Integrand size = 24, antiderivative size = 482 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {15 b^2 d^4 n^2 \left (d+e x^{2/3}\right )^2}{8 e^6}-\frac {10 b^2 d^3 n^2 \left (d+e x^{2/3}\right )^3}{9 e^6}+\frac {15 b^2 d^2 n^2 \left (d+e x^{2/3}\right )^4}{32 e^6}-\frac {3 b^2 d n^2 \left (d+e x^{2/3}\right )^5}{25 e^6}+\frac {b^2 n^2 \left (d+e x^{2/3}\right )^6}{72 e^6}-\frac {3 b^2 d^5 n^2 x^{2/3}}{e^5}+\frac {b^2 d^6 n^2 \log ^2\left (d+e x^{2/3}\right )}{4 e^6}+\frac {3 b d^5 n \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{e^6}-\frac {15 b d^4 n \left (d+e x^{2/3}\right )^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{4 e^6}+\frac {10 b d^3 n \left (d+e x^{2/3}\right )^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 e^6}-\frac {15 b d^2 n \left (d+e x^{2/3}\right )^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{8 e^6}+\frac {3 b d n \left (d+e x^{2/3}\right )^5 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 e^6}-\frac {b n \left (d+e x^{2/3}\right )^6 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{12 e^6}-\frac {b d^6 n \log \left (d+e x^{2/3}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{2 e^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \] Output:
15/8*b^2*d^4*n^2*(d+e*x^(2/3))^2/e^6-10/9*b^2*d^3*n^2*(d+e*x^(2/3))^3/e^6+ 15/32*b^2*d^2*n^2*(d+e*x^(2/3))^4/e^6-3/25*b^2*d*n^2*(d+e*x^(2/3))^5/e^6+1 /72*b^2*n^2*(d+e*x^(2/3))^6/e^6-3*b^2*d^5*n^2*x^(2/3)/e^5+1/4*b^2*d^6*n^2* ln(d+e*x^(2/3))^2/e^6+3*b*d^5*n*(d+e*x^(2/3))*(a+b*ln(c*(d+e*x^(2/3))^n))/ e^6-15/4*b*d^4*n*(d+e*x^(2/3))^2*(a+b*ln(c*(d+e*x^(2/3))^n))/e^6+10/3*b*d^ 3*n*(d+e*x^(2/3))^3*(a+b*ln(c*(d+e*x^(2/3))^n))/e^6-15/8*b*d^2*n*(d+e*x^(2 /3))^4*(a+b*ln(c*(d+e*x^(2/3))^n))/e^6+3/5*b*d*n*(d+e*x^(2/3))^5*(a+b*ln(c *(d+e*x^(2/3))^n))/e^6-1/12*b*n*(d+e*x^(2/3))^6*(a+b*ln(c*(d+e*x^(2/3))^n) )/e^6-1/2*b*d^6*n*ln(d+e*x^(2/3))*(a+b*ln(c*(d+e*x^(2/3))^n))/e^6+1/4*x^4* (a+b*ln(c*(d+e*x^(2/3))^n))^2
Time = 0.40 (sec) , antiderivative size = 323, normalized size of antiderivative = 0.67 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {e x^{2/3} \left (1800 a^2 e^5 x^{10/3}+60 a b n \left (60 d^5-30 d^4 e x^{2/3}+20 d^3 e^2 x^{4/3}-15 d^2 e^3 x^2+12 d e^4 x^{8/3}-10 e^5 x^{10/3}\right )+b^2 n^2 \left (-8820 d^5+2610 d^4 e x^{2/3}-1140 d^3 e^2 x^{4/3}+555 d^2 e^3 x^2-264 d e^4 x^{8/3}+100 e^5 x^{10/3}\right )\right )+180 b d^6 n (-20 a+49 b n) \log \left (d+e x^{2/3}\right )-60 b e x^{2/3} \left (-60 a e^5 x^{10/3}+b n \left (-60 d^5+30 d^4 e x^{2/3}-20 d^3 e^2 x^{4/3}+15 d^2 e^3 x^2-12 d e^4 x^{8/3}+10 e^5 x^{10/3}\right )\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )-1800 b^2 \left (d^6-e^6 x^4\right ) \log ^2\left (c \left (d+e x^{2/3}\right )^n\right )}{7200 e^6} \] Input:
Integrate[x^3*(a + b*Log[c*(d + e*x^(2/3))^n])^2,x]
Output:
