Integrand size = 24, antiderivative size = 449 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx=-\frac {9 b^3 d n^3 \left (d+\frac {e}{x^{2/3}}\right )^2}{8 e^3}+\frac {b^3 n^3 \left (d+\frac {e}{x^{2/3}}\right )^3}{9 e^3}-\frac {9 a b^2 d^2 n^2}{e^2 x^{2/3}}+\frac {9 b^3 d^2 n^3}{e^2 x^{2/3}}-\frac {9 b^3 d^2 n^2 \left (d+\frac {e}{x^{2/3}}\right ) \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{e^3}+\frac {9 b^2 d n^2 \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{4 e^3}-\frac {b^2 n^2 \left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 e^3}+\frac {9 b d^2 n \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{2 e^3}-\frac {9 b d n \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{4 e^3}+\frac {b n \left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{2 e^3}-\frac {3 d^2 \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{2 e^3}+\frac {3 d \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{2 e^3}-\frac {\left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{2 e^3} \] Output:
-9/8*b^3*d*n^3*(d+e/x^(2/3))^2/e^3+1/9*b^3*n^3*(d+e/x^(2/3))^3/e^3-9*a*b^2 *d^2*n^2/e^2/x^(2/3)+9*b^3*d^2*n^3/e^2/x^(2/3)-9*b^3*d^2*n^2*(d+e/x^(2/3)) *ln(c*(d+e/x^(2/3))^n)/e^3+9/4*b^2*d*n^2*(d+e/x^(2/3))^2*(a+b*ln(c*(d+e/x^ (2/3))^n))/e^3-1/3*b^2*n^2*(d+e/x^(2/3))^3*(a+b*ln(c*(d+e/x^(2/3))^n))/e^3 +9/2*b*d^2*n*(d+e/x^(2/3))*(a+b*ln(c*(d+e/x^(2/3))^n))^2/e^3-9/4*b*d*n*(d+ e/x^(2/3))^2*(a+b*ln(c*(d+e/x^(2/3))^n))^2/e^3+1/2*b*n*(d+e/x^(2/3))^3*(a+ b*ln(c*(d+e/x^(2/3))^n))^2/e^3-3/2*d^2*(d+e/x^(2/3))*(a+b*ln(c*(d+e/x^(2/3 ))^n))^3/e^3+3/2*d*(d+e/x^(2/3))^2*(a+b*ln(c*(d+e/x^(2/3))^n))^3/e^3-1/2*( d+e/x^(2/3))^3*(a+b*ln(c*(d+e/x^(2/3))^n))^3/e^3
Time = 1.36 (sec) , antiderivative size = 692, normalized size of antiderivative = 1.54 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx=\frac {-36 a^3 e^3+36 a^2 b e^3 n-24 a b^2 e^3 n^2+8 b^3 e^3 n^3-54 a^2 b d e^2 n x^{2/3}+90 a b^2 d e^2 n^2 x^{2/3}-57 b^3 d e^2 n^3 x^{2/3}+108 a^2 b d^2 e n x^{4/3}-396 a b^2 d^2 e n^2 x^{4/3}+510 b^3 d^2 e n^3 x^{4/3}+72 b^3 d^3 n^3 x^2 \log ^3\left (d+\frac {e}{x^{2/3}}\right )-36 b^3 e^3 \log ^3\left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )-108 a^2 b d^3 n x^2 \log \left (e+d x^{2/3}\right )+396 a b^2 d^3 n^2 x^2 \log \left (e+d x^{2/3}\right )-510 b^3 d^3 n^3 x^2 \log \left (e+d x^{2/3}\right )+12 b^2 d^3 n^2 x^2 \log \left (d+\frac {e}{x^{2/3}}\right ) \left (6 a-11 b n+6 b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \left (3 \log \left (e+d x^{2/3}\right )-2 \log (x)\right )+72 a^2 b d^3 n x^2 \log (x)-264 a b^2 d^3 n^2 x^2 \log (x)+340 b^3 d^3 n^3 x^2 \log (x)-18 b^2 d^3 n^2 x^2 \log ^2\left (d+\frac {e}{x^{2/3}}\right ) \left (6 a-11 b n+6 b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )+6 b n \log \left (e+d x^{2/3}\right )-4 b n \log (x)\right )+18 b^2 \log ^2\left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right ) \left (e \left (-6 a e^2+2 b e^2 n-3 b d e n x^{2/3}+6 b d^2 n x^{4/3}\right )-6 b d^3 n x^2 \log \left (e+d x^{2/3}\right )+4 b d^3 n x^2 \log (x)\right )-6 b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right ) \left (18 a^2 e^3-6 a b e n \left (2 e^2-3 d e x^{2/3}+6 d^2 x^{4/3}\right )+b^2 e n^2 \left (4 e^2-15 d e x^{2/3}+66 d^2 x^{4/3}\right )+6 b d^3 n (6 a-11 b n) x^2 \log \left (e+d x^{2/3}\right )+4 b d^3 n (-6 a+11 b n) x^2 \log (x)\right )}{72 e^3 x^2} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(2/3))^n])^3/x^3,x]
Output:
(-36*a^3*e^3 + 36*a^2*b*e^3*n - 24*a*b^2*e^3*n^2 + 8*b^3*e^3*n^3 - 54*a^2* b*d*e^2*n*x^(2/3) + 90*a*b^2*d*e^2*n^2*x^(2/3) - 57*b^3*d*e^2*n^3*x^(2/3) + 108*a^2*b*d^2*e*n*x^(4/3) - 396*a*b^2*d^2*e*n^2*x^(4/3) + 510*b^3*d^2*e* n^3*x^(4/3) + 72*b^3*d^3*n^3*x^2*Log[d + e/x^(2/3)]^3 - 36*b^3*e^3*Log[c*( d + e/x^(2/3))^n]^3 - 108*a^2*b*d^3*n*x^2*Log[e + d*x^(2/3)] + 396*a*b^2*d ^3*n^2*x^2*Log[e + d*x^(2/3)] - 510*b^3*d^3*n^3*x^2*Log[e + d*x^(2/3)] + 1 2*b^2*d^3*n^2*x^2*Log[d + e/x^(2/3)]*(6*a - 11*b*n + 6*b*Log[c*(d + e/x^(2 /3))^n])*(3*Log[e + d*x^(2/3)] - 2*Log[x]) + 72*a^2*b*d^3*n*x^2*Log[x] - 2 64*a*b^2*d^3*n^2*x^2*Log[x] + 340*b^3*d^3*n^3*x^2*Log[x] - 18*b^2*d^3*n^2* x^2*Log[d + e/x^(2/3)]^2*(6*a - 11*b*n + 6*b*Log[c*(d + e/x^(2/3))^n] + 6* b*n*Log[e + d*x^(2/3)] - 4*b*n*Log[x]) + 18*b^2*Log[c*(d + e/x^(2/3))^n]^2 *(e*(-6*a*e^2 + 2*b*e^2*n - 3*b*d*e*n*x^(2/3) + 6*b*d^2*n*x^(4/3)) - 6*b*d ^3*n*x^2*Log[e + d*x^(2/3)] + 4*b*d^3*n*x^2*Log[x]) - 6*b*Log[c*(d + e/x^( 2/3))^n]*(18*a^2*e^3 - 6*a*b*e*n*(2*e^2 - 3*d*e*x^(2/3) + 6*d^2*x^(4/3)) + b^2*e*n^2*(4*e^2 - 15*d*e*x^(2/3) + 66*d^2*x^(4/3)) + 6*b*d^3*n*(6*a - 11 *b*n)*x^2*Log[e + d*x^(2/3)] + 4*b*d^3*n*(-6*a + 11*b*n)*x^2*Log[x]))/(72* e^3*x^2)
Time = 1.25 (sec) , antiderivative size = 446, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2904, 2848, 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.
