\(\int x^2 (a+b \log (c (d+e \sqrt [3]{x})^n)) \, dx\) [443]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=-\frac {b d^8 n \sqrt [3]{x}}{3 e^8}+\frac {b d^7 n x^{2/3}}{6 e^7}-\frac {b d^6 n x}{9 e^6}+\frac {b d^5 n x^{4/3}}{12 e^5}-\frac {b d^4 n x^{5/3}}{15 e^4}+\frac {b d^3 n x^2}{18 e^3}-\frac {b d^2 n x^{7/3}}{21 e^2}+\frac {b d n x^{8/3}}{24 e}-\frac {1}{27} b n x^3+\frac {b d^9 n \log \left (d+e \sqrt [3]{x}\right )}{3 e^9}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \] Output:

-1/3*b*d^8*n*x^(1/3)/e^8+1/6*b*d^7*n*x^(2/3)/e^7-1/9*b*d^6*n*x/e^6+1/12*b* 
d^5*n*x^(4/3)/e^5-1/15*b*d^4*n*x^(5/3)/e^4+1/18*b*d^3*n*x^2/e^3-1/21*b*d^2 
*n*x^(7/3)/e^2+1/24*b*d*n*x^(8/3)/e-1/27*b*n*x^3+1/3*b*d^9*n*ln(d+e*x^(1/3 
))/e^9+1/3*x^3*(a+b*ln(c*(d+e*x^(1/3))^n))
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.94 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {a x^3}{3}-\frac {1}{9} b e n \left (\frac {3 d^8 \sqrt [3]{x}}{e^9}-\frac {3 d^7 x^{2/3}}{2 e^8}+\frac {d^6 x}{e^7}-\frac {3 d^5 x^{4/3}}{4 e^6}+\frac {3 d^4 x^{5/3}}{5 e^5}-\frac {d^3 x^2}{2 e^4}+\frac {3 d^2 x^{7/3}}{7 e^3}-\frac {3 d x^{8/3}}{8 e^2}+\frac {x^3}{3 e}-\frac {3 d^9 \log \left (d+e \sqrt [3]{x}\right )}{e^{10}}\right )+\frac {1}{3} b x^3 \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right ) \] Input:

Integrate[x^2*(a + b*Log[c*(d + e*x^(1/3))^n]),x]
 

Output:

(a*x^3)/3 - (b*e*n*((3*d^8*x^(1/3))/e^9 - (3*d^7*x^(2/3))/(2*e^8) + (d^6*x 
)/e^7 - (3*d^5*x^(4/3))/(4*e^6) + (3*d^4*x^(5/3))/(5*e^5) - (d^3*x^2)/(2*e 
^4) + (3*d^2*x^(7/3))/(7*e^3) - (3*d*x^(8/3))/(8*e^2) + x^3/(3*e) - (3*d^9 
*Log[d + e*x^(1/3)])/e^10))/9 + (b*x^3*Log[c*(d + e*x^(1/3))^n])/3
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2904, 2842, 49, 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.

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

\(\Big \downarrow \) 2904

\(\displaystyle 3 \int x^{8/3} \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )d\sqrt [3]{x}\)

\(\Big \downarrow \) 2842

\(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )-\frac {1}{9} b e n \int \frac {x^3}{d+e \sqrt [3]{x}}d\sqrt [3]{x}\right )\)

\(\Big \downarrow \) 49

\(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )-\frac {1}{9} b e n \int \left (-\frac {d^9}{e^9 \left (d+e \sqrt [3]{x}\right )}+\frac {d^8}{e^9}-\frac {\sqrt [3]{x} d^7}{e^8}+\frac {x^{2/3} d^6}{e^7}-\frac {x d^5}{e^6}+\frac {x^{4/3} d^4}{e^5}-\frac {x^{5/3} d^3}{e^4}+\frac {x^2 d^2}{e^3}-\frac {x^{7/3} d}{e^2}+\frac {x^{8/3}}{e}\right )d\sqrt [3]{x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )-\frac {1}{9} b e n \left (-\frac {d^9 \log \left (d+e \sqrt [3]{x}\right )}{e^{10}}+\frac {d^8 \sqrt [3]{x}}{e^9}-\frac {d^7 x^{2/3}}{2 e^8}+\frac {d^6 x}{3 e^7}-\frac {d^5 x^{4/3}}{4 e^6}+\frac {d^4 x^{5/3}}{5 e^5}-\frac {d^3 x^2}{6 e^4}+\frac {d^2 x^{7/3}}{7 e^3}-\frac {d x^{8/3}}{8 e^2}+\frac {x^3}{9 e}\right )\right )\)