(e*x^(2/3)*(1800*a^2*e^5*x^(10/3) + 60*a*b*n*(60*d^5 - 30*d^4*e*x^(2/3) + 20*d^3*e^2*x^(4/3) - 15*d^2*e^3*x^2 + 12*d*e^4*x^(8/3) - 10*e^5*x^(10/3)) + b^2*n^2*(-8820*d^5 + 2610*d^4*e*x^(2/3) - 1140*d^3*e^2*x^(4/3) + 555*d^2 *e^3*x^2 - 264*d*e^4*x^(8/3) + 100*e^5*x^(10/3))) + 180*b*d^6*n*(-20*a + 4 9*b*n)*Log[d + e*x^(2/3)] - 60*b*e*x^(2/3)*(-60*a*e^5*x^(10/3) + b*n*(-60* d^5 + 30*d^4*e*x^(2/3) - 20*d^3*e^2*x^(4/3) + 15*d^2*e^3*x^2 - 12*d*e^4*x^ (8/3) + 10*e^5*x^(10/3)))*Log[c*(d + e*x^(2/3))^n] - 1800*b^2*(d^6 - e^6*x ^4)*Log[c*(d + e*x^(2/3))^n]^2)/(7200*e^6)
Time = 0.97 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.64, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2904, 2845, 2858, 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.
| \(\displaystyle \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle \frac {3}{2} \int x^{10/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2dx^{2/3}\) |
| \(\Big \downarrow \) 2845 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {1}{3} b e n \int \frac {x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d+e x^{2/3}}dx^{2/3}\right )\) |
| \(\Big \downarrow \) 2858 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {1}{3} b n \int x^{10/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )d\left (d+e x^{2/3}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {b n \int e^6 x^{10/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )d\left (d+e x^{2/3}\right )}{3 e^6}\right )\) |
| \(\Big \downarrow \) 2772 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {b n \left (-b n \int \left (\frac {\log \left (d+e x^{2/3}\right ) d^6}{x^{2/3}}-6 d^5+\frac {15}{2} \left (d+e x^{2/3}\right ) d^4-\frac {20}{3} x^{4/3} d^3+\frac {15 x^2 d^2}{4}-\frac {6}{5} x^{8/3} d+\frac {x^{10/3}}{6}\right )d\left (d+e x^{2/3}\right )+d^6 \log \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )-6 d^5 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {15}{2} d^4 x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )-\frac {20}{3} d^3 x^2 \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {15}{4} d^2 x^{8/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )-\frac {6}{5} d x^{10/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {1}{6} x^4 \left (a+b \log \left (c x^{2 n/3}\right )\right )\right )}{3 e^6}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{2} \left (\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {b n \left (d^6 \log \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )-6 d^5 \left (d+e x^{2/3}\right ) \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {15}{2} d^4 x^{4/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )-\frac {20}{3} d^3 x^2 \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {15}{4} d^2 x^{8/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )-\frac {6}{5} d x^{10/3} \left (a+b \log \left (c x^{2 n/3}\right )\right )+\frac {1}{6} x^4 \left (a+b \log \left (c x^{2 n/3}\right )\right )-b n \left (\frac {1}{2} d^6 \log ^2\left (d+e x^{2/3}\right )-6 d^5 \left (d+e x^{2/3}\right )+\frac {15}{4} d^4 x^{4/3}-\frac {20 d^3 x^2}{9}+\frac {15}{16} d^2 x^{8/3}-\frac {6}{25} d x^{10/3}+\frac {x^4}{36}\right )\right )}{3 e^6}\right )\) |
