\(\int x (a+b \log (c (d+e x^{2/3})^n))^3 \, dx\) [482]

Optimal result

Integrand size = 22, antiderivative size = 449 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\frac {9 b^3 d n^3 \left (d+e x^{2/3}\right )^2}{8 e^3}-\frac {b^3 n^3 \left (d+e x^{2/3}\right )^3}{9 e^3}+\frac {9 a b^2 d^2 n^2 x^{2/3}}{e^2}-\frac {9 b^3 d^2 n^3 x^{2/3}}{e^2}+\frac {9 b^3 d^2 n^2 \left (d+e x^{2/3}\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )}{e^3}-\frac {9 b^2 d n^2 \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{4 e^3}+\frac {b^2 n^2 \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 e^3}-\frac {9 b d^2 n \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {9 b d n \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{4 e^3}-\frac {b n \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {3 d^2 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}-\frac {3 d \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}+\frac {\left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3} \] 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*x^(2/3)/e^2-9*b^3*d^2*n^3*x^(2/3)/e^2+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
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 428, normalized size of antiderivative = 0.95 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\frac {36 a^3 d^3-198 a^2 b d^3 n-108 a^2 b d^2 e n x^{2/3}+396 a b^2 d^2 e n^2 x^{2/3}-510 b^3 d^2 e n^3 x^{2/3}+54 a^2 b d e^2 n x^{4/3}-90 a b^2 d e^2 n^2 x^{4/3}+57 b^3 d e^2 n^3 x^{4/3}+36 a^3 e^3 x^2-36 a^2 b e^3 n x^2+24 a b^2 e^3 n^2 x^2-8 b^3 e^3 n^3 x^2+114 b^3 d^3 n^3 \log \left (d+e x^{2/3}\right )+6 b \left (18 a^2 \left (d^3+e^3 x^2\right )-6 a b n \left (11 d^3+6 d^2 e x^{2/3}-3 d e^2 x^{4/3}+2 e^3 x^2\right )+b^2 n^2 \left (66 d^3+66 d^2 e x^{2/3}-15 d e^2 x^{4/3}+4 e^3 x^2\right )\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )+18 b^2 \left (6 a \left (d^3+e^3 x^2\right )-b n \left (11 d^3+6 d^2 e x^{2/3}-3 d e^2 x^{4/3}+2 e^3 x^2\right )\right ) \log ^2\left (c \left (d+e x^{2/3}\right )^n\right )+36 b^3 \left (d^3+e^3 x^2\right ) \log ^3\left (c \left (d+e x^{2/3}\right )^n\right )}{72 e^3} \] Input:

Integrate[x*(a + b*Log[c*(d + e*x^(2/3))^n])^3,x]
 

Output:

(36*a^3*d^3 - 198*a^2*b*d^3*n - 108*a^2*b*d^2*e*n*x^(2/3) + 396*a*b^2*d^2* 
e*n^2*x^(2/3) - 510*b^3*d^2*e*n^3*x^(2/3) + 54*a^2*b*d*e^2*n*x^(4/3) - 90* 
a*b^2*d*e^2*n^2*x^(4/3) + 57*b^3*d*e^2*n^3*x^(4/3) + 36*a^3*e^3*x^2 - 36*a 
^2*b*e^3*n*x^2 + 24*a*b^2*e^3*n^2*x^2 - 8*b^3*e^3*n^3*x^2 + 114*b^3*d^3*n^ 
3*Log[d + e*x^(2/3)] + 6*b*(18*a^2*(d^3 + e^3*x^2) - 6*a*b*n*(11*d^3 + 6*d 
^2*e*x^(2/3) - 3*d*e^2*x^(4/3) + 2*e^3*x^2) + b^2*n^2*(66*d^3 + 66*d^2*e*x 
^(2/3) - 15*d*e^2*x^(4/3) + 4*e^3*x^2))*Log[c*(d + e*x^(2/3))^n] + 18*b^2* 
(6*a*(d^3 + e^3*x^2) - b*n*(11*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]^2 + 36*b^3*(d^3 + e^3*x^2)*Log[c*(d + 
e*x^(2/3))^n]^3)/(72*e^3)
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 446, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, 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.