Input:

Int[x^2*(a + b*Log[c*(d + e*x^(1/3))^n]),x]
 

Output:

3*(-1/9*(b*e*n*((d^8*x^(1/3))/e^9 - (d^7*x^(2/3))/(2*e^8) + (d^6*x)/(3*e^7 
) - (d^5*x^(4/3))/(4*e^6) + (d^4*x^(5/3))/(5*e^5) - (d^3*x^2)/(6*e^4) + (d 
^2*x^(7/3))/(7*e^3) - (d*x^(8/3))/(8*e^2) + x^3/(9*e) - (d^9*Log[d + e*x^( 
1/3)])/e^10)) + (x^3*(a + b*Log[c*(d + e*x^(1/3))^n]))/9)
 

Defintions of rubi rules used

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009

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

rule 2842

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

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]

\[\int x^{2} \left (a +b \ln \left (c \left (d +e \,x^{\frac {1}{3}}\right )^{n}\right )\right )d x\]

Input:

int(x^2*(a+b*ln(c*(d+e*x^(1/3))^n)),x)
 

Output:

int(x^2*(a+b*ln(c*(d+e*x^(1/3))^n)),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.87 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {2520 \, b e^{9} x^{3} \log \left (c\right ) + 420 \, b d^{3} e^{6} n x^{2} - 840 \, b d^{6} e^{3} n x - 280 \, {\left (b e^{9} n - 9 \, a e^{9}\right )} x^{3} + 2520 \, {\left (b e^{9} n x^{3} + b d^{9} n\right )} \log \left (e x^{\frac {1}{3}} + d\right ) + 63 \, {\left (5 \, b d e^{8} n x^{2} - 8 \, b d^{4} e^{5} n x + 20 \, b d^{7} e^{2} n\right )} x^{\frac {2}{3}} - 90 \, {\left (4 \, b d^{2} e^{7} n x^{2} - 7 \, b d^{5} e^{4} n x + 28 \, b d^{8} e n\right )} x^{\frac {1}{3}}}{7560 \, e^{9}} \] Input:

integrate(x^2*(a+b*log(c*(d+e*x^(1/3))^n)),x, algorithm="fricas")
 

Output:

1/7560*(2520*b*e^9*x^3*log(c) + 420*b*d^3*e^6*n*x^2 - 840*b*d^6*e^3*n*x - 
280*(b*e^9*n - 9*a*e^9)*x^3 + 2520*(b*e^9*n*x^3 + b*d^9*n)*log(e*x^(1/3) + 
 d) + 63*(5*b*d*e^8*n*x^2 - 8*b*d^4*e^5*n*x + 20*b*d^7*e^2*n)*x^(2/3) - 90 
*(4*b*d^2*e^7*n*x^2 - 7*b*d^5*e^4*n*x + 28*b*d^8*e*n)*x^(1/3))/e^9
 

Sympy [A] (verification not implemented)

Time = 6.97 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.94 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {a x^{3}}{3} + b \left (- \frac {e n \left (- \frac {3 d^{9} \left (\begin {cases} \frac {\sqrt [3]{x}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e \sqrt [3]{x} \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{9}} + \frac {3 d^{8} \sqrt [3]{x}}{e^{9}} - \frac {3 d^{7} x^{\frac {2}{3}}}{2 e^{8}} + \frac {d^{6} x}{e^{7}} - \frac {3 d^{5} x^{\frac {4}{3}}}{4 e^{6}} + \frac {3 d^{4} x^{\frac {5}{3}}}{5 e^{5}} - \frac {d^{3} x^{2}}{2 e^{4}} + \frac {3 d^{2} x^{\frac {7}{3}}}{7 e^{3}} - \frac {3 d x^{\frac {8}{3}}}{8 e^{2}} + \frac {x^{3}}{3 e}\right )}{9} + \frac {x^{3} \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}}{3}\right ) \] Input:

integrate(x**2*(a+b*ln(c*(d+e*x**(1/3))**n)),x)
 

