\(\int x^3 (a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)) \, dx\) [489]

Optimal result

Integrand size = 22, antiderivative size = 239 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {b e^{11} n \sqrt [3]{x}}{4 d^{11}}-\frac {b e^{10} n x^{2/3}}{8 d^{10}}+\frac {b e^9 n x}{12 d^9}-\frac {b e^8 n x^{4/3}}{16 d^8}+\frac {b e^7 n x^{5/3}}{20 d^7}-\frac {b e^6 n x^2}{24 d^6}+\frac {b e^5 n x^{7/3}}{28 d^5}-\frac {b e^4 n x^{8/3}}{32 d^4}+\frac {b e^3 n x^3}{36 d^3}-\frac {b e^2 n x^{10/3}}{40 d^2}+\frac {b e n x^{11/3}}{44 d}-\frac {b e^{12} n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{4 d^{12}}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {b e^{12} n \log (x)}{12 d^{12}} \] Output:

1/4*b*e^11*n*x^(1/3)/d^11-1/8*b*e^10*n*x^(2/3)/d^10+1/12*b*e^9*n*x/d^9-1/1 
6*b*e^8*n*x^(4/3)/d^8+1/20*b*e^7*n*x^(5/3)/d^7-1/24*b*e^6*n*x^2/d^6+1/28*b 
*e^5*n*x^(7/3)/d^5-1/32*b*e^4*n*x^(8/3)/d^4+1/36*b*e^3*n*x^3/d^3-1/40*b*e^ 
2*n*x^(10/3)/d^2+1/44*b*e*n*x^(11/3)/d-1/4*b*e^12*n*ln(d+e/x^(1/3))/d^12+1 
/4*x^4*(a+b*ln(c*(d+e/x^(1/3))^n))-1/12*b*e^12*n*ln(x)/d^12
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} b x^4 \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {1}{12} b e n \left (\frac {3 e^{10} \sqrt [3]{x}}{d^{11}}-\frac {3 e^9 x^{2/3}}{2 d^{10}}+\frac {e^8 x}{d^9}-\frac {3 e^7 x^{4/3}}{4 d^8}+\frac {3 e^6 x^{5/3}}{5 d^7}-\frac {e^5 x^2}{2 d^6}+\frac {3 e^4 x^{7/3}}{7 d^5}-\frac {3 e^3 x^{8/3}}{8 d^4}+\frac {e^2 x^3}{3 d^3}-\frac {3 e x^{10/3}}{10 d^2}+\frac {3 x^{11/3}}{11 d}-\frac {3 e^{11} \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^{12}}-\frac {e^{11} \log (x)}{d^{12}}\right ) \] Input:

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

Output:

(a*x^4)/4 + (b*x^4*Log[c*(d + e/x^(1/3))^n])/4 + (b*e*n*((3*e^10*x^(1/3))/ 
d^11 - (3*e^9*x^(2/3))/(2*d^10) + (e^8*x)/d^9 - (3*e^7*x^(4/3))/(4*d^8) + 
(3*e^6*x^(5/3))/(5*d^7) - (e^5*x^2)/(2*d^6) + (3*e^4*x^(7/3))/(7*d^5) - (3 
*e^3*x^(8/3))/(8*d^4) + (e^2*x^3)/(3*d^3) - (3*e*x^(10/3))/(10*d^2) + (3*x 
^(11/3))/(11*d) - (3*e^11*Log[d + e/x^(1/3)])/d^12 - (e^11*Log[x])/d^12))/ 
12
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2904, 2842, 54, 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.

