772 |
| \(\displaystyle -\frac {3}{2} \left (\frac {2 b n \left (-b n \int \left (x^{2/3} \log \left (d+\frac {e}{x^{2/3}}\right ) d^3-3 d^2+\frac {3}{2} \left (d+\frac {e}{x^{2/3}}\right ) d-\frac {1}{3 x^{4/3}}\right )d\left (d+\frac {e}{x^{2/3}}\right )+d^3 \log \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c x^{-2 n/3}\right )\right )-3 d^2 \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c x^{-2 n/3}\right )\right )+\frac {3 d \left (a+b \log \left (c x^{-2 n/3}\right )\right )}{2 x^{4/3}}-\frac {a+b \log \left (c x^{-2 n/3}\right )}{3 x^2}\right )}{3 e^3}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 x^2}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3}{2} \left (\frac {2 b n \left (d^3 \log \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c x^{-2 n/3}\right )\right )-3 d^2 \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c x^{-2 n/3}\right )\right )+\frac {3 d \left (a+b \log \left (c x^{-2 n/3}\right )\right )}{2 x^{4/3}}-\frac {a+b \log \left (c x^{-2 n/3}\right )}{3 x^2}-b n \left (\frac {1}{2} d^3 \log ^2\left (d+\frac {e}{x^{2/3}}\right )-3 d^2 \left (d+\frac {e}{x^{2/3}}\right )+\frac {3 d}{4 x^{4/3}}-\frac {1}{9 x^2}\right )\right )}{3 e^3}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 x^2}\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(2/3))^n])^2/x^3,x]
Output:
(-3*((a + b*Log[c*(d + e/x^(2/3))^n])^2/(3*x^2) + (2*b*n*(-(b*n*(-3*d^2*(d + e/x^(2/3)) - 1/(9*x^2) + (3*d)/(4*x^(4/3)) + (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)]) - (a + b*Log[c/x^( (2*n)/3)])/(3*x^2) + (3*d*(a + b*Log[c/x^((2*n)/3)]))/(2*x^(4/3)) + d^3*Lo g[d + e/x^(2/3)]*(a + b*Log[c/x^((2*n)/3)])))/(3*e^3)))/2
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])
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )}^{2}}{x^{3}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(2/3))^n))^2/x^3,x)
Output:
int((a+b*ln(c*(d+e/x^(2/3))^n))^2/x^3,x)
Time = 0.11 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=-\frac {4 \, b^{2} e^{3} n^{2} + 18 \, b^{2} e^{3} \log \left (c\right )^{2} - 12 \, a b e^{3} n + 18 \, a^{2} e^{3} + 18 \, {\left (b^{2} d^{3} n^{2} x^{2} + b^{2} e^{3} n^{2}\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )^{2} - 12 \, {\left (b^{2} e^{3} n - 3 \, a b e^{3}\right )} \log \left (c\right ) - 6 \, {\left (6 \, b^{2} d^{2} e n^{2} x^{\frac {4}{3}} - 3 \, b^{2} d e^{2} n^{2} x^{\frac {2}{3}} + 2 \, b^{2} e^{3} n^{2} - 6 \, a b e^{3} n + {\left (11 \, b^{2} d^{3} n^{2} - 6 \, a b d^{3} n\right )} x^{2} - 6 \, {\left (b^{2} d^{3} n x^{2} + b^{2} e^{3} n\right )} \log \left (c\right )\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) - 3 \, {\left (5 \, b^{2} d e^{2} n^{2} - 6 \, b^{2} d e^{2} n \log \left (c\right ) - 6 \, a b d e^{2} n\right )} x^{\frac {2}{3}} - 6 \, {\left (6 \, b^{2} d^{2} e n x \log \left (c\right ) - {\left (11 \, b^{2} d^{2} e n^{2} - 6 \, a b d^{2} e n\right )} x\right )} x^{\frac {1}{3}}}{36 \, e^{3} x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^2/x^3,x, algorithm="fricas")