Input:
Int[x^3*(a + b*Log[c*(d + e*x^(2/3))^n])^2,x]
Output:
(3*((x^4*(a + b*Log[c*(d + e*x^(2/3))^n])^2)/6 - (b*n*(-(b*n*(-6*d^5*(d + e*x^(2/3)) + (15*d^4*x^(4/3))/4 - (20*d^3*x^2)/9 + (15*d^2*x^(8/3))/16 - ( 6*d*x^(10/3))/25 + x^4/36 + (d^6*Log[d + e*x^(2/3)]^2)/2)) - 6*d^5*(d + e* x^(2/3))*(a + b*Log[c*x^((2*n)/3)]) + (15*d^4*x^(4/3)*(a + b*Log[c*x^((2*n )/3)]))/2 - (20*d^3*x^2*(a + b*Log[c*x^((2*n)/3)]))/3 + (15*d^2*x^(8/3)*(a + b*Log[c*x^((2*n)/3)]))/4 - (6*d*x^(10/3)*(a + b*Log[c*x^((2*n)/3)]))/5 + (x^4*(a + b*Log[c*x^((2*n)/3)]))/6 + d^6*Log[d + e*x^(2/3)]*(a + b*Log[c *x^((2*n)/3)])))/(3*e^6)))/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])
\[\int x^{3} {\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{2}d x\]
Input:
int(x^3*(a+b*ln(c*(d+e*x^(2/3))^n))^2,x)
Output:
int(x^3*(a+b*ln(c*(d+e*x^(2/3))^n))^2,x)
Time = 0.17 (sec) , antiderivative size = 508, normalized size of antiderivative = 1.05 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {1800 \, b^{2} e^{6} x^{4} \log \left (c\right )^{2} + 100 \, {\left (b^{2} e^{6} n^{2} - 6 \, a b e^{6} n + 18 \, a^{2} e^{6}\right )} x^{4} - 60 \, {\left (19 \, b^{2} d^{3} e^{3} n^{2} - 20 \, a b d^{3} e^{3} n\right )} x^{2} + 1800 \, {\left (b^{2} e^{6} n^{2} x^{4} - b^{2} d^{6} n^{2}\right )} \log \left (e x^{\frac {2}{3}} + d\right )^{2} + 60 \, {\left (20 \, b^{2} d^{3} e^{3} n^{2} x^{2} + 147 \, b^{2} d^{6} n^{2} - 60 \, a b d^{6} n - 10 \, {\left (b^{2} e^{6} n^{2} - 6 \, a b e^{6} n\right )} x^{4} + 60 \, {\left (b^{2} e^{6} n x^{4} - b^{2} d^{6} n\right )} \log \left (c\right ) - 15 \, {\left (b^{2} d^{2} e^{4} n^{2} x^{2} - 4 \, b^{2} d^{5} e n^{2}\right )} x^{\frac {2}{3}} + 6 \, {\left (2 \, b^{2} d e^{5} n^{2} x^{3} - 5 \, b^{2} d^{4} e^{2} n^{2} x\right )} x^{\frac {1}{3}}\right )} \log \left (e x^{\frac {2}{3}} + d\right ) + 600 \, {\left (2 \, b^{2} d^{3} e^{3} n x^{2} - {\left (b^{2} e^{6} n - 6 \, a b e^{6}\right )} x^{4}\right )} \log \left (c\right ) - 15 \, {\left (588 \, b^{2} d^{5} e n^{2} - 240 \, a b d^{5} e n - {\left (37 \, b^{2} d^{2} e^{4} n^{2} - 60 \, a b d^{2} e^{4} n\right )} x^{2} + 60 \, {\left (b^{2} d^{2} e^{4} n x^{2} - 4 \, b^{2} d^{5} e n\right )} \log \left (c\right )\right )} x^{\frac {2}{3}} - 6 \, {\left (4 \, {\left (11 \, b^{2} d e^{5} n^{2} - 30 \, a b d e^{5} n\right )} x^{3} - 15 \, {\left (29 \, b^{2} d^{4} e^{2} n^{2} - 20 \, a b d^{4} e^{2} n\right )} x - 60 \, {\left (2 \, b^{2} d e^{5} n x^{3} - 5 \, b^{2} d^{4} e^{2} n x\right )} \log \left (c\right )\right )} x^{\frac {1}{3}}}{7200 \, e^{6}} \] Input:
integrate(x^3*(a+b*log(c*(d+e*x^(2/3))^n))^2,x, algorithm="fricas")
Output:
1/7200*(1800*b^2*e^6*x^4*log(c)^2 + 100*(b^2*e^6*n^2 - 6*a*b*e^6*n + 18*a^ 2*e^6)*x^4 - 60*(19*b^2*d^3*e^3*n^2 - 20*a*b*d^3*e^3*n)*x^2 + 1800*(b^2*e^ 6*n^2*x^4 - b^2*d^6*n^2)*log(e*x^(2/3) + d)^2 + 60*(20*b^2*d^3*e^3*n^2*x^2 + 147*b^2*d^6*n^2 - 60*a*b*d^6*n - 10*(b^2*e^6*n^2 - 6*a*b*e^6*n)*x^4 + 6 0*(b^2*e^6*n*x^4 - b^2*d^6*n)*log(c) - 15*(b^2*d^2*e^4*n^2*x^2 - 4*b^2*d^5 *e*n^2)*x^(2/3) + 6*(2*b^2*d*e^5*n^2*x^3 - 5*b^2*d^4*e^2*n^2*x)*x^(1/3))*l og(e*x^(2/3) + d) + 600*(2*b^2*d^3*e^3*n*x^2 - (b^2*e^6*n - 6*a*b*e^6)*x^4 )*log(c) - 15*(588*b^2*d^5*e*n^2 - 240*a*b*d^5*e*n - (37*b^2*d^2*e^4*n^2 - 60*a*b*d^2*e^4*n)*x^2 + 60*(b^2*d^2*e^4*n*x^2 - 4*b^2*d^5*e*n)*log(c))*x^ (2/3) - 6*(4*(11*b^2*d*e^5*n^2 - 30*a*b*d*e^5*n)*x^3 - 15*(29*b^2*d^4*e^2* n^2 - 20*a*b*d^4*e^2*n)*x - 60*(2*b^2*d*e^5*n*x^3 - 5*b^2*d^4*e^2*n*x)*log (c))*x^(1/3))/e^6
Timed out. \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\text {Timed out} \] Input:
integrate(x**3*(a+b*ln(c*(d+e*x**(2/3))**n))**2,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.68 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )^{2} + \frac {1}{2} \, a b x^{4} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {1}{4} \, a^{2} x^{4} - \frac {1}{120} \, a b e n {\left (\frac {60 \, d^{6} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{7}} + \frac {10 \, e^{5} x^{4} - 12 \, d e^{4} x^{\frac {10}{3}} + 15 \, d^{2} e^{3} x^{\frac {8}{3}} - 20 \, d^{3} e^{2} x^{2} + 30 \, d^{4} e x^{\frac {4}{3}} - 60 \, d^{5} x^{\frac {2}{3}}}{e^{6}}\right )} - \frac {1}{7200} \, {\left (60 \, e n {\left (\frac {60 \, d^{6} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{7}} + \frac {10 \, e^{5} x^{4} - 12 \, d e^{4} x^{\frac {10}{3}} + 15 \, d^{2} e^{3} x^{\frac {8}{3}} - 20 \, d^{3} e^{2} x^{2} + 30 \, d^{4} e x^{\frac {4}{3}} - 60 \, d^{5} x^{\frac {2}{3}}}{e^{6}}\right )} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) - \frac {{\left (100 \, e^{6} x^{4} - 264 \, d e^{5} x^{\frac {10}{3}} + 555 \, d^{2} e^{4} x^{\frac {8}{3}} - 1140 \, d^{3} e^{3} x^{2} + 1800 \, d^{6} \log \left (e x^{\frac {2}{3}} + d\right )^{2} + 2610 \, d^{4} e^{2} x^{\frac {4}{3}} + 8820 \, d^{6} \log \left (e x^{\frac {2}{3}} + d\right ) - 8820 \, d^{5} e x^{\frac {2}{3}}\right )} n^{2}}{e^{6}}\right )} b^{2} \] Input:
integrate(x^3*(a+b*log(c*(d+e*x^(2/3))^n))^2,x, algorithm="maxima")
Output:
1/4*b^2*x^4*log((e*x^(2/3) + d)^n*c)^2 + 1/2*a*b*x^4*log((e*x^(2/3) + d)^n *c) + 1/4*a^2*x^4 - 1/120*a*b*e*n*(60*d^6*log(e*x^(2/3) + d)/e^7 + (10*e^5 *x^4 - 12*d*e^4*x^(10/3) + 15*d^2*e^3*x^(8/3) - 20*d^3*e^2*x^2 + 30*d^4*e* x^(4/3) - 60*d^5*x^(2/3))/e^6) - 1/7200*(60*e*n*(60*d^6*log(e*x^(2/3) + d) /e^7 + (10*e^5*x^4 - 12*d*e^4*x^(10/3) + 15*d^2*e^3*x^(8/3) - 20*d^3*e^2*x ^2 + 30*d^4*e*x^(4/3) - 60*d^5*x^(2/3))/e^6)*log((e*x^(2/3) + d)^n*c) - (1 00*e^6*x^4 - 264*d*e^5*x^(10/3) + 555*d^2*e^4*x^(8/3) - 1140*d^3*e^3*x^2 + 1800*d^6*log(e*x^(2/3) + d)^2 + 2610*d^4*e^2*x^(4/3) + 8820*d^6*log(e*x^( 2/3) + d) - 8820*d^5*e*x^(2/3))*n^2/e^6)*b^2
Leaf count of result is larger than twice the leaf count of optimal. 905 vs. \(2 (412) = 824\).