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

\(\Big \downarrow \) 2904

\(\displaystyle -3 \int x^{13/3} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )d\frac {1}{\sqrt [3]{x}}\)

\(\Big \downarrow \) 2842

\(\displaystyle -3 \left (\frac {1}{12} b e n \int \frac {x^4}{d+\frac {e}{\sqrt [3]{x}}}d\frac {1}{\sqrt [3]{x}}-\frac {1}{12} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )\right )\)

\(\Big \downarrow \) 54

\(\displaystyle -3 \left (\frac {1}{12} b e n \int \left (\frac {e^{12}}{d^{12} \left (d+\frac {e}{\sqrt [3]{x}}\right )}-\frac {\sqrt [3]{x} e^{11}}{d^{12}}+\frac {x^{2/3} e^{10}}{d^{11}}-\frac {x e^9}{d^{10}}+\frac {x^{4/3} e^8}{d^9}-\frac {x^{5/3} e^7}{d^8}+\frac {x^2 e^6}{d^7}-\frac {x^{7/3} e^5}{d^6}+\frac {x^{8/3} e^4}{d^5}-\frac {x^3 e^3}{d^4}+\frac {x^{10/3} e^2}{d^3}-\frac {x^{11/3} e}{d^2}+\frac {x^4}{d}\right )d\frac {1}{\sqrt [3]{x}}-\frac {1}{12} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 \left (\frac {1}{12} b e n \left (\frac {e^{11} \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^{12}}-\frac {e^{11} \log \left (\frac {1}{\sqrt [3]{x}}\right )}{d^{12}}-\frac {e^{10} \sqrt [3]{x}}{d^{11}}+\frac {e^9 x^{2/3}}{2 d^{10}}-\frac {e^8 x}{3 d^9}+\frac {e^7 x^{4/3}}{4 d^8}-\frac {e^6 x^{5/3}}{5 d^7}+\frac {e^5 x^2}{6 d^6}-\frac {e^4 x^{7/3}}{7 d^5}+\frac {e^3 x^{8/3}}{8 d^4}-\frac {e^2 x^3}{9 d^3}+\frac {e x^{10/3}}{10 d^2}-\frac {x^{11/3}}{11 d}\right )-\frac {1}{12} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )\right )\)

Input:

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

Output:

-3*(-1/12*(x^4*(a + b*Log[c*(d + e/x^(1/3))^n])) + (b*e*n*(-((e^10*x^(1/3) 
)/d^11) + (e^9*x^(2/3))/(2*d^10) - (e^8*x)/(3*d^9) + (e^7*x^(4/3))/(4*d^8) 
 - (e^6*x^(5/3))/(5*d^7) + (e^5*x^2)/(6*d^6) - (e^4*x^(7/3))/(7*d^5) + (e^ 
3*x^(8/3))/(8*d^4) - (e^2*x^3)/(9*d^3) + (e*x^(10/3))/(10*d^2) - x^(11/3)/ 
(11*d) + (e^11*Log[d + e/x^(1/3)])/d^12 - (e^11*Log[x^(-1/3)])/d^12))/12)
 

Defintions of rubi rules used

rule 54

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

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.97 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {27720 \, b d^{12} x^{4} \log \left (c\right ) + 3080 \, b d^{9} e^{3} n x^{3} + 27720 \, a d^{12} x^{4} - 4620 \, b d^{6} e^{6} n x^{2} + 9240 \, b d^{3} e^{9} n x - 27720 \, b d^{12} n \log \left (x^{\frac {1}{3}}\right ) + 27720 \, {\left (b d^{12} - b e^{12}\right )} n \log \left (d x^{\frac {1}{3}} + e\right ) + 27720 \, {\left (b d^{12} n x^{4} - b d^{12} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 63 \, {\left (40 \, b d^{11} e n x^{3} - 55 \, b d^{8} e^{4} n x^{2} + 88 \, b d^{5} e^{7} n x - 220 \, b d^{2} e^{10} n\right )} x^{\frac {2}{3}} - 198 \, {\left (14 \, b d^{10} e^{2} n x^{3} - 20 \, b d^{7} e^{5} n x^{2} + 35 \, b d^{4} e^{8} n x - 140 \, b d e^{11} n\right )} x^{\frac {1}{3}}}{110880 \, d^{12}} \] Input:

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

Output:

1/110880*(27720*b*d^12*x^4*log(c) + 3080*b*d^9*e^3*n*x^3 + 27720*a*d^12*x^ 
4 - 4620*b*d^6*e^6*n*x^2 + 9240*b*d^3*e^9*n*x - 27720*b*d^12*n*log(x^(1/3) 
) + 27720*(b*d^12 - b*e^12)*n*log(d*x^(1/3) + e) + 27720*(b*d^12*n*x^4 - b 
*d^12*n)*log((d*x + e*x^(2/3))/x) + 63*(40*b*d^11*e*n*x^3 - 55*b*d^8*e^4*n 
*x^2 + 88*b*d^5*e^7*n*x - 220*b*d^2*e^10*n)*x^(2/3) - 198*(14*b*d^10*e^2*n 
*x^3 - 20*b*d^7*e^5*n*x^2 + 35*b*d^4*e^8*n*x - 140*b*d*e^11*n)*x^(1/3))/d^ 
12
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.68 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{110880} \, b e n {\left (\frac {27720 \, e^{11} \log \left (d x^{\frac {1}{3}} + e\right )}{d^{12}} - \frac {2520 \, d^{10} x^{\frac {11}{3}} - 2772 \, d^{9} e x^{\frac {10}{3}} + 3080 \, d^{8} e^{2} x^{3} - 3465 \, d^{7} e^{3} x^{\frac {8}{3}} + 3960 \, d^{6} e^{4} x^{\frac {7}{3}} - 4620 \, d^{5} e^{5} x^{2} + 5544 \, d^{4} e^{6} x^{\frac {5}{3}} - 6930 \, d^{3} e^{7} x^{\frac {4}{3}} + 9240 \, d^{2} e^{8} x - 13860 \, d e^{9} x^{\frac {2}{3}} + 27720 \, e^{10} x^{\frac {1}{3}}}{d^{11}}\right )} \] Input:

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

Output:

1/4*b*x^4*log(c*(d + e/x^(1/3))^n) + 1/4*a*x^4 - 1/110880*b*e*n*(27720*e^1 
1*log(d*x^(1/3) + e)/d^12 - (2520*d^10*x^(11/3) - 2772*d^9*e*x^(10/3) + 30 
80*d^8*e^2*x^3 - 3465*d^7*e^3*x^(8/3) + 3960*d^6*e^4*x^(7/3) - 4620*d^5*e^ 
5*x^2 + 5544*d^4*e^6*x^(5/3) - 6930*d^3*e^7*x^(4/3) + 9240*d^2*e^8*x - 138 
60*d*e^9*x^(2/3) + 27720*e^10*x^(1/3))/d^11)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.71 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} + \frac {1}{110880} \, {\left (27720 \, x^{4} \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right ) - e {\left (\frac {27720 \, e^{11} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{d^{12}} - \frac {2520 \, d^{10} x^{\frac {11}{3}} - 2772 \, d^{9} e x^{\frac {10}{3}} + 3080 \, d^{8} e^{2} x^{3} - 3465 \, d^{7} e^{3} x^{\frac {8}{3}} + 3960 \, d^{6} e^{4} x^{\frac {7}{3}} - 4620 \, d^{5} e^{5} x^{2} + 5544 \, d^{4} e^{6} x^{\frac {5}{3}} - 6930 \, d^{3} e^{7} x^{\frac {4}{3}} + 9240 \, d^{2} e^{8} x - 13860 \, d e^{9} x^{\frac {2}{3}} + 27720 \, e^{10} x^{\frac {1}{3}}}{d^{11}}\right )}\right )} b n \] Input:

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

Output:

1/4*b*x^4*log(c) + 1/4*a*x^4 + 1/110880*(27720*x^4*log(d + e/x^(1/3)) - e* 
(27720*e^11*log(abs(d*x^(1/3) + e))/d^12 - (2520*d^10*x^(11/3) - 2772*d^9* 
e*x^(10/3) + 3080*d^8*e^2*x^3 - 3465*d^7*e^3*x^(8/3) + 3960*d^6*e^4*x^(7/3 
) - 4620*d^5*e^5*x^2 + 5544*d^4*e^6*x^(5/3) - 6930*d^3*e^7*x^(4/3) + 9240* 
d^2*e^8*x - 13860*d*e^9*x^(2/3) + 27720*e^10*x^(1/3))/d^11))*b*n
 