Output:

a*x**3/3 + b*(-e*n*(-3*d**9*Piecewise((x**(1/3)/d, Eq(e, 0)), (log(d + e*x 
**(1/3))/e, True))/e**9 + 3*d**8*x**(1/3)/e**9 - 3*d**7*x**(2/3)/(2*e**8) 
+ d**6*x/e**7 - 3*d**5*x**(4/3)/(4*e**6) + 3*d**4*x**(5/3)/(5*e**5) - d**3 
*x**2/(2*e**4) + 3*d**2*x**(7/3)/(7*e**3) - 3*d*x**(8/3)/(8*e**2) + x**3/( 
3*e))/9 + x**3*log(c*(d + e*x**(1/3))**n)/3)
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.76 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{7560} \, b e n {\left (\frac {2520 \, d^{9} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{10}} - \frac {280 \, e^{8} x^{3} - 315 \, d e^{7} x^{\frac {8}{3}} + 360 \, d^{2} e^{6} x^{\frac {7}{3}} - 420 \, d^{3} e^{5} x^{2} + 504 \, d^{4} e^{4} x^{\frac {5}{3}} - 630 \, d^{5} e^{3} x^{\frac {4}{3}} + 840 \, d^{6} e^{2} x - 1260 \, d^{7} e x^{\frac {2}{3}} + 2520 \, d^{8} x^{\frac {1}{3}}}{e^{9}}\right )} \] Input:

integrate(x^2*(a+b*log(c*(d+e*x^(1/3))^n)),x, algorithm="maxima")
 

Output:

1/3*b*x^3*log((e*x^(1/3) + d)^n*c) + 1/3*a*x^3 + 1/7560*b*e*n*(2520*d^9*lo 
g(e*x^(1/3) + d)/e^10 - (280*e^8*x^3 - 315*d*e^7*x^(8/3) + 360*d^2*e^6*x^( 
7/3) - 420*d^3*e^5*x^2 + 504*d^4*e^4*x^(5/3) - 630*d^5*e^3*x^(4/3) + 840*d 
^6*e^2*x - 1260*d^7*e*x^(2/3) + 2520*d^8*x^(1/3))/e^9)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (147) = 294\).

Time = 0.13 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.11 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {2520 \, b e x^{3} \log \left (c\right ) + 2520 \, a e x^{3} + {\left (\frac {2520 \, {\left (e x^{\frac {1}{3}} + d\right )}^{9} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} - \frac {22680 \, {\left (e x^{\frac {1}{3}} + d\right )}^{8} d \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} + \frac {90720 \, {\left (e x^{\frac {1}{3}} + d\right )}^{7} d^{2} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} - \frac {211680 \, {\left (e x^{\frac {1}{3}} + d\right )}^{6} d^{3} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} + \frac {317520 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} d^{4} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} - \frac {317520 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} d^{5} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} + \frac {211680 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{6} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} - \frac {90720 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{7} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} + \frac {22680 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{8} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{8}} - \frac {280 \, {\left (e x^{\frac {1}{3}} + d\right )}^{9}}{e^{8}} + \frac {2835 \, {\left (e x^{\frac {1}{3}} + d\right )}^{8} d}{e^{8}} - \frac {12960 \, {\left (e x^{\frac {1}{3}} + d\right )}^{7} d^{2}}{e^{8}} + \frac {35280 \, {\left (e x^{\frac {1}{3}} + d\right )}^{6} d^{3}}{e^{8}} - \frac {63504 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} d^{4}}{e^{8}} + \frac {79380 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} d^{5}}{e^{8}} - \frac {70560 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{6}}{e^{8}} + \frac {45360 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{7}}{e^{8}} - \frac {22680 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{8}}{e^{8}}\right )} b n}{7560 \, e} \] Input:

integrate(x^2*(a+b*log(c*(d+e*x^(1/3))^n)),x, algorithm="giac")
 

Output:

1/7560*(2520*b*e*x^3*log(c) + 2520*a*e*x^3 + (2520*(e*x^(1/3) + d)^9*log(e 
*x^(1/3) + d)/e^8 - 22680*(e*x^(1/3) + d)^8*d*log(e*x^(1/3) + d)/e^8 + 907 
20*(e*x^(1/3) + d)^7*d^2*log(e*x^(1/3) + d)/e^8 - 211680*(e*x^(1/3) + d)^6 
*d^3*log(e*x^(1/3) + d)/e^8 + 317520*(e*x^(1/3) + d)^5*d^4*log(e*x^(1/3) + 
 d)/e^8 - 317520*(e*x^(1/3) + d)^4*d^5*log(e*x^(1/3) + d)/e^8 + 211680*(e* 
x^(1/3) + d)^3*d^6*log(e*x^(1/3) + d)/e^8 - 90720*(e*x^(1/3) + d)^2*d^7*lo 
g(e*x^(1/3) + d)/e^8 + 22680*(e*x^(1/3) + d)*d^8*log(e*x^(1/3) + d)/e^8 - 
280*(e*x^(1/3) + d)^9/e^8 + 2835*(e*x^(1/3) + d)^8*d/e^8 - 12960*(e*x^(1/3 
) + d)^7*d^2/e^8 + 35280*(e*x^(1/3) + d)^6*d^3/e^8 - 63504*(e*x^(1/3) + d) 
^5*d^4/e^8 + 79380*(e*x^(1/3) + d)^4*d^5/e^8 - 70560*(e*x^(1/3) + d)^3*d^6 
/e^8 + 45360*(e*x^(1/3) + d)^2*d^7/e^8 - 22680*(e*x^(1/3) + d)*d^8/e^8)*b* 
n)/e
 

Mupad [B] (verification not implemented)

Time = 14.77 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.81 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {a\,x^3}{3}-\frac {b\,n\,x^3}{27}+\frac {b\,x^3\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}{3}+\frac {b\,d\,n\,x^{8/3}}{24\,e}-\frac {b\,d^6\,n\,x}{9\,e^6}+\frac {b\,d^9\,n\,\ln \left (d+e\,x^{1/3}\right )}{3\,e^9}+\frac {b\,d^3\,n\,x^2}{18\,e^3}-\frac {b\,d^2\,n\,x^{7/3}}{21\,e^2}-\frac {b\,d^4\,n\,x^{5/3}}{15\,e^4}+\frac {b\,d^5\,n\,x^{4/3}}{12\,e^5}+\frac {b\,d^7\,n\,x^{2/3}}{6\,e^7}-\frac {b\,d^8\,n\,x^{1/3}}{3\,e^8} \] Input:

int(x^2*(a + b*log(c*(d + e*x^(1/3))^n)),x)
 

Output:

(a*x^3)/3 - (b*n*x^3)/27 + (b*x^3*log(c*(d + e*x^(1/3))^n))/3 + (b*d*n*x^( 
8/3))/(24*e) - (b*d^6*n*x)/(9*e^6) + (b*d^9*n*log(d + e*x^(1/3)))/(3*e^9) 
+ (b*d^3*n*x^2)/(18*e^3) - (b*d^2*n*x^(7/3))/(21*e^2) - (b*d^4*n*x^(5/3))/ 
(15*e^4) + (b*d^5*n*x^(4/3))/(12*e^5) + (b*d^7*n*x^(2/3))/(6*e^7) - (b*d^8 
*n*x^(1/3))/(3*e^8)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {1260 x^{\frac {2}{3}} b \,d^{7} e^{2} n -504 x^{\frac {5}{3}} b \,d^{4} e^{5} n +315 x^{\frac {8}{3}} b d \,e^{8} n -2520 x^{\frac {1}{3}} b \,d^{8} e n +630 x^{\frac {4}{3}} b \,d^{5} e^{4} n -360 x^{\frac {7}{3}} b \,d^{2} e^{7} n +2520 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,d^{9}+2520 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,e^{9} x^{3}+2520 a \,e^{9} x^{3}-840 b \,d^{6} e^{3} n x +420 b \,d^{3} e^{6} n \,x^{2}-280 b \,e^{9} n \,x^{3}}{7560 e^{9}} \] Input:

int(x^2*(a+b*log(c*(d+e*x^(1/3))^n)),x)
 

Output:

(1260*x**(2/3)*b*d**7*e**2*n - 504*x**(2/3)*b*d**4*e**5*n*x + 315*x**(2/3) 
*b*d*e**8*n*x**2 - 2520*x**(1/3)*b*d**8*e*n + 630*x**(1/3)*b*d**5*e**4*n*x 
 - 360*x**(1/3)*b*d**2*e**7*n*x**2 + 2520*log((x**(1/3)*e + d)**n*c)*b*d** 
9 + 2520*log((x**(1/3)*e + d)**n*c)*b*e**9*x**3 + 2520*a*e**9*x**3 - 840*b 
*d**6*e**3*n*x + 420*b*d**3*e**6*n*x**2 - 280*b*e**9*n*x**3)/(7560*e**9)