Output:
-1/36*(4*b^2*e^3*n^2 + 18*b^2*e^3*log(c)^2 - 12*a*b*e^3*n + 18*a^2*e^3 + 1 8*(b^2*d^3*n^2*x^2 + b^2*e^3*n^2)*log((d*x + e*x^(1/3))/x)^2 - 12*(b^2*e^3 *n - 3*a*b*e^3)*log(c) - 6*(6*b^2*d^2*e*n^2*x^(4/3) - 3*b^2*d*e^2*n^2*x^(2 /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^2 - 6 *(b^2*d^3*n*x^2 + b^2*e^3*n)*log(c))*log((d*x + e*x^(1/3))/x) - 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^(2/3) - 6*(6*b^2*d^2*e* n*x*log(c) - (11*b^2*d^2*e*n^2 - 6*a*b*d^2*e*n)*x)*x^(1/3))/(e^3*x^2)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(2/3))**n))**2/x**3,x)
Output:
Timed out
Time = 0.07 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=-\frac {1}{6} \, a b e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {2}{3}} + e\right )}{e^{4}} - \frac {6 \, d^{3} \log \left (x^{\frac {2}{3}}\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {4}{3}} - 3 \, d e x^{\frac {2}{3}} + 2 \, e^{2}}{e^{3} x^{2}}\right )} - \frac {1}{36} \, {\left (6 \, e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {2}{3}} + e\right )}{e^{4}} - \frac {6 \, d^{3} \log \left (x^{\frac {2}{3}}\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {4}{3}} - 3 \, d e x^{\frac {2}{3}} + 2 \, e^{2}}{e^{3} x^{2}}\right )} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) - \frac {{\left (18 \, d^{3} x^{2} \log \left (d x^{\frac {2}{3}} + e\right )^{2} + 8 \, d^{3} x^{2} \log \left (x\right )^{2} - 44 \, d^{3} x^{2} \log \left (x\right ) - 66 \, d^{2} e x^{\frac {4}{3}} + 15 \, d e^{2} x^{\frac {2}{3}} - 4 \, e^{3} - 6 \, {\left (4 \, d^{3} x^{2} \log \left (x\right ) - 11 \, d^{3} x^{2}\right )} \log \left (d x^{\frac {2}{3}} + e\right )\right )} n^{2}}{e^{3} x^{2}}\right )} b^{2} - \frac {b^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )^{2}}{2 \, x^{2}} - \frac {a b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )}{x^{2}} - \frac {a^{2}}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^2/x^3,x, algorithm="maxima")
Output:
-1/6*a*b*e*n*(6*d^3*log(d*x^(2/3) + e)/e^4 - 6*d^3*log(x^(2/3))/e^4 - (6*d ^2*x^(4/3) - 3*d*e*x^(2/3) + 2*e^2)/(e^3*x^2)) - 1/36*(6*e*n*(6*d^3*log(d* x^(2/3) + e)/e^4 - 6*d^3*log(x^(2/3))/e^4 - (6*d^2*x^(4/3) - 3*d*e*x^(2/3) + 2*e^2)/(e^3*x^2))*log(c*(d + e/x^(2/3))^n) - (18*d^3*x^2*log(d*x^(2/3) + e)^2 + 8*d^3*x^2*log(x)^2 - 44*d^3*x^2*log(x) - 66*d^2*e*x^(4/3) + 15*d* e^2*x^(2/3) - 4*e^3 - 6*(4*d^3*x^2*log(x) - 11*d^3*x^2)*log(d*x^(2/3) + e) )*n^2/(e^3*x^2))*b^2 - 1/2*b^2*log(c*(d + e/x^(2/3))^n)^2/x^2 - a*b*log(c* (d + e/x^(2/3))^n)/x^2 - 1/2*a^2/x^2
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{2}}{x^{3}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^2/x^3,x, algorithm="giac")
Output:
integrate((b*log(c*(d + e/x^(2/3))^n) + a)^2/x^3, x)