Time = 0.34 (sec) , antiderivative size = 905, normalized size of antiderivative = 1.88 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\text {Too large to display} \] Input:
integrate(x^3*(a+b*log(c*(d+e*x^(2/3))^n))^2,x, algorithm="giac")
Output:
1/4*b^2*x^4*log(c)^2 + 1/2*a*b*x^4*log(c) + 1/4*a^2*x^4 + 1/7200*(1800*(e* x^(2/3) + d)^6*log(e*x^(2/3) + d)^2/e^6 - 10800*(e*x^(2/3) + d)^5*d*log(e* x^(2/3) + d)^2/e^6 + 27000*(e*x^(2/3) + d)^4*d^2*log(e*x^(2/3) + d)^2/e^6 - 36000*(e*x^(2/3) + d)^3*d^3*log(e*x^(2/3) + d)^2/e^6 + 27000*(e*x^(2/3) + d)^2*d^4*log(e*x^(2/3) + d)^2/e^6 - 600*(e*x^(2/3) + d)^6*log(e*x^(2/3) + d)/e^6 + 4320*(e*x^(2/3) + d)^5*d*log(e*x^(2/3) + d)/e^6 - 13500*(e*x^(2 /3) + d)^4*d^2*log(e*x^(2/3) + d)/e^6 + 24000*(e*x^(2/3) + d)^3*d^3*log(e* x^(2/3) + d)/e^6 - 27000*(e*x^(2/3) + d)^2*d^4*log(e*x^(2/3) + d)/e^6 + 10 0*(e*x^(2/3) + d)^6/e^6 - 864*(e*x^(2/3) + d)^5*d/e^6 + 3375*(e*x^(2/3) + d)^4*d^2/e^6 - 8000*(e*x^(2/3) + d)^3*d^3/e^6 + 13500*(e*x^(2/3) + d)^2*d^ 4/e^6 - 10800*((e*x^(2/3) + d)*log(e*x^(2/3) + d)^2 - 2*(e*x^(2/3) + d)*lo g(e*x^(2/3) + d) + 2*e*x^(2/3) + 2*d)*d^5/e^6)*b^2*n^2 + 1/120*b^2*n*(60*( e*x^(2/3) + d)^6*log(e*x^(2/3) + d)/e^6 - 360*(e*x^(2/3) + d)^5*d*log(e*x^ (2/3) + d)/e^6 + 900*(e*x^(2/3) + d)^4*d^2*log(e*x^(2/3) + d)/e^6 - 1200*( e*x^(2/3) + d)^3*d^3*log(e*x^(2/3) + d)/e^6 + 900*(e*x^(2/3) + d)^2*d^4*lo g(e*x^(2/3) + d)/e^6 - 10*(e*x^(2/3) + d)^6/e^6 + 72*(e*x^(2/3) + d)^5*d/e ^6 - 225*(e*x^(2/3) + d)^4*d^2/e^6 + 400*(e*x^(2/3) + d)^3*d^3/e^6 - 450*( e*x^(2/3) + d)^2*d^4/e^6 - 360*((e*x^(2/3) + d)*log(e*x^(2/3) + d) - e*x^( 2/3) - d)*d^5/e^6)*log(c) + 1/120*a*b*n*(60*(e*x^(2/3) + d)^6*log(e*x^(2/3 ) + d)/e^6 - 360*(e*x^(2/3) + d)^5*d*log(e*x^(2/3) + d)/e^6 + 900*(e*x^...
