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

Optimal. Leaf size=449 \[ \frac {b^2 n^2 \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 e^3}-\frac {9 b^2 d n^2 \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{4 e^3}+\frac {9 a b^2 d^2 n^2 x^{2/3}}{e^2}-\frac {9 b d^2 n \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {3 d^2 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}-\frac {b n \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {9 b d n \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{4 e^3}+\frac {\left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}-\frac {3 d \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}+\frac {9 b^3 d^2 n^2 \left (d+e x^{2/3}\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )}{e^3}-\frac {9 b^3 d^2 n^3 x^{2/3}}{e^2}-\frac {b^3 n^3 \left (d+e x^{2/3}\right )^3}{9 e^3}+\frac {9 b^3 d n^3 \left (d+e x^{2/3}\right )^2}{8 e^3} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304} \[ \frac {b^2 n^2 \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 e^3}-\frac {9 b^2 d n^2 \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{4 e^3}+\frac {9 a b^2 d^2 n^2 x^{2/3}}{e^2}-\frac {9 b d^2 n \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {3 d^2 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}-\frac {b n \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{2 e^3}+\frac {9 b d n \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{4 e^3}+\frac {\left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}-\frac {3 d \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{2 e^3}+\frac {9 b^3 d^2 n^2 \left (d+e x^{2/3}\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )}{e^3}-\frac {9 b^3 d^2 n^3 x^{2/3}}{e^2}-\frac {b^3 n^3 \left (d+e x^{2/3}\right )^3}{9 e^3}+\frac {9 b^3 d n^3 \left (d+e x^{2/3}\right )^2}{8 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*b^3*d*n^3*(d + e*x^(2/3))^2)/(8*e^3) - (b^3*n^3*(d + e*x^(2/3))^3)/(9*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))*Log[c*(d + e*x^(2/3))^n])/e^3 - (9*b^2*d*n^2*(d
+ e*x^(2/3))^2*(a + b*Log[c*(d + e*x^(2/3))^n]))/(4*e^3) + (b^2*n^2*(d + e*x^(2/3))^3*(a + b*Log[c*(d + e*x^(
2/3))^n]))/(3*e^3) - (9*b*d^2*n*(d + e*x^(2/3))*(a + b*Log[c*(d + e*x^(2/3))^n])^2)/(2*e^3) + (9*b*d*n*(d + e*
x^(2/3))^2*(a + b*Log[c*(d + e*x^(2/3))^n])^2)/(4*e^3) - (b*n*(d + e*x^(2/3))^3*(a + b*Log[c*(d + e*x^(2/3))^n
])^2)/(2*e^3) + (3*d^2*(d + e*x^(2/3))*(a + b*Log[c*(d + e*x^(2/3))^n])^3)/(2*e^3) - (3*d*(d + e*x^(2/3))^2*(a
+ b*Log[c*(d + e*x^(2/3))^n])^3)/(2*e^3) + ((d + e*x^(2/3))^3*(a + b*Log[c*(d + e*x^(2/3))^n])^3)/(2*e^3)

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(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 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rubi steps