Time = 25.55 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=\frac {\frac {d\,\left (\frac {3\,a^2}{2}-a\,b\,n+\frac {b^2\,n^2}{3}\right )}{2\,e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{4\,e}}{x^{4/3}}-{\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}^2\,\left (\frac {b^2}{2\,x^2}+\frac {b^2\,d^3}{2\,e^3}\right )-\frac {\frac {a^2}{2}-\frac {a\,b\,n}{3}+\frac {b^2\,n^2}{9}}{x^2}-\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )\,\left (\frac {b\,\left (3\,a-b\,n\right )}{3\,x^2}-\frac {\frac {b\,d\,\left (3\,a-b\,n\right )}{2\,e}-\frac {3\,a\,b\,d}{2\,e}}{x^{4/3}}+\frac {d\,\left (\frac {b\,d\,\left (3\,a-b\,n\right )}{e}-\frac {3\,a\,b\,d}{e}\right )}{e\,x^{2/3}}\right )-\frac {\frac {d\,\left (\frac {d\,\left (\frac {3\,a^2}{2}-a\,b\,n+\frac {b^2\,n^2}{3}\right )}{e}-\frac {d\,\left (3\,a^2-b^2\,n^2\right )}{2\,e}\right )}{e}+\frac {b^2\,d^2\,n^2}{e^2}}{x^{2/3}}+\frac {\ln \left (d+\frac {e}{x^{2/3}}\right )\,\left (11\,b^2\,d^3\,n^2-6\,a\,b\,d^3\,n\right )}{6\,e^3} \] Input:
int((a + b*log(c*(d + e/x^(2/3))^n))^2/x^3,x)
Output:
((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^(4/3) - log(c*(d + e/x^(2/3))^n)^2*(b^2/(2*x^2) + (b^2*d^3)/(2*e^3)) - (a^2/2 + (b^2*n^2)/9 - (a*b*n)/3)/x^2 - log(c*(d + e/x^(2/3))^n)*((b*(3*a - b*n))/(3*x^2) - ((b*d*(3*a - b*n))/(2*e) - (3*a*b*d)/(2*e))/x^(4/3) + ( d*((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)/x^(2/3) + (log(d + e/x^(2/3))*(11*b^2*d^3*n^2 - 6*a*b*d^3*n))/(6*e^3 )
Time = 0.16 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^3} \, dx=\frac {-18 x^{\frac {2}{3}} \mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b^{2} d \,e^{2} n -18 x^{\frac {2}{3}} a b d \,e^{2} n +15 x^{\frac {2}{3}} b^{2} d \,e^{2} n^{2}+36 x^{\frac {4}{3}} \mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b^{2} d^{2} e n +36 x^{\frac {4}{3}} a b \,d^{2} e n -66 x^{\frac {4}{3}} b^{2} d^{2} e \,n^{2}-18 {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{2} b^{2} d^{3} x^{2}-18 {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{2} b^{2} e^{3}-36 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) a b \,d^{3} x^{2}-36 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) a b \,e^{3}+66 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b^{2} d^{3} n \,x^{2}+12 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b^{2} e^{3} n -18 a^{2} e^{3}+12 a b \,e^{3} n -4 b^{2} e^{3} n^{2}}{36 e^{3} x^{2}} \] Input:
int((a+b*log(c*(d+e/x^(2/3))^n))^2/x^3,x)
Output:
( - 18*x**(2/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b**2*d*e**2*n - 18*x**(2/3)*a*b*d*e**2*n + 15*x**(2/3)*b**2*d*e**2*n**2 + 36*x**(1/3)*log( ((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b**2*d**2*e*n*x + 36*x**(1/3)*a*b*d* *2*e*n*x - 66*x**(1/3)*b**2*d**2*e*n**2*x - 18*log(((x**(2/3)*d + e)**n*c) /x**((2*n)/3))**2*b**2*d**3*x**2 - 18*log(((x**(2/3)*d + e)**n*c)/x**((2*n )/3))**2*b**2*e**3 - 36*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*a*b*d**3 *x**2 - 36*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*a*b*e**3 + 66*log(((x **(2/3)*d + e)**n*c)/x**((2*n)/3))*b**2*d**3*n*x**2 + 12*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b**2*e**3*n - 18*a**2*e**3 + 12*a*b*e**3*n - 4*b* *2*e**3*n**2)/(36*e**3*x**2)