\(\int \frac {a+b \log (c (d+e \sqrt [3]{x})^n)}{x^4} \, dx\) [449]

Optimal result

Integrand size = 22, antiderivative size = 192 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=-\frac {b e n}{24 d x^{8/3}}+\frac {b e^2 n}{21 d^2 x^{7/3}}-\frac {b e^3 n}{18 d^3 x^2}+\frac {b e^4 n}{15 d^4 x^{5/3}}-\frac {b e^5 n}{12 d^5 x^{4/3}}+\frac {b e^6 n}{9 d^6 x}-\frac {b e^7 n}{6 d^7 x^{2/3}}+\frac {b e^8 n}{3 d^8 \sqrt [3]{x}}-\frac {b e^9 n \log \left (d+e \sqrt [3]{x}\right )}{3 d^9}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{3 x^3}+\frac {b e^9 n \log (x)}{9 d^9} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{3 x^3}+\frac {1}{9} b e n \left (-\frac {3}{8 d x^{8/3}}+\frac {3 e}{7 d^2 x^{7/3}}-\frac {e^2}{2 d^3 x^2}+\frac {3 e^3}{5 d^4 x^{5/3}}-\frac {3 e^4}{4 d^5 x^{4/3}}+\frac {e^5}{d^6 x}-\frac {3 e^6}{2 d^7 x^{2/3}}+\frac {3 e^7}{d^8 \sqrt [3]{x}}-\frac {3 e^8 \log \left (d+e \sqrt [3]{x}\right )}{d^9}+\frac {e^8 \log (x)}{d^9}\right ) \] Input:

Integrate[(a + b*Log[c*(d + e*x^(1/3))^n])/x^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 175, 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.

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

\(\Big \downarrow \) 2904

\(\displaystyle 3 \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^{10/3}}d\sqrt [3]{x}\)

\(\Big \downarrow \) 2842

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

\(\Big \downarrow \) 54

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

\(\Big \downarrow \) 2009

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

Input:

Int[(a + b*Log[c*(d + e*x^(1/3))^n])/x^4,x]
 

Output:

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

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]

\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {1}{3}}\right )^{n}\right )}{x^{4}}d x\]

Input:

int((a+b*ln(c*(d+e*x^(1/3))^n))/x^4,x)
 

Output:

int((a+b*ln(c*(d+e*x^(1/3))^n))/x^4,x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=\frac {840 \, b e^{9} n x^{3} \log \left (x^{\frac {1}{3}}\right ) + 280 \, b d^{3} e^{6} n x^{2} - 140 \, b d^{6} e^{3} n x - 840 \, b d^{9} \log \left (c\right ) - 840 \, a d^{9} - 840 \, {\left (b e^{9} n x^{3} + b d^{9} n\right )} \log \left (e x^{\frac {1}{3}} + d\right ) + 30 \, {\left (28 \, b d e^{8} n x^{2} - 7 \, b d^{4} e^{5} n x + 4 \, b d^{7} e^{2} n\right )} x^{\frac {2}{3}} - 21 \, {\left (20 \, b d^{2} e^{7} n x^{2} - 8 \, b d^{5} e^{4} n x + 5 \, b d^{8} e n\right )} x^{\frac {1}{3}}}{2520 \, d^{9} x^{3}} \] Input:

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^4,x, algorithm="fricas")
 

Output:

1/2520*(840*b*e^9*n*x^3*log(x^(1/3)) + 280*b*d^3*e^6*n*x^2 - 140*b*d^6*e^3 
*n*x - 840*b*d^9*log(c) - 840*a*d^9 - 840*(b*e^9*n*x^3 + b*d^9*n)*log(e*x^ 
(1/3) + d) + 30*(28*b*d*e^8*n*x^2 - 7*b*d^4*e^5*n*x + 4*b*d^7*e^2*n)*x^(2/ 
3) - 21*(20*b*d^2*e^7*n*x^2 - 8*b*d^5*e^4*n*x + 5*b*d^8*e*n)*x^(1/3))/(d^9 
*x^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*(d+e*x**(1/3))**n))/x**4,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.72 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=-\frac {1}{2520} \, b e n {\left (\frac {840 \, e^{8} \log \left (e x^{\frac {1}{3}} + d\right )}{d^{9}} - \frac {280 \, e^{8} \log \left (x\right )}{d^{9}} - \frac {840 \, e^{7} x^{\frac {7}{3}} - 420 \, d e^{6} x^{2} + 280 \, d^{2} e^{5} x^{\frac {5}{3}} - 210 \, d^{3} e^{4} x^{\frac {4}{3}} + 168 \, d^{4} e^{3} x - 140 \, d^{5} e^{2} x^{\frac {2}{3}} + 120 \, d^{6} e x^{\frac {1}{3}} - 105 \, d^{7}}{d^{8} x^{\frac {8}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \] Input:

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^4,x, algorithm="maxima")
 

Output:

-1/2520*b*e*n*(840*e^8*log(e*x^(1/3) + d)/d^9 - 280*e^8*log(x)/d^9 - (840* 
e^7*x^(7/3) - 420*d*e^6*x^2 + 280*d^2*e^5*x^(5/3) - 210*d^3*e^4*x^(4/3) + 
168*d^4*e^3*x - 140*d^5*e^2*x^(2/3) + 120*d^6*e*x^(1/3) - 105*d^7)/(d^8*x^ 
(8/3))) - 1/3*b*log((e*x^(1/3) + d)^n*c)/x^3 - 1/3*a/x^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 489 vs. \(2 (154) = 308\).

Time = 0.14 (sec) , antiderivative size = 489, normalized size of antiderivative = 2.55 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=-\frac {\frac {840 \, b e^{10} n \log \left (e x^{\frac {1}{3}} + d\right )}{{\left (e x^{\frac {1}{3}} + d\right )}^{9} - 9 \, {\left (e x^{\frac {1}{3}} + d\right )}^{8} d + 36 \, {\left (e x^{\frac {1}{3}} + d\right )}^{7} d^{2} - 84 \, {\left (e x^{\frac {1}{3}} + d\right )}^{6} d^{3} + 126 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} d^{4} - 126 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} d^{5} + 84 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{6} - 36 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{7} + 9 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{8} - d^{9}} + \frac {840 \, b e^{10} n \log \left (e x^{\frac {1}{3}} + d\right )}{d^{9}} - \frac {840 \, b e^{10} n \log \left (e x^{\frac {1}{3}}\right )}{d^{9}} - \frac {840 \, {\left (e x^{\frac {1}{3}} + d\right )}^{8} b e^{10} n - 7140 \, {\left (e x^{\frac {1}{3}} + d\right )}^{7} b d e^{10} n + 26740 \, {\left (e x^{\frac {1}{3}} + d\right )}^{6} b d^{2} e^{10} n - 57750 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} b d^{3} e^{10} n + 78918 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} b d^{4} e^{10} n - 70252 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} b d^{5} e^{10} n + 40188 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} b d^{6} e^{10} n - 13827 \, {\left (e x^{\frac {1}{3}} + d\right )} b d^{7} e^{10} n + 2283 \, b d^{8} e^{10} n - 840 \, b d^{8} e^{10} \log \left (c\right ) - 840 \, a d^{8} e^{10}}{{\left (e x^{\frac {1}{3}} + d\right )}^{9} d^{8} - 9 \, {\left (e x^{\frac {1}{3}} + d\right )}^{8} d^{9} + 36 \, {\left (e x^{\frac {1}{3}} + d\right )}^{7} d^{10} - 84 \, {\left (e x^{\frac {1}{3}} + d\right )}^{6} d^{11} + 126 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} d^{12} - 126 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} d^{13} + 84 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{14} - 36 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{15} + 9 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{16} - d^{17}}}{2520 \, e} \] Input:

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^4,x, algorithm="giac")
 

Output:

-1/2520*(840*b*e^10*n*log(e*x^(1/3) + d)/((e*x^(1/3) + d)^9 - 9*(e*x^(1/3) 
 + d)^8*d + 36*(e*x^(1/3) + d)^7*d^2 - 84*(e*x^(1/3) + d)^6*d^3 + 126*(e*x 
^(1/3) + d)^5*d^4 - 126*(e*x^(1/3) + d)^4*d^5 + 84*(e*x^(1/3) + d)^3*d^6 - 
 36*(e*x^(1/3) + d)^2*d^7 + 9*(e*x^(1/3) + d)*d^8 - d^9) + 840*b*e^10*n*lo 
g(e*x^(1/3) + d)/d^9 - 840*b*e^10*n*log(e*x^(1/3))/d^9 - (840*(e*x^(1/3) + 
 d)^8*b*e^10*n - 7140*(e*x^(1/3) + d)^7*b*d*e^10*n + 26740*(e*x^(1/3) + d) 
^6*b*d^2*e^10*n - 57750*(e*x^(1/3) + d)^5*b*d^3*e^10*n + 78918*(e*x^(1/3) 
+ d)^4*b*d^4*e^10*n - 70252*(e*x^(1/3) + d)^3*b*d^5*e^10*n + 40188*(e*x^(1 
/3) + d)^2*b*d^6*e^10*n - 13827*(e*x^(1/3) + d)*b*d^7*e^10*n + 2283*b*d^8* 
e^10*n - 840*b*d^8*e^10*log(c) - 840*a*d^8*e^10)/((e*x^(1/3) + d)^9*d^8 - 
9*(e*x^(1/3) + d)^8*d^9 + 36*(e*x^(1/3) + d)^7*d^10 - 84*(e*x^(1/3) + d)^6 
*d^11 + 126*(e*x^(1/3) + d)^5*d^12 - 126*(e*x^(1/3) + d)^4*d^13 + 84*(e*x^ 
(1/3) + d)^3*d^14 - 36*(e*x^(1/3) + d)^2*d^15 + 9*(e*x^(1/3) + d)*d^16 - d 
^17))/e
 

Mupad [B] (verification not implemented)

Time = 14.72 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=-\frac {\frac {a\,d^9}{3}+\frac {b\,d^9\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}{3}+\frac {b\,d^6\,e^3\,n\,x}{18}+\frac {b\,d^8\,e\,n\,x^{1/3}}{24}-\frac {b\,d\,e^8\,n\,x^{8/3}}{3}-\frac {b\,d^3\,e^6\,n\,x^2}{9}-\frac {b\,d^7\,e^2\,n\,x^{2/3}}{21}-\frac {b\,d^5\,e^4\,n\,x^{4/3}}{15}+\frac {b\,d^4\,e^5\,n\,x^{5/3}}{12}+\frac {b\,d^2\,e^7\,n\,x^{7/3}}{6}}{d^9\,x^3}-\frac {2\,b\,e^9\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{1/3}}{d}+1\right )}{3\,d^9} \] Input:

int((a + b*log(c*(d + e*x^(1/3))^n))/x^4,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^4} \, dx=\frac {120 x^{\frac {2}{3}} b \,d^{7} e^{2} n -210 x^{\frac {5}{3}} b \,d^{4} e^{5} n +840 x^{\frac {8}{3}} b d \,e^{8} n -105 x^{\frac {1}{3}} b \,d^{8} e n +168 x^{\frac {4}{3}} b \,d^{5} e^{4} n -420 x^{\frac {7}{3}} b \,d^{2} e^{7} n +840 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{9} n \,x^{3}-840 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,d^{9}-840 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,e^{9} x^{3}-840 a \,d^{9}-140 b \,d^{6} e^{3} n x +280 b \,d^{3} e^{6} n \,x^{2}}{2520 d^{9} x^{3}} \] Input:

int((a+b*log(c*(d+e*x^(1/3))^n))/x^4,x)
 

Output:

(120*x**(2/3)*b*d**7*e**2*n - 210*x**(2/3)*b*d**4*e**5*n*x + 840*x**(2/3)* 
b*d*e**8*n*x**2 - 105*x**(1/3)*b*d**8*e*n + 168*x**(1/3)*b*d**5*e**4*n*x - 
 420*x**(1/3)*b*d**2*e**7*n*x**2 + 840*log(x**(1/3))*b*e**9*n*x**3 - 840*l 
og((x**(1/3)*e + d)**n*c)*b*d**9 - 840*log((x**(1/3)*e + d)**n*c)*b*e**9*x 
**3 - 840*a*d**9 - 140*b*d**6*e**3*n*x + 280*b*d**3*e**6*n*x**2)/(2520*d** 
9*x**3)