Time = 15.88 (sec) , antiderivative size = 440, normalized size of antiderivative = 0.91 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}+\frac {b^2\,x^4\,{\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^2}{4}+\frac {b^2\,n^2\,x^4}{72}+\frac {a\,b\,x^4\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2}-\frac {b^2\,d^6\,{\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}^2}{4\,e^6}-\frac {a\,b\,n\,x^4}{12}-\frac {b^2\,n\,x^4\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{12}+\frac {49\,b^2\,d^6\,n^2\,\ln \left (d+e\,x^{2/3}\right )}{40\,e^6}-\frac {19\,b^2\,d^3\,n^2\,x^2}{120\,e^3}+\frac {37\,b^2\,d^2\,n^2\,x^{8/3}}{480\,e^2}+\frac {29\,b^2\,d^4\,n^2\,x^{4/3}}{80\,e^4}-\frac {49\,b^2\,d^5\,n^2\,x^{2/3}}{40\,e^5}-\frac {11\,b^2\,d\,n^2\,x^{10/3}}{300\,e}+\frac {b^2\,d^3\,n\,x^2\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{6\,e^3}-\frac {b^2\,d^2\,n\,x^{8/3}\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{8\,e^2}-\frac {b^2\,d^4\,n\,x^{4/3}\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{4\,e^4}+\frac {b^2\,d^5\,n\,x^{2/3}\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2\,e^5}+\frac {a\,b\,d\,n\,x^{10/3}}{10\,e}-\frac {a\,b\,d^6\,n\,\ln \left (d+e\,x^{2/3}\right )}{2\,e^6}+\frac {b^2\,d\,n\,x^{10/3}\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{10\,e}+\frac {a\,b\,d^3\,n\,x^2}{6\,e^3}-\frac {a\,b\,d^2\,n\,x^{8/3}}{8\,e^2}-\frac {a\,b\,d^4\,n\,x^{4/3}}{4\,e^4}+\frac {a\,b\,d^5\,n\,x^{2/3}}{2\,e^5} \] Input:
int(x^3*(a + b*log(c*(d + e*x^(2/3))^n))^2,x)
Output:
(a^2*x^4)/4 + (b^2*x^4*log(c*(d + e*x^(2/3))^n)^2)/4 + (b^2*n^2*x^4)/72 + (a*b*x^4*log(c*(d + e*x^(2/3))^n))/2 - (b^2*d^6*log(c*(d + e*x^(2/3))^n)^2 )/(4*e^6) - (a*b*n*x^4)/12 - (b^2*n*x^4*log(c*(d + e*x^(2/3))^n))/12 + (49 *b^2*d^6*n^2*log(d + e*x^(2/3)))/(40*e^6) - (19*b^2*d^3*n^2*x^2)/(120*e^3) + (37*b^2*d^2*n^2*x^(8/3))/(480*e^2) + (29*b^2*d^4*n^2*x^(4/3))/(80*e^4) - (49*b^2*d^5*n^2*x^(2/3))/(40*e^5) - (11*b^2*d*n^2*x^(10/3))/(300*e) + (b ^2*d^3*n*x^2*log(c*(d + e*x^(2/3))^n))/(6*e^3) - (b^2*d^2*n*x^(8/3)*log(c* (d + e*x^(2/3))^n))/(8*e^2) - (b^2*d^4*n*x^(4/3)*log(c*(d + e*x^(2/3))^n)) /(4*e^4) + (b^2*d^5*n*x^(2/3)*log(c*(d + e*x^(2/3))^n))/(2*e^5) + (a*b*d*n *x^(10/3))/(10*e) - (a*b*d^6*n*log(d + e*x^(2/3)))/(2*e^6) + (b^2*d*n*x^(1 0/3)*log(c*(d + e*x^(2/3))^n))/(10*e) + (a*b*d^3*n*x^2)/(6*e^3) - (a*b*d^2 *n*x^(8/3))/(8*e^2) - (a*b*d^4*n*x^(4/3))/(4*e^4) + (a*b*d^5*n*x^(2/3))/(2 *e^5)
