\(\int \frac {a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)}{x^4} \, dx\) [496]

Optimal result

Integrand size = 22, antiderivative size = 187 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=\frac {b n}{27 x^3}-\frac {b d n}{24 e x^{8/3}}+\frac {b d^2 n}{21 e^2 x^{7/3}}-\frac {b d^3 n}{18 e^3 x^2}+\frac {b d^4 n}{15 e^4 x^{5/3}}-\frac {b d^5 n}{12 e^5 x^{4/3}}+\frac {b d^6 n}{9 e^6 x}-\frac {b d^7 n}{6 e^7 x^{2/3}}+\frac {b d^8 n}{3 e^8 \sqrt [3]{x}}-\frac {b d^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 e^9}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{3 x^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.95 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {1}{9} b e n \left (-\frac {1}{3 e x^3}+\frac {3 d}{8 e^2 x^{8/3}}-\frac {3 d^2}{7 e^3 x^{7/3}}+\frac {d^3}{2 e^4 x^2}-\frac {3 d^4}{5 e^5 x^{5/3}}+\frac {3 d^5}{4 e^6 x^{4/3}}-\frac {d^6}{e^7 x}+\frac {3 d^7}{2 e^8 x^{2/3}}-\frac {3 d^8}{e^9 \sqrt [3]{x}}+\frac {3 d^9 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^{10}}\right )-\frac {b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{3 x^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 175, 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.

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

\(\Big \downarrow \) 2904

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

\(\Big \downarrow \) 2842

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

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 = 0.13 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.14 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=\frac {840 \, b d^{6} e^{3} n x^{2} - 420 \, b d^{3} e^{6} n x + 280 \, b e^{9} n - 2520 \, a e^{9} + 140 \, {\left (18 \, a e^{9} - {\left (6 \, b d^{6} e^{3} - 3 \, b d^{3} e^{6} + 2 \, b e^{9}\right )} n\right )} x^{3} + 2520 \, {\left (b e^{9} x^{3} - b e^{9}\right )} \log \left (c\right ) - 2520 \, {\left (b d^{9} n x^{3} + b e^{9} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 90 \, {\left (28 \, b d^{8} e n x^{2} - 7 \, b d^{5} e^{4} n x + 4 \, b d^{2} e^{7} n\right )} x^{\frac {2}{3}} - 63 \, {\left (20 \, b d^{7} e^{2} n x^{2} - 8 \, b d^{4} e^{5} n x + 5 \, b d e^{8} n\right )} x^{\frac {1}{3}}}{7560 \, e^{9} x^{3}} \] Input:

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

Output:

1/7560*(840*b*d^6*e^3*n*x^2 - 420*b*d^3*e^6*n*x + 280*b*e^9*n - 2520*a*e^9 
 + 140*(18*a*e^9 - (6*b*d^6*e^3 - 3*b*d^3*e^6 + 2*b*e^9)*n)*x^3 + 2520*(b* 
e^9*x^3 - b*e^9)*log(c) - 2520*(b*d^9*n*x^3 + b*e^9*n)*log((d*x + e*x^(2/3 
))/x) + 90*(28*b*d^8*e*n*x^2 - 7*b*d^5*e^4*n*x + 4*b*d^2*e^7*n)*x^(2/3) - 
63*(20*b*d^7*e^2*n*x^2 - 8*b*d^4*e^5*n*x + 5*b*d*e^8*n)*x^(1/3))/(e^9*x^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=-\frac {1}{7560} \, b e n {\left (\frac {2520 \, d^{9} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{10}} - \frac {840 \, d^{9} \log \left (x\right )}{e^{10}} - \frac {2520 \, d^{8} x^{\frac {8}{3}} - 1260 \, d^{7} e x^{\frac {7}{3}} + 840 \, d^{6} e^{2} x^{2} - 630 \, d^{5} e^{3} x^{\frac {5}{3}} + 504 \, d^{4} e^{4} x^{\frac {4}{3}} - 420 \, d^{3} e^{5} x + 360 \, d^{2} e^{6} x^{\frac {2}{3}} - 315 \, d e^{7} x^{\frac {1}{3}} + 280 \, e^{8}}{e^{9} x^{3}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=-\frac {1}{7560} \, {\left (e {\left (\frac {2520 \, d^{9} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{e^{10}} - \frac {840 \, d^{9} \log \left ({\left | x \right |}\right )}{e^{10}} - \frac {2520 \, d^{8} e x^{\frac {8}{3}} - 1260 \, d^{7} e^{2} x^{\frac {7}{3}} + 840 \, d^{6} e^{3} x^{2} - 630 \, d^{5} e^{4} x^{\frac {5}{3}} + 504 \, d^{4} e^{5} x^{\frac {4}{3}} - 420 \, d^{3} e^{6} x + 360 \, d^{2} e^{7} x^{\frac {2}{3}} - 315 \, d e^{8} x^{\frac {1}{3}} + 280 \, e^{9}}{e^{10} x^{3}}\right )} + \frac {2520 \, \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )}{x^{3}}\right )} b n - \frac {b \log \left (c\right )}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 25.72 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.81 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=\frac {b\,n}{27\,x^3}-\frac {a}{3\,x^3}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{3\,x^3}-\frac {b\,d\,n}{24\,e\,x^{8/3}}-\frac {b\,d^9\,n\,\ln \left (d+\frac {e}{x^{1/3}}\right )}{3\,e^9}-\frac {b\,d^3\,n}{18\,e^3\,x^2}+\frac {b\,d^6\,n}{9\,e^6\,x}+\frac {b\,d^2\,n}{21\,e^2\,x^{7/3}}+\frac {b\,d^4\,n}{15\,e^4\,x^{5/3}}-\frac {b\,d^5\,n}{12\,e^5\,x^{4/3}}-\frac {b\,d^7\,n}{6\,e^7\,x^{2/3}}+\frac {b\,d^8\,n}{3\,e^8\,x^{1/3}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^4} \, dx=\frac {2520 x^{\frac {8}{3}} b \,d^{8} e n -630 x^{\frac {5}{3}} b \,d^{5} e^{4} n +360 x^{\frac {2}{3}} b \,d^{2} e^{7} n -1260 x^{\frac {7}{3}} b \,d^{7} e^{2} n +504 x^{\frac {4}{3}} b \,d^{4} e^{5} n -315 x^{\frac {1}{3}} b d \,e^{8} n -2520 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,d^{9} x^{3}-2520 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,e^{9}-2520 a \,e^{9}+840 b \,d^{6} e^{3} n \,x^{2}-420 b \,d^{3} e^{6} n x +280 b \,e^{9} n}{7560 e^{9} x^{3}} \] Input:

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

Output:

(2520*x**(2/3)*b*d**8*e*n*x**2 - 630*x**(2/3)*b*d**5*e**4*n*x + 360*x**(2/ 
3)*b*d**2*e**7*n - 1260*x**(1/3)*b*d**7*e**2*n*x**2 + 504*x**(1/3)*b*d**4* 
e**5*n*x - 315*x**(1/3)*b*d*e**8*n - 2520*log(((x**(1/3)*d + e)**n*c)/x**( 
n/3))*b*d**9*x**3 - 2520*log(((x**(1/3)*d + e)**n*c)/x**(n/3))*b*e**9 - 25 
20*a*e**9 + 840*b*d**6*e**3*n*x**2 - 420*b*d**3*e**6*n*x + 280*b*e**9*n)/( 
7560*e**9*x**3)