\begin {align*} \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx &=\frac {3}{2} \operatorname {Subst}\left (\int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3 \, dx,x,x^{2/3}\right )\\ &=\frac {3}{2} \operatorname {Subst}\left (\int \left (\frac {d^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{e^2}-\frac {2 d (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{e^2}+\frac {(d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3}{e^2}\right ) \, dx,x,x^{2/3}\right )\\ &=\frac {3 \operatorname {Subst}\left (\int (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^3 \, dx,x,x^{2/3}\right )}{2 e^2}-\frac {(3 d) \operatorname {Subst}\left (\int (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^3 \, dx,x,x^{2/3}\right )}{e^2}+\frac {\left (3 d^2\right ) \operatorname {Subst}\left (\int \left (a+b \log \left (c (d+e x)^n\right )\right )^3 \, dx,x,x^{2/3}\right )}{2 e^2}\\ &=\frac {3 \operatorname {Subst}\left (\int x^2 \left (a+b \log \left (c x^n\right )\right )^3 \, dx,x,d+e x^{2/3}\right )}{2 e^3}-\frac {(3 d) \operatorname {Subst}\left (\int x \left (a+b \log \left (c x^n\right )\right )^3 \, dx,x,d+e x^{2/3}\right )}{e^3}+\frac {\left (3 d^2\right ) \operatorname {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^3 \, dx,x,d+e x^{2/3}\right )}{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}-\frac {(3 b n) \operatorname {Subst}\left (\int x^2 \left (a+b \log \left (c x^n\right )\right )^2 \, dx,x,d+e x^{2/3}\right )}{2 e^3}+\frac {(9 b d n) \operatorname {Subst}\left (\int x \left (a+b \log \left (c x^n\right )\right )^2 \, dx,x,d+e x^{2/3}\right )}{2 e^3}-\frac {\left (9 b d^2 n\right ) \operatorname {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^2 \, dx,x,d+e x^{2/3}\right )}{2 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}+\frac {\left (b^2 n^2\right ) \operatorname {Subst}\left (\int x^2 \left (a+b \log \left (c x^n\right )\right ) \, dx,x,d+e x^{2/3}\right )}{e^3}-\frac {\left (9 b^2 d n^2\right ) \operatorname {Subst}\left (\int x \left (a+b \log \left (c x^n\right )\right ) \, dx,x,d+e x^{2/3}\right )}{2 e^3}+\frac {\left (9 b^2 d^2 n^2\right ) \operatorname {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right ) \, dx,x,d+e x^{2/3}\right )}{e^3}\\ &=\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^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}+\frac {\left (9 b^3 d^2 n^2\right ) \operatorname {Subst}\left (\int \log \left (c x^n\right ) \, dx,x,d+e x^{2/3}\right )}{e^3}\\ &=\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}\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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fricas [A]  time = 0.52, size = 720, normalized size = 1.60 \[ \frac {36 \, b^{3} e^{3} x^{2} \log \relax (c)^{3} - 36 \, {\left (b^{3} e^{3} n - 3 \, a b^{2} e^{3}\right )} x^{2} \log \relax (c)^{2} + 36 \, {\left (b^{3} e^{3} n^{3} x^{2} + b^{3} d^{3} n^{3}\right )} \log \left (e x^{\frac {2}{3}} + d\right )^{3} + 12 \, {\left (2 \, b^{3} e^{3} n^{2} - 6 \, a b^{2} e^{3} n + 9 \, a^{2} b e^{3}\right )} x^{2} \log \relax (c) - 4 \, {\left (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}\right )} x^{2} + 18 \, {\left (3 \, b^{3} d e^{2} n^{3} x^{\frac {4}{3}} - 6 \, b^{3} d^{2} e n^{3} x^{\frac {2}{3}} - 11 \, b^{3} d^{3} n^{3} + 6 \, a b^{2} d^{3} n^{2} - 2 \, {\left (b^{3} e^{3} n^{3} - 3 \, a b^{2} e^{3} n^{2}\right )} x^{2} + 6 \, {\left (b^{3} e^{3} n^{2} x^{2} + b^{3} d^{3} n^{2}\right )} \log \relax (c)\right )} \log \left (e x^{\frac {2}{3}} + d\right )^{2} + 6 \, {\left (85 \, b^{3} d^{3} n^{3} - 66 \, a b^{2} d^{3} n^{2} + 18 \, a^{2} b d^{3} n + 2 \, {\left (2 \, b^{3} e^{3} n^{3} - 6 \, a b^{2} e^{3} n^{2} + 9 \, a^{2} b e^{3} n\right )} x^{2} + 18 \, {\left (b^{3} e^{3} n x^{2} + b^{3} d^{3} n\right )} \log \relax (c)^{2} - 6 \, {\left (11 \, b^{3} d^{3} n^{2} - 6 \, a b^{2} d^{3} n + 2 \, {\left (b^{3} e^{3} n^{2} - 3 \, a b^{2} e^{3} n\right )} x^{2}\right )} \log \relax (c) + 6 \, {\left (11 \, b^{3} d^{2} e n^{3} - 6 \, b^{3} d^{2} e n^{2} \log \relax (c) - 6 \, a b^{2} d^{2} e n^{2}\right )} x^{\frac {2}{3}} + 3 \, {\left (6 \, b^{3} d e^{2} n^{2} x \log \relax (c) - {\left (5 \, b^{3} d e^{2} n^{3} - 6 \, a b^{2} d e^{2} n^{2}\right )} x\right )} x^{\frac {1}{3}}\right )} \log \left (e x^{\frac {2}{3}} + d\right ) - 6 \, {\left (85 \, b^{3} d^{2} e n^{3} + 18 \, b^{3} d^{2} e n \log \relax (c)^{2} - 66 \, a b^{2} d^{2} e n^{2} + 18 \, a^{2} b d^{2} e n - 6 \, {\left (11 \, b^{3} d^{2} e n^{2} - 6 \, a b^{2} d^{2} e n\right )} \log \relax (c)\right )} x^{\frac {2}{3}} + 3 \, {\left (18 \, b^{3} d e^{2} n x \log \relax (c)^{2} - 6 \, {\left (5 \, b^{3} d e^{2} n^{2} - 6 \, a b^{2} d e^{2} n\right )} x \log \relax (c) + {\left (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\right )} x\right )} x^{\frac {1}{3}}}{72 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))*l
og(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*log(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

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giac [B]  time = 1.87, size = 778, normalized size = 1.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^3*x^2*log(c)^3 + 1/72*(36*x^2*log(x^(2/3)*e + d)^3 + (36*d^3*log(x^(2/3)*e + d)^3 - 36*(x^(2/3)*e + d)^3
*log(x^(2/3)*e + d)^2 + 162*(x^(2/3)*e + d)^2*d*log(x^(2/3)*e + d)^2 - 324*(x^(2/3)*e + d)*d^2*log(x^(2/3)*e +
d)^2 + 24*(x^(2/3)*e + d)^3*log(x^(2/3)*e + d) - 162*(x^(2/3)*e + d)^2*d*log(x^(2/3)*e + d) + 648*(x^(2/3)*e
+ d)*d^2*log(x^(2/3)*e + d) - 8*(x^(2/3)*e + d)^3 + 81*(x^(2/3)*e + d)^2*d - 648*(x^(2/3)*e + d)*d^2)*e^(-3))*
b^3*n^3 + 1/12*(18*x^2*log(x^(2/3)*e + d)^2 + (18*d^3*log(x^(2/3)*e + d)^2 - 12*(x^(2/3)*e + d)^3*log(x^(2/3)*
e + d) + 54*(x^(2/3)*e + d)^2*d*log(x^(2/3)*e + d) - 108*(x^(2/3)*e + d)*d^2*log(x^(2/3)*e + d) + 4*(x^(2/3)*e
+ d)^3 - 27*(x^(2/3)*e + d)^2*d + 108*(x^(2/3)*e + d)*d^2)*e^(-3))*b^3*n^2*log(c) + 1/4*(6*x^2*log(x^(2/3)*e
+ d) + (6*d^3*e^(-4)*log(abs(x^(2/3)*e + d)) + (3*d*x^(4/3)*e - 2*x^2*e^2 - 6*d^2*x^(2/3))*e^(-3))*e)*b^3*n*lo
g(c)^2 + 3/2*a*b^2*x^2*log(c)^2 + 1/12*(18*x^2*log(x^(2/3)*e + d)^2 + (18*d^3*log(x^(2/3)*e + d)^2 - 12*(x^(2/