Time = 0.17 (sec) , antiderivative size = 453, normalized size of antiderivative = 0.94 \[ \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {3600 x^{\frac {2}{3}} \mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{5} e n -900 x^{\frac {8}{3}} \mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{2} e^{4} n +3600 x^{\frac {2}{3}} a b \,d^{5} e n -900 x^{\frac {8}{3}} a b \,d^{2} e^{4} n -8820 x^{\frac {2}{3}} b^{2} d^{5} e \,n^{2}+555 x^{\frac {8}{3}} b^{2} d^{2} e^{4} n^{2}-1800 x^{\frac {4}{3}} \mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{4} e^{2} n +720 x^{\frac {10}{3}} \mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d \,e^{5} n -1800 x^{\frac {4}{3}} a b \,d^{4} e^{2} n +720 x^{\frac {10}{3}} a b d \,e^{5} n +2610 x^{\frac {4}{3}} b^{2} d^{4} e^{2} n^{2}-264 x^{\frac {10}{3}} b^{2} d \,e^{5} n^{2}-1800 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} b^{2} d^{6}+1800 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} b^{2} e^{6} x^{4}-3600 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) a b \,d^{6}+3600 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) a b \,e^{6} x^{4}+8820 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{6} n +1200 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{3} e^{3} n \,x^{2}-600 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} e^{6} n \,x^{4}+1800 a^{2} e^{6} x^{4}+1200 a b \,d^{3} e^{3} n \,x^{2}-600 a b \,e^{6} n \,x^{4}-1140 b^{2} d^{3} e^{3} n^{2} x^{2}+100 b^{2} e^{6} n^{2} x^{4}}{7200 e^{6}} \] Input:
int(x^3*(a+b*log(c*(d+e*x^(2/3))^n))^2,x)
Output:
(3600*x**(2/3)*log((x**(2/3)*e + d)**n*c)*b**2*d**5*e*n - 900*x**(2/3)*log ((x**(2/3)*e + d)**n*c)*b**2*d**2*e**4*n*x**2 + 3600*x**(2/3)*a*b*d**5*e*n - 900*x**(2/3)*a*b*d**2*e**4*n*x**2 - 8820*x**(2/3)*b**2*d**5*e*n**2 + 55 5*x**(2/3)*b**2*d**2*e**4*n**2*x**2 - 1800*x**(1/3)*log((x**(2/3)*e + d)** n*c)*b**2*d**4*e**2*n*x + 720*x**(1/3)*log((x**(2/3)*e + d)**n*c)*b**2*d*e **5*n*x**3 - 1800*x**(1/3)*a*b*d**4*e**2*n*x + 720*x**(1/3)*a*b*d*e**5*n*x **3 + 2610*x**(1/3)*b**2*d**4*e**2*n**2*x - 264*x**(1/3)*b**2*d*e**5*n**2* x**3 - 1800*log((x**(2/3)*e + d)**n*c)**2*b**2*d**6 + 1800*log((x**(2/3)*e + d)**n*c)**2*b**2*e**6*x**4 - 3600*log((x**(2/3)*e + d)**n*c)*a*b*d**6 + 3600*log((x**(2/3)*e + d)**n*c)*a*b*e**6*x**4 + 8820*log((x**(2/3)*e + d) **n*c)*b**2*d**6*n + 1200*log((x**(2/3)*e + d)**n*c)*b**2*d**3*e**3*n*x**2 - 600*log((x**(2/3)*e + d)**n*c)*b**2*e**6*n*x**4 + 1800*a**2*e**6*x**4 + 1200*a*b*d**3*e**3*n*x**2 - 600*a*b*e**6*n*x**4 - 1140*b**2*d**3*e**3*n** 2*x**2 + 100*b**2*e**6*n**2*x**4)/(7200*e**6)