Integrand size = 22, antiderivative size = 138 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {b n}{12 x^2}-\frac {b d n}{10 e x^{5/3}}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \] Output:
1/12*b*n/x^2-1/10*b*d*n/e/x^(5/3)+1/8*b*d^2*n/e^2/x^(4/3)-1/6*b*d^3*n/e^3/ x+1/4*b*d^4*n/e^4/x^(2/3)-1/2*b*d^5*n/e^5/x^(1/3)+1/2*b*d^6*n*ln(d+e/x^(1/ 3))/e^6-1/2*(a+b*ln(c*(d+e/x^(1/3))^n))/x^2
Time = 0.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {1}{6} b e n \left (-\frac {1}{2 e x^2}+\frac {3 d}{5 e^2 x^{5/3}}-\frac {3 d^2}{4 e^3 x^{4/3}}+\frac {d^3}{e^4 x}-\frac {3 d^4}{2 e^5 x^{2/3}}+\frac {3 d^5}{e^6 \sqrt [3]{x}}-\frac {3 d^6 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^7}\right )-\frac {b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(1/3))^n])/x^3,x]
Output:
-1/2*a/x^2 - (b*e*n*(-1/2*1/(e*x^2) + (3*d)/(5*e^2*x^(5/3)) - (3*d^2)/(4*e ^3*x^(4/3)) + d^3/(e^4*x) - (3*d^4)/(2*e^5*x^(2/3)) + (3*d^5)/(e^6*x^(1/3) ) - (3*d^6*Log[d + e/x^(1/3)])/e^7))/6 - (b*Log[c*(d + e/x^(1/3))^n])/(2*x ^2)
Time = 0.53 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -3 \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^{5/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -3 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{6 x^2}-\frac {1}{6} b e n \int \frac {1}{\left (d+\frac {e}{\sqrt [3]{x}}\right ) x^2}d\frac {1}{\sqrt [3]{x}}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -3 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{6 x^2}-\frac {1}{6} b e n \int \left (\frac {d^6}{e^6 \left (d+\frac {e}{\sqrt [3]{x}}\right )}-\frac {d^5}{e^6}+\frac {d^4}{e^5 \sqrt [3]{x}}-\frac {d^3}{e^4 x^{2/3}}+\frac {d^2}{e^3 x}-\frac {d}{e^2 x^{4/3}}+\frac {1}{e x^{5/3}}\right )d\frac {1}{\sqrt [3]{x}}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{6 x^2}-\frac {1}{6} b e n \left (\frac {d^6 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^7}-\frac {d^5}{e^6 \sqrt [3]{x}}+\frac {d^4}{2 e^5 x^{2/3}}-\frac {d^3}{3 e^4 x}+\frac {d^2}{4 e^3 x^{4/3}}-\frac {d}{5 e^2 x^{5/3}}+\frac {1}{6 e x^2}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(1/3))^n])/x^3,x]
Output:
-3*(-1/6*(b*e*n*(1/(6*e*x^2) - d/(5*e^2*x^(5/3)) + d^2/(4*e^3*x^(4/3)) - d ^3/(3*e^4*x) + d^4/(2*e^5*x^(2/3)) - d^5/(e^6*x^(1/3)) + (d^6*Log[d + e/x^ (1/3)])/e^7)) + (a + b*Log[c*(d + e/x^(1/3))^n])/(6*x^2))
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]
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]
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])
\[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )}{x^{3}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(1/3))^n))/x^3,x)
Output:
int((a+b*ln(c*(d+e/x^(1/3))^n))/x^3,x)
Time = 0.09 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=-\frac {20 \, b d^{3} e^{3} n x - 10 \, b e^{6} n + 60 \, a e^{6} - 10 \, {\left (6 \, a e^{6} + {\left (2 \, b d^{3} e^{3} - b e^{6}\right )} n\right )} x^{2} - 60 \, {\left (b e^{6} x^{2} - b e^{6}\right )} \log \left (c\right ) - 60 \, {\left (b d^{6} n x^{2} - b e^{6} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 15 \, {\left (4 \, b d^{5} e n x - b d^{2} e^{4} n\right )} x^{\frac {2}{3}} - 6 \, {\left (5 \, b d^{4} e^{2} n x - 2 \, b d e^{5} n\right )} x^{\frac {1}{3}}}{120 \, e^{6} x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x, algorithm="fricas")
Output:
-1/120*(20*b*d^3*e^3*n*x - 10*b*e^6*n + 60*a*e^6 - 10*(6*a*e^6 + (2*b*d^3* e^3 - b*e^6)*n)*x^2 - 60*(b*e^6*x^2 - b*e^6)*log(c) - 60*(b*d^6*n*x^2 - b* e^6*n)*log((d*x + e*x^(2/3))/x) + 15*(4*b*d^5*e*n*x - b*d^2*e^4*n)*x^(2/3) - 6*(5*b*d^4*e^2*n*x - 2*b*d*e^5*n)*x^(1/3))/(e^6*x^2)
Timed out. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(1/3))**n))/x**3,x)
Output:
Timed out
Time = 0.06 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {1}{120} \, b e n {\left (\frac {60 \, d^{6} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{7}} - \frac {20 \, d^{6} \log \left (x\right )}{e^{7}} - \frac {60 \, d^{5} x^{\frac {5}{3}} - 30 \, d^{4} e x^{\frac {4}{3}} + 20 \, d^{3} e^{2} x - 15 \, d^{2} e^{3} x^{\frac {2}{3}} + 12 \, d e^{4} x^{\frac {1}{3}} - 10 \, e^{5}}{e^{6} x^{2}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x, algorithm="maxima")
Output:
1/120*b*e*n*(60*d^6*log(d*x^(1/3) + e)/e^7 - 20*d^6*log(x)/e^7 - (60*d^5*x ^(5/3) - 30*d^4*e*x^(4/3) + 20*d^3*e^2*x - 15*d^2*e^3*x^(2/3) + 12*d*e^4*x ^(1/3) - 10*e^5)/(e^6*x^2)) - 1/2*b*log(c*(d + e/x^(1/3))^n)/x^2 - 1/2*a/x ^2
Time = 0.18 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {1}{120} \, {\left (e {\left (\frac {60 \, d^{6} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{e^{7}} - \frac {20 \, d^{6} \log \left ({\left | x \right |}\right )}{e^{7}} - \frac {60 \, d^{5} e x^{\frac {5}{3}} - 30 \, d^{4} e^{2} x^{\frac {4}{3}} + 20 \, d^{3} e^{3} x - 15 \, d^{2} e^{4} x^{\frac {2}{3}} + 12 \, d e^{5} x^{\frac {1}{3}} - 10 \, e^{6}}{e^{7} x^{2}}\right )} - \frac {60 \, \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )}{x^{2}}\right )} b n - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x, algorithm="giac")
Output:
1/120*(e*(60*d^6*log(abs(d*x^(1/3) + e))/e^7 - 20*d^6*log(abs(x))/e^7 - (6 0*d^5*e*x^(5/3) - 30*d^4*e^2*x^(4/3) + 20*d^3*e^3*x - 15*d^2*e^4*x^(2/3) + 12*d*e^5*x^(1/3) - 10*e^6)/(e^7*x^2)) - 60*log(d + e/x^(1/3))/x^2)*b*n - 1/2*b*log(c)/x^2 - 1/2*a/x^2
Time = 25.53 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {b\,n}{12\,x^2}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{2\,x^2}-\frac {b\,d\,n}{10\,e\,x^{5/3}}+\frac {b\,d^6\,n\,\ln \left (d+\frac {e}{x^{1/3}}\right )}{2\,e^6}-\frac {b\,d^3\,n}{6\,e^3\,x}+\frac {b\,d^2\,n}{8\,e^2\,x^{4/3}}+\frac {b\,d^4\,n}{4\,e^4\,x^{2/3}}-\frac {b\,d^5\,n}{2\,e^5\,x^{1/3}} \] Input:
int((a + b*log(c*(d + e/x^(1/3))^n))/x^3,x)
Output:
(b*n)/(12*x^2) - a/(2*x^2) - (b*log(c*(d + e/x^(1/3))^n))/(2*x^2) - (b*d*n )/(10*e*x^(5/3)) + (b*d^6*n*log(d + e/x^(1/3)))/(2*e^6) - (b*d^3*n)/(6*e^3 *x) + (b*d^2*n)/(8*e^2*x^(4/3)) + (b*d^4*n)/(4*e^4*x^(2/3)) - (b*d^5*n)/(2 *e^5*x^(1/3))
Time = 0.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.97 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {-60 x^{\frac {5}{3}} b \,d^{5} e n +15 x^{\frac {2}{3}} b \,d^{2} e^{4} n +30 x^{\frac {4}{3}} b \,d^{4} e^{2} n -12 x^{\frac {1}{3}} b d \,e^{5} n +60 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,d^{6} x^{2}-60 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,e^{6}-60 a \,e^{6}-20 b \,d^{3} e^{3} n x +10 b \,e^{6} n}{120 e^{6} x^{2}} \] Input:
int((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x)
Output:
( - 60*x**(2/3)*b*d**5*e*n*x + 15*x**(2/3)*b*d**2*e**4*n + 30*x**(1/3)*b*d **4*e**2*n*x - 12*x**(1/3)*b*d*e**5*n + 60*log(((x**(1/3)*d + e)**n*c)/x** (n/3))*b*d**6*x**2 - 60*log(((x**(1/3)*d + e)**n*c)/x**(n/3))*b*e**6 - 60* a*e**6 - 20*b*d**3*e**3*n*x + 10*b*e**6*n)/(120*e**6*x**2)