\(\displaystyle \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx\)

\(\Big \downarrow \) 2904

\(\displaystyle \frac {3}{2} \int x^{4/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3dx^{2/3}\)

\(\Big \downarrow \) 2848

\(\displaystyle \frac {3}{2} \int \left (\frac {\left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{e^2}-\frac {2 d \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{e^2}+\frac {d^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{e^2}\right )dx^{2/3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3}{2} \left (\frac {2 b^2 n^2 \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{9 e^3}-\frac {3 b^2 d n^2 \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{2 e^3}+\frac {6 a b^2 d^2 n^2 x^{2/3}}{e^2}-\frac {3 b d^2 n \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{e^3}+\frac {d^2 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{e^3}-\frac {b n \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{3 e^3}+\frac {3 b d n \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {\left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{3 e^3}-\frac {d \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{e^3}+\frac {6 b^3 d^2 n^2 \left (d+e x^{2/3}\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )}{e^3}-\frac {6 b^3 d^2 n^3 x^{2/3}}{e^2}-\frac {2 b^3 n^3 \left (d+e x^{2/3}\right )^3}{27 e^3}+\frac {3 b^3 d n^3 \left (d+e x^{2/3}\right )^2}{4 e^3}\right )\)

Input:

Int[x*(a + b*Log[c*(d + e*x^(2/3))^n])^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*x^(2/3))/e^2 - (6*b^3*d^2*n^3*x^(2/3))/e^2 + 
 (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
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2848

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]
 

rule 2904

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])
 

Maple [F]

\[\int x {\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{3}d x\]

Input:

int(x*(a+b*ln(c*(d+e*x^(2/3))^n))^3,x)
 

Output:

int(x*(a+b*ln(c*(d+e*x^(2/3))^n))^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 720, normalized size of antiderivative = 1.60 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx =\text {Too large to display} \] Input:

integrate(x*(a+b*log(c*(d+e*x^(2/3))^n))^3,x, algorithm="fricas")
 

Output:

1/72*(36*b^3*e^3*x^2*log(c)^3 - 36*(b^3*e^3*n - 3*a*b^2*e^3)*x^2*log(c)^2 
+ 36*(b^3*e^3*n^3*x^2 + b^3*d^3*n^3)*log(e*x^(2/3) + d)^3 + 12*(2*b^3*e^3* 
n^2 - 6*a*b^2*e^3*n + 9*a^2*b*e^3)*x^2*log(c) - 4*(2*b^3*e^3*n^3 - 6*a*b^2 
*e^3*n^2 + 9*a^2*b*e^3*n - 9*a^3*e^3)*x^2 + 18*(3*b^3*d*e^2*n^3*x^(4/3) - 
6*b^3*d^2*e*n^3*x^(2/3) - 11*b^3*d^3*n^3 + 6*a*b^2*d^3*n^2 - 2*(b^3*e^3*n^ 
3 - 3*a*b^2*e^3*n^2)*x^2 + 6*(b^3*e^3*n^2*x^2 + b^3*d^3*n^2)*log(c))*log(e 
*x^(2/3) + d)^2 + 6*(85*b^3*d^3*n^3 - 66*a*b^2*d^3*n^2 + 18*a^2*b*d^3*n + 
2*(2*b^3*e^3*n^3 - 6*a*b^2*e^3*n^2 + 9*a^2*b*e^3*n)*x^2 + 18*(b^3*e^3*n*x^ 
2 + b^3*d^3*n)*log(c)^2 - 6*(11*b^3*d^3*n^2 - 6*a*b^2*d^3*n + 2*(b^3*e^3*n 
^2 - 3*a*b^2*e^3*n)*x^2)*log(c) + 6*(11*b^3*d^2*e*n^3 - 6*b^3*d^2*e*n^2*lo 
g(c) - 6*a*b^2*d^2*e*n^2)*x^(2/3) + 3*(6*b^3*d*e^2*n^2*x*log(c) - (5*b^3*d 
*e^2*n^3 - 6*a*b^2*d*e^2*n^2)*x)*x^(1/3))*log(e*x^(2/3) + d) - 6*(85*b^3*d 
^2*e*n^3 + 18*b^3*d^2*e*n*log(c)^2 - 66*a*b^2*d^2*e*n^2 + 18*a^2*b*d^2*e*n 
 - 6*(11*b^3*d^2*e*n^2 - 6*a*b^2*d^2*e*n)*log(c))*x^(2/3) + 3*(18*b^3*d*e^ 
2*n*x*log(c)^2 - 6*(5*b^3*d*e^2*n^2 - 6*a*b^2*d*e^2*n)*x*log(c) + (19*b^3* 
d*e^2*n^3 - 30*a*b^2*d*e^2*n^2 + 18*a^2*b*d*e^2*n)*x)*x^(1/3))/e^3
 

Sympy [F(-1)]