| \(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle -\frac {3}{2} \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^{4/3}}d\frac {1}{x^{2/3}}\) |
| \(\Big \downarrow \) 2848 |
| \(\displaystyle -\frac {3}{2} \int \left (\frac {\left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{e^2}-\frac {2 d \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{e^2}+\frac {d^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{e^2}\right )d\frac {1}{x^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3}{2} \left (\frac {2 b^2 n^2 \left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{9 e^3}-\frac {3 b^2 d n^2 \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{2 e^3}+\frac {6 a b^2 d^2 n^2}{e^2 x^{2/3}}-\frac {3 b d^2 n \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e^3}+\frac {d^2 \left (d+\frac {e}{x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{e^3}-\frac {b n \left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 e^3}+\frac {3 b d n \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{2 e^3}+\frac {\left (d+\frac {e}{x^{2/3}}\right )^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{3 e^3}-\frac {d \left (d+\frac {e}{x^{2/3}}\right )^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{e^3}+\frac {6 b^3 d^2 n^2 \left (d+\frac {e}{x^{2/3}}\right ) \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{e^3}-\frac {6 b^3 d^2 n^3}{e^2 x^{2/3}}-\frac {2 b^3 n^3 \left (d+\frac {e}{x^{2/3}}\right )^3}{27 e^3}+\frac {3 b^3 d n^3 \left (d+\frac {e}{x^{2/3}}\right )^2}{4 e^3}\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(2/3))^n])^3/x^3,x]
Output:
(-3*((3*b^3*d*n^3*(d + e/x^(2/3))^2)/(4*e^3) - (2*b^3*n^3*(d + e/x^(2/3))^ 3)/(27*e^3) + (6*a*b^2*d^2*n^2)/(e^2*x^(2/3)) - (6*b^3*d^2*n^3)/(e^2*x^(2/ 3)) + (6*b^3*d^2*n^2*(d + e/x^(2/3))*Log[c*(d + e/x^(2/3))^n])/e^3 - (3*b^ 2*d*n^2*(d + e/x^(2/3))^2*(a + b*Log[c*(d + e/x^(2/3))^n]))/(2*e^3) + (2*b ^2*n^2*(d + e/x^(2/3))^3*(a + b*Log[c*(d + e/x^(2/3))^n]))/(9*e^3) - (3*b* d^2*n*(d + e/x^(2/3))*(a + b*Log[c*(d + e/x^(2/3))^n])^2)/e^3 + (3*b*d*n*( d + e/x^(2/3))^2*(a + b*Log[c*(d + e/x^(2/3))^n])^2)/(2*e^3) - (b*n*(d + e /x^(2/3))^3*(a + b*Log[c*(d + e/x^(2/3))^n])^2)/(3*e^3) + (d^2*(d + e/x^(2 /3))*(a + b*Log[c*(d + e/x^(2/3))^n])^3)/e^3 - (d*(d + e/x^(2/3))^2*(a + b *Log[c*(d + e/x^(2/3))^n])^3)/e^3 + ((d + e/x^(2/3))^3*(a + b*Log[c*(d + e /x^(2/3))^n])^3)/(3*e^3)))/2
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[e*f - d*g, 0] && IGtQ[q, 0]
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])
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )}^{3}}{x^{3}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(2/3))^n))^3/x^3,x)
Output:
int((a+b*ln(c*(d+e/x^(2/3))^n))^3/x^3,x)
Time = 0.10 (sec) , antiderivative size = 725, normalized size of antiderivative = 1.61 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx =\text {Too large to display} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^3,x, algorithm="fricas")
Output:
1/72*(8*b^3*e^3*n^3 - 36*b^3*e^3*log(c)^3 - 24*a*b^2*e^3*n^2 + 36*a^2*b*e^ 3*n - 36*a^3*e^3 - 36*(b^3*d^3*n^3*x^2 + b^3*e^3*n^3)*log((d*x + e*x^(1/3) )/x)^3 + 36*(b^3*e^3*n - 3*a*b^2*e^3)*log(c)^2 + 18*(6*b^3*d^2*e*n^3*x^(4/ 3) - 3*b^3*d*e^2*n^3*x^(2/3) + 2*b^3*e^3*n^3 - 6*a*b^2*e^3*n^2 + (11*b^3*d ^3*n^3 - 6*a*b^2*d^3*n^2)*x^2 - 6*(b^3*d^3*n^2*x^2 + b^3*e^3*n^2)*log(c))* log((d*x + e*x^(1/3))/x)^2 - 12*(2*b^3*e^3*n^2 - 6*a*b^2*e^3*n + 9*a^2*b*e ^3)*log(c) - 6*(4*b^3*e^3*n^3 - 12*a*b^2*e^3*n^2 + 18*a^2*b*e^3*n + (85*b^ 3*d^3*n^3 - 66*a*b^2*d^3*n^2 + 18*a^2*b*d^3*n)*x^2 + 18*(b^3*d^3*n*x^2 + b ^3*e^3*n)*log(c)^2 - 6*(2*b^3*e^3*n^2 - 6*a*b^2*e^3*n + (11*b^3*d^3*n^2 - 6*a*b^2*d^3*n)*x^2)*log(c) - 3*(5*b^3*d*e^2*n^3 - 6*b^3*d*e^2*n^2*log(c) - 6*a*b^2*d*e^2*n^2)*x^(2/3) - 6*(6*b^3*d^2*e*n^2*x*log(c) - (11*b^3*d^2*e* n^3 - 6*a*b^2*d^2*e*n^2)*x)*x^(1/3))*log((d*x + e*x^(1/3))/x) - 3*(19*b^3* d*e^2*n^3 + 18*b^3*d*e^2*n*log(c)^2 - 30*a*b^2*d*e^2*n^2 + 18*a^2*b*d*e^2* n - 6*(5*b^3*d*e^2*n^2 - 6*a*b^2*d*e^2*n)*log(c))*x^(2/3) + 6*(18*b^3*d^2* e*n*x*log(c)^2 - 6*(11*b^3*d^2*e*n^2 - 6*a*b^2*d^2*e*n)*x*log(c) + (85*b^3 *d^2*e*n^3 - 66*a*b^2*d^2*e*n^2 + 18*a^2*b*d^2*e*n)*x)*x^(1/3))/(e^3*x^2)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(2/3))**n))**3/x**3,x)
Output:
Timed out
Time = 0.08 (sec) , antiderivative size = 684, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx =\text {Too large to display} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^3,x, algorithm="maxima")
Output:
-1/4*a^2*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)) - 1/12*(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))*a*b^2 - 1/216*(54*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)^2 + e*n*((108*d^3*x^2*log(d*x^(2/3) + e)^3 - 32* d^3*x^2*log(x)^3 + 264*d^3*x^2*log(x)^2 - 1020*d^3*x^2*log(x) - 1530*d^2*e *x^(4/3) + 171*d*e^2*x^(2/3) - 24*e^3 - 54*(4*d^3*x^2*log(x) - 11*d^3*x^2) *log(d*x^(2/3) + e)^2 + 18*(8*d^3*x^2*log(x)^2 - 44*d^3*x^2*log(x) + 85*d^ 3*x^2)*log(d*x^(2/3) + e))*n^2/(e^4*x^2) - 18*(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*log(c*(d + e/x^(2/3))^n)/(e^4*x^2)))*b^3 - 1/2*b^3*log(c*(d + e/x^(2/3)) ^n)^3/x^2 - 3/2*a*b^2*log(c*(d + e/x^(2/3))^n)^2/x^2 - 3/2*a^2*b*log(c*(d + e/x^(2/3))^n)/x^2 - 1/2*a^3/x^2
\[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{3}}{x^{3}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^3,x, algorithm="giac")
Output:
integrate((b*log(c*(d + e/x^(2/3))^n) + a)^3/x^3, x)
Time = 25.63 (sec) , antiderivative size = 578, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx=\frac {\frac {d\,\left (\frac {3\,a^3}{2}-\frac {3\,a^2\,b\,n}{2}+a\,b^2\,n^2-\frac {b^3\,n^3}{3}\right )}{2\,e}-\frac {d\,\left (6\,a^3-6\,a\,b^2\,n^2+5\,b^3\,n^3\right )}{8\,e}}{x^{4/3}}-{\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}^3\,\left (\frac {b^3}{2\,x^2}+\frac {b^3\,d^3}{2\,e^3}\right )-{\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}^2\,\left (\frac {b^2\,\left (3\,a-b\,n\right )}{2\,x^2}-\frac {\frac {3\,b^2\,d\,\left (3\,a-b\,n\right )}{2\,e}-\frac {9\,a\,b^2\,d}{2\,e}}{2\,x^{4/3}}+\frac {d\,\left (6\,a\,b^2\,d^2-11\,b^3\,d^2\,n\right )}{4\,e^3}+\frac {d\,\left (\frac {6\,b^2\,d\,\left (3\,a-b\,n\right )}{e}-\frac {18\,a\,b^2\,d}{e}\right )}{4\,e\,x^{2/3}}\right )-\frac {\frac {d\,\left (\frac {d\,\left (\frac {3\,a^3}{2}-\frac {3\,a^2\,b\,n}{2}+a\,b^2\,n^2-\frac {b^3\,n^3}{3}\right )}{e}-\frac {d\,\left (6\,a^3-6\,a\,b^2\,n^2+5\,b^3\,n^3\right )}{4\,e}\right )}{e}+\frac {b^2\,d^2\,n^2\,\left (6\,a-11\,b\,n\right )}{2\,e^2}}{x^{2/3}}-\frac {\frac {a^3}{2}-\frac {a^2\,b\,n}{2}+\frac {a\,b^2\,n^2}{3}-\frac {b^3\,n^3}{9}}{x^2}-\frac {\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )\,\left (\frac {\frac {d\,\left (2\,b\,d\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )-6\,b\,d\,e\,\left (3\,a^2-b^2\,n^2\right )\right )}{e}+12\,b^3\,d^2\,n^2}{2\,e\,x^{2/3}}-\frac {2\,b\,d\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )-6\,b\,d\,e\,\left (3\,a^2-b^2\,n^2\right )}{4\,e\,x^{4/3}}+\frac {b\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )}{3\,x^2}\right )}{2\,e}-\frac {\ln \left (d+\frac {e}{x^{2/3}}\right )\,\left (18\,a^2\,b\,d^3\,n-66\,a\,b^2\,d^3\,n^2+85\,b^3\,d^3\,n^3\right )}{12\,e^3} \] Input:
int((a + b*log(c*(d + e/x^(2/3))^n))^3/x^3,x)
Output:
((d*((3*a^3)/2 - (b^3*n^3)/3 + a*b^2*n^2 - (3*a^2*b*n)/2))/(2*e) - (d*(6*a ^3 + 5*b^3*n^3 - 6*a*b^2*n^2))/(8*e))/x^(4/3) - log(c*(d + e/x^(2/3))^n)^3 *(b^3/(2*x^2) + (b^3*d^3)/(2*e^3)) - log(c*(d + e/x^(2/3))^n)^2*((b^2*(3*a - b*n))/(2*x^2) - ((3*b^2*d*(3*a - b*n))/(2*e) - (9*a*b^2*d)/(2*e))/(2*x^ (4/3)) + (d*(6*a*b^2*d^2 - 11*b^3*d^2*n))/(4*e^3) + (d*((6*b^2*d*(3*a - b* n))/e - (18*a*b^2*d)/e))/(4*e*x^(2/3))) - ((d*((d*((3*a^3)/2 - (b^3*n^3)/3 + a*b^2*n^2 - (3*a^2*b*n)/2))/e - (d*(6*a^3 + 5*b^3*n^3 - 6*a*b^2*n^2))/( 4*e)))/e + (b^2*d^2*n^2*(6*a - 11*b*n))/(2*e^2))/x^(2/3) - (a^3/2 - (b^3*n ^3)/9 + (a*b^2*n^2)/3 - (a^2*b*n)/2)/x^2 - (log(c*(d + e/x^(2/3))^n)*(((d* (2*b*d*e*(9*a^2 + 2*b^2*n^2 - 6*a*b*n) - 6*b*d*e*(3*a^2 - b^2*n^2)))/e + 1 2*b^3*d^2*n^2)/(2*e*x^(2/3)) - (2*b*d*e*(9*a^2 + 2*b^2*n^2 - 6*a*b*n) - 6* b*d*e*(3*a^2 - b^2*n^2))/(4*e*x^(4/3)) + (b*e*(9*a^2 + 2*b^2*n^2 - 6*a*b*n ))/(3*x^2)))/(2*e) - (log(d + e/x^(2/3))*(85*b^3*d^3*n^3 - 66*a*b^2*d^3*n^ 2 + 18*a^2*b*d^3*n))/(12*e^3)
Time = 0.17 (sec) , antiderivative size = 712, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^3} \, dx =\text {Too large to display} \] Input:
int((a+b*log(c*(d+e/x^(2/3))^n))^3/x^3,x)
Output:
( - 54*x**(2/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**2*b**3*d*e**2*n - 108*x**(2/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*a*b**2*d*e**2*n + 90*x**(2/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b**3*d*e**2*n**2 - 54*x**(2/3)*a**2*b*d*e**2*n + 90*x**(2/3)*a*b**2*d*e**2*n**2 - 57*x**(2/3 )*b**3*d*e**2*n**3 + 108*x**(1/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3) )**2*b**3*d**2*e*n*x + 216*x**(1/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/ 3))*a*b**2*d**2*e*n*x - 396*x**(1/3)*log(((x**(2/3)*d + e)**n*c)/x**((2*n) /3))*b**3*d**2*e*n**2*x + 108*x**(1/3)*a**2*b*d**2*e*n*x - 396*x**(1/3)*a* b**2*d**2*e*n**2*x + 510*x**(1/3)*b**3*d**2*e*n**3*x - 36*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**3*b**3*d**3*x**2 - 36*log(((x**(2/3)*d + e)**n* c)/x**((2*n)/3))**3*b**3*e**3 - 108*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/ 3))**2*a*b**2*d**3*x**2 - 108*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**2 *a*b**2*e**3 + 198*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**2*b**3*d**3* n*x**2 + 36*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**2*b**3*e**3*n - 108 *log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*a**2*b*d**3*x**2 - 108*log(((x* *(2/3)*d + e)**n*c)/x**((2*n)/3))*a**2*b*e**3 + 396*log(((x**(2/3)*d + e)* *n*c)/x**((2*n)/3))*a*b**2*d**3*n*x**2 + 72*log(((x**(2/3)*d + e)**n*c)/x* *((2*n)/3))*a*b**2*e**3*n - 510*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))* b**3*d**3*n**2*x**2 - 24*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b**3*e* *3*n**2 - 36*a**3*e**3 + 36*a**2*b*e**3*n - 24*a*b**2*e**3*n**2 + 8*b**...