Mupad [B] (verification not implemented)

Time = 25.72 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.80 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {\frac {a\,d^{12}\,x^4}{4}-\frac {b\,e^{12}\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{1/3}}+1\right )}{2}+\frac {b\,d^{12}\,x^4\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{4}+\frac {b\,d^3\,e^9\,n\,x}{12}+\frac {b\,d\,e^{11}\,n\,x^{1/3}}{4}+\frac {b\,d^{11}\,e\,n\,x^{11/3}}{44}-\frac {b\,d^6\,e^6\,n\,x^2}{24}+\frac {b\,d^9\,e^3\,n\,x^3}{36}-\frac {b\,d^2\,e^{10}\,n\,x^{2/3}}{8}-\frac {b\,d^4\,e^8\,n\,x^{4/3}}{16}+\frac {b\,d^5\,e^7\,n\,x^{5/3}}{20}+\frac {b\,d^7\,e^5\,n\,x^{7/3}}{28}-\frac {b\,d^8\,e^4\,n\,x^{8/3}}{32}-\frac {b\,d^{10}\,e^2\,n\,x^{10/3}}{40}}{d^{12}} \] Input:

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

Output:

((a*d^12*x^4)/4 - (b*e^12*n*atanh((2*e)/(d*x^(1/3)) + 1))/2 + (b*d^12*x^4* 
log(c*(d + e/x^(1/3))^n))/4 + (b*d^3*e^9*n*x)/12 + (b*d*e^11*n*x^(1/3))/4 
+ (b*d^11*e*n*x^(11/3))/44 - (b*d^6*e^6*n*x^2)/24 + (b*d^9*e^3*n*x^3)/36 - 
 (b*d^2*e^10*n*x^(2/3))/8 - (b*d^4*e^8*n*x^(4/3))/16 + (b*d^5*e^7*n*x^(5/3 
))/20 + (b*d^7*e^5*n*x^(7/3))/28 - (b*d^8*e^4*n*x^(8/3))/32 - (b*d^10*e^2* 
n*x^(10/3))/40)/d^12
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {2520 x^{\frac {11}{3}} b \,d^{11} e n -3465 x^{\frac {8}{3}} b \,d^{8} e^{4} n +5544 x^{\frac {5}{3}} b \,d^{5} e^{7} n -13860 x^{\frac {2}{3}} b \,d^{2} e^{10} n -2772 x^{\frac {10}{3}} b \,d^{10} e^{2} n +3960 x^{\frac {7}{3}} b \,d^{7} e^{5} n -6930 x^{\frac {4}{3}} b \,d^{4} e^{8} n +27720 x^{\frac {1}{3}} b d \,e^{11} n -27720 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{12} n +27720 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,d^{12} x^{4}-27720 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,e^{12}+27720 a \,d^{12} x^{4}+3080 b \,d^{9} e^{3} n \,x^{3}-4620 b \,d^{6} e^{6} n \,x^{2}+9240 b \,d^{3} e^{9} n x}{110880 d^{12}} \] Input:

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

Output:

(2520*x**(2/3)*b*d**11*e*n*x**3 - 3465*x**(2/3)*b*d**8*e**4*n*x**2 + 5544* 
x**(2/3)*b*d**5*e**7*n*x - 13860*x**(2/3)*b*d**2*e**10*n - 2772*x**(1/3)*b 
*d**10*e**2*n*x**3 + 3960*x**(1/3)*b*d**7*e**5*n*x**2 - 6930*x**(1/3)*b*d* 
*4*e**8*n*x + 27720*x**(1/3)*b*d*e**11*n - 27720*log(x**(1/3))*b*e**12*n + 
 27720*log(((x**(1/3)*d + e)**n*c)/x**(n/3))*b*d**12*x**4 - 27720*log(((x* 
*(1/3)*d + e)**n*c)/x**(n/3))*b*e**12 + 27720*a*d**12*x**4 + 3080*b*d**9*e 
**3*n*x**3 - 4620*b*d**6*e**6*n*x**2 + 9240*b*d**3*e**9*n*x)/(110880*d**12 
)