3)*e + d)^3*log(x^(2/3)*e + d) + 54*(x^(2/3)*e + d)^2*d*log(x^(2/3)*e + d) - 108*(x^(2/3)*e + d)*d^2*log(x^(2/
3)*e + d) + 4*(x^(2/3)*e + d)^3 - 27*(x^(2/3)*e + d)^2*d + 108*(x^(2/3)*e + d)*d^2)*e^(-3))*a*b^2*n^2 + 1/2*(6
*x^2*log(x^(2/3)*e + d) + (6*d^3*e^(-4)*log(abs(x^(2/3)*e + d)) + (3*d*x^(4/3)*e - 2*x^2*e^2 - 6*d^2*x^(2/3))*
e^(-3))*e)*a*b^2*n*log(c) + 3/2*a^2*b*x^2*log(c) + 1/4*(6*x^2*log(x^(2/3)*e + d) + (6*d^3*e^(-4)*log(abs(x^(2/
3)*e + d)) + (3*d*x^(4/3)*e - 2*x^2*e^2 - 6*d^2*x^(2/3))*e^(-3))*e)*a^2*b*n + 1/2*a^3*x^2

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \left (e \,x^{\frac {2}{3}}+d \right )^{n}\right )+a \right )^{3} x\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 484, normalized size = 1.08 \[ \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*l
og((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)*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

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mupad [B]  time = 0.72, size = 575, normalized size = 1.28 \[ {\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^3\,\left (\frac {b^3\,x^2}{2}+\frac {b^3\,d^3}{2\,e^3}\right )-x^{4/3}\,\left (\frac {d\,\left (\frac {3\,a^3}{2}-\frac {3\,a^2\,b\,n}{2}+a\,b^2\,n^2-\frac {b^3\,n^3}{3}\right )}{2\,e}-\frac {d\,\left (6\,a^3-6\,a\,b^2\,n^2+5\,b^3\,n^3\right )}{8\,e}\right )+{\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^2\,\left (\frac {b^2\,x^2\,\left (3\,a-b\,n\right )}{2}-\frac {x^{4/3}\,\left (\frac {3\,b^2\,d\,\left (3\,a-b\,n\right )}{2\,e}-\frac {9\,a\,b^2\,d}{2\,e}\right )}{2}+\frac {d\,\left (6\,a\,b^2\,d^2-11\,b^3\,d^2\,n\right )}{4\,e^3}+\frac {d\,x^{2/3}\,\left (\frac {6\,b^2\,d\,\left (3\,a-b\,n\right )}{e}-\frac {18\,a\,b^2\,d}{e}\right )}{4\,e}\right )+x^{2/3}\,\left (\frac {d\,\left (\frac {d\,\left (\frac {3\,a^3}{2}-\frac {3\,a^2\,b\,n}{2}+a\,b^2\,n^2-\frac {b^3\,n^3}{3}\right )}{e}-\frac {d\,\left (6\,a^3-6\,a\,b^2\,n^2+5\,b^3\,n^3\right )}{4\,e}\right )}{e}+\frac {b^2\,d^2\,n^2\,\left (6\,a-11\,b\,n\right )}{2\,e^2}\right )+x^2\,\left (\frac {a^3}{2}-\frac {a^2\,b\,n}{2}+\frac {a\,b^2\,n^2}{3}-\frac {b^3\,n^3}{9}\right )+\frac {\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\,\left (\frac {x^{2/3}\,\left (\frac {d\,\left (2\,b\,d\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )-6\,b\,d\,e\,\left (3\,a^2-b^2\,n^2\right )\right )}{e}+12\,b^3\,d^2\,n^2\right )}{2\,e}-\frac {x^{4/3}\,\left (2\,b\,d\,e\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )-6\,b\,d\,e\,\left (3\,a^2-b^2\,n^2\right )\right )}{4\,e}+\frac {b\,e\,x^2\,\left (9\,a^2-6\,a\,b\,n+2\,b^2\,n^2\right )}{3}\right )}{2\,e}+\frac {\ln \left (d+e\,x^{2/3}\right )\,\left (18\,a^2\,b\,d^3\,n-66\,a\,b^2\,d^3\,n^2+85\,b^3\,d^3\,n^3\right )}{12\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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
)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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