Timed out. \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\text {Timed out} \] Input:

integrate(x*(a+b*ln(c*(d+e*x**(2/3))**n))**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.08 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\frac {1}{2} \, b^{3} x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )^{3} + \frac {3}{2} \, a b^{2} x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )^{2} + \frac {1}{4} \, a^{2} 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 )} + \frac {3}{2} \, a^{2} b x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {1}{2} \, a^{3} x^{2} + \frac {1}{12} \, {\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 )} a b^{2} + \frac {1}{72} \, {\left (18 \, 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 )^{2} + e n {\left (\frac {{\left (36 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )^{3} - 8 \, e^{3} x^{2} + 198 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )^{2} + 57 \, d e^{2} x^{\frac {4}{3}} + 510 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right ) - 510 \, d^{2} e x^{\frac {2}{3}}\right )} n^{2}}{e^{4}} + \frac {6 \, {\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 \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )}{e^{4}}\right )}\right )} b^{3} \] Input:

integrate(x*(a+b*log(c*(d+e*x^(2/3))^n))^3,x, algorithm="maxima")
 

Output:

1/2*b^3*x^2*log((e*x^(2/3) + d)^n*c)^3 + 3/2*a*b^2*x^2*log((e*x^(2/3) + d) 
^n*c)^2 + 1/4*a^2*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) + 3/2*a^2*b*x^2*log((e*x^(2/3) + d)^n*c) + 
1/2*a^3*x^2 + 1/12*(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 - 1 
8*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)*a*b^2 + 1/72*(18*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)^2 + e*n*((36*d^3*log(e*x^(2/3) + d)^3 - 8*e^3*x^2 + 198*d^3*log(e* 
x^(2/3) + d)^2 + 57*d*e^2*x^(4/3) + 510*d^3*log(e*x^(2/3) + d) - 510*d^2*e 
*x^(2/3))*n^2/e^4 + 6*(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*log((e*x^(2/3) + 
 d)^n*c)/e^4))*b^3
 

Giac [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 755, normalized size of antiderivative = 1.68 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\text {Too large to display} \] Input:

integrate(x*(a+b*log(c*(d+e*x^(2/3))^n))^3,x, algorithm="giac")
 

Output:

1/2*b^3*x^2*log(c)^3 + 1/72*(36*x^2*log(e*x^(2/3) + d)^3 - (18*(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)*lo 
g(e*x^(2/3) + d)^2 - 6*(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)*log(e*x^(2/3) + d) - 36*d^3*log(e*x^(2/3 
) + d)^3/e^4 + 8*(e*x^(2/3) + d)^3/e^4 - 81*(e*x^(2/3) + d)^2*d/e^4 + 648* 
(e*x^(2/3) + d)*d^2/e^4)*e)*b^3*n^3 + 1/12*(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^3*n^2*log(c) + 1/4*(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^3* 
n*log(c)^2 + 3/2*a*b^2*x^2*log(c)^2 + 1/12*(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)*a*b^2*n^2 + 1/2*(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^2*n*l 
og(c) + 3/2*a^2*b*x^2*log(c) + 1/4*(6*x^2*log(e*x^(2/3) + d) + e*(6*d^3*lo 
g(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^2*b*n + 1/2*a^3*x^2
 

Mupad [B] (verification not implemented)

Time = 25.85 (sec) , antiderivative size = 575, normalized size of antiderivative = 1.28 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx={\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^3\,\left (\frac {b^3\,x^2}{2}+\frac {b^3\,d^3}{2\,e^3}\right )-x^{4/3}\,\left (\frac {d\,\left (\frac {3\,a^3}{2}-\frac {3\,a^2\,b\,n}{2}+a\,b^2\,n^2-\frac {b^3\,n^3}{3}\right )}{2\,e}-\frac {d\,\left (6\,a^3-6\,a\,b^2\,n^2+5\,b^3\,n^3\right )}{8\,e}\right )+{\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^2\,\left (\frac {b^2\,x^2\,\left (3\,a-b\,n\right )}{2}-\frac {x^{4/3}\,\left (\frac {3\,b^2\,d\,\left (3\,a-b\,n\right )}{2\,e}-\frac {9\,a\,b^2\,d}{2\,e}\right )}{2}+\frac {d\,\left (6\,a\,b^2\,d^2-11\,b^3\,d^2\,n\right )}{4\,e^3}+\frac {d\,x^{2/3}\,\left (\frac {6\,b^2\,d\,\left (3\,a-b\,n\right )}{e}-\frac {18\,a\,b^2\,d}{e}\right )}{4\,e}\right )+x^{2/3}\,\left (\frac {d\,\left (\frac {d\,\left (\frac {3\,a^3}{2}-\frac {3\,a^2\,b\,n}{2}+a\,b^2\,n^2-\frac {b^3\,n^3}{3}\right )}{e}-\frac {d\,\left (6\,a^3-6\,a\,b^2\,n^2+5\,b^3\,n^3\right )}{4\,e}\right )}{e}+\frac {b^2\,d^2\,n^2\,\left (6\,a-11\,b\,n\right )}{2\,e^2}\right )+x^2\,\left (\frac {a^3}{2}-\frac {a^2\,b\,n}{2}+\frac {a\,b^2\,n^2}{3}-\frac {b^3\,n^3}{9}\right )+\frac {\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\,\left (\frac {x^{2/3}\,\left (\frac {d\,\left (2\,b\,d\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )-6\,b\,d\,e\,\left (3\,a^2-b^2\,n^2\right )\right )}{e}+12\,b^3\,d^2\,n^2\right )}{2\,e}-\frac {x^{4/3}\,\left (2\,b\,d\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )-6\,b\,d\,e\,\left (3\,a^2-b^2\,n^2\right )\right )}{4\,e}+\frac {b\,e\,x^2\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )}{3}\right )}{2\,e}+\frac {\ln \left (d+e\,x^{2/3}\right )\,\left (18\,a^2\,b\,d^3\,n-66\,a\,b^2\,d^3\,n^2+85\,b^3\,d^3\,n^3\right )}{12\,e^3} \] Input:

int(x*(a + b*log(c*(d + e*x^(2/3))^n))^3,x)
 

Output:

log(c*(d + e*x^(2/3))^n)^3*((b^3*x^2)/2 + (b^3*d^3)/(2*e^3)) - x^(4/3)*((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)) + log(c*(d + e*x^(2/3))^n)^2*((b^2*x^2* 
(3*a - b*n))/2 - (x^(4/3)*((3*b^2*d*(3*a - b*n))/(2*e) - (9*a*b^2*d)/(2*e) 
))/2 + (d*(6*a*b^2*d^2 - 11*b^3*d^2*n))/(4*e^3) + (d*x^(2/3)*((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*(a^3/2 - (b^3 
*n^3)/9 + (a*b^2*n^2)/3 - (a^2*b*n)/2) + (log(c*(d + e*x^(2/3))^n)*((x^(2/ 
3)*((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 + 12*b^3*d^2*n^2))/(2*e) - (x^(4/3)*(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) + (b*e*x^2*(9*a^2 + 2*b^2*n^2 - 6 
*a*b*n))/3))/(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)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 595, normalized size of antiderivative = 1.33 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx =\text {Too large to display} \] Input:

int(x*(a+b*log(c*(d+e*x^(2/3))^n))^3,x)
 

Output:

( - 108*x**(2/3)*log((x**(2/3)*e + d)**n*c)**2*b**3*d**2*e*n - 216*x**(2/3 
)*log((x**(2/3)*e + d)**n*c)*a*b**2*d**2*e*n + 396*x**(2/3)*log((x**(2/3)* 
e + d)**n*c)*b**3*d**2*e*n**2 - 108*x**(2/3)*a**2*b*d**2*e*n + 396*x**(2/3 
)*a*b**2*d**2*e*n**2 - 510*x**(2/3)*b**3*d**2*e*n**3 + 54*x**(1/3)*log((x* 
*(2/3)*e + d)**n*c)**2*b**3*d*e**2*n*x + 108*x**(1/3)*log((x**(2/3)*e + d) 
**n*c)*a*b**2*d*e**2*n*x - 90*x**(1/3)*log((x**(2/3)*e + d)**n*c)*b**3*d*e 
**2*n**2*x + 54*x**(1/3)*a**2*b*d*e**2*n*x - 90*x**(1/3)*a*b**2*d*e**2*n** 
2*x + 57*x**(1/3)*b**3*d*e**2*n**3*x + 36*log((x**(2/3)*e + d)**n*c)**3*b* 
*3*d**3 + 36*log((x**(2/3)*e + d)**n*c)**3*b**3*e**3*x**2 + 108*log((x**(2 
/3)*e + d)**n*c)**2*a*b**2*d**3 + 108*log((x**(2/3)*e + d)**n*c)**2*a*b**2 
*e**3*x**2 - 198*log((x**(2/3)*e + d)**n*c)**2*b**3*d**3*n - 36*log((x**(2 
/3)*e + d)**n*c)**2*b**3*e**3*n*x**2 + 108*log((x**(2/3)*e + d)**n*c)*a**2 
*b*d**3 + 108*log((x**(2/3)*e + d)**n*c)*a**2*b*e**3*x**2 - 396*log((x**(2 
/3)*e + d)**n*c)*a*b**2*d**3*n - 72*log((x**(2/3)*e + d)**n*c)*a*b**2*e**3 
*n*x**2 + 510*log((x**(2/3)*e + d)**n*c)*b**3*d**3*n**2 + 24*log((x**(2/3) 
*e + d)**n*c)*b**3*e**3*n**2*x**2 + 36*a**3*e**3*x**2 - 36*a**2*b*e**3*n*x 
**2 + 24*a*b**2*e**3*n**2*x**2 - 8*b**3*e**3*n**3*x**2)/(72*e**3)