Integrand size = 22, antiderivative size = 143 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=-\frac {b e n}{10 d x^{5/3}}+\frac {b e^2 n}{8 d^2 x^{4/3}}-\frac {b e^3 n}{6 d^3 x}+\frac {b e^4 n}{4 d^4 x^{2/3}}-\frac {b e^5 n}{2 d^5 \sqrt [3]{x}}+\frac {b e^6 n \log \left (d+e \sqrt [3]{x}\right )}{2 d^6}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{2 x^2}-\frac {b e^6 n \log (x)}{6 d^6} \] Output:
-1/10*b*e*n/d/x^(5/3)+1/8*b*e^2*n/d^2/x^(4/3)-1/6*b*e^3*n/d^3/x+1/4*b*e^4* n/d^4/x^(2/3)-1/2*b*e^5*n/d^5/x^(1/3)+1/2*b*e^6*n*ln(d+e*x^(1/3))/d^6-1/2* (a+b*ln(c*(d+e*x^(1/3))^n))/x^2-1/6*b*e^6*n*ln(x)/d^6
Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{2 x^2}+\frac {1}{6} b e n \left (-\frac {3}{5 d x^{5/3}}+\frac {3 e}{4 d^2 x^{4/3}}-\frac {e^2}{d^3 x}+\frac {3 e^3}{2 d^4 x^{2/3}}-\frac {3 e^4}{d^5 \sqrt [3]{x}}+\frac {3 e^5 \log \left (d+e \sqrt [3]{x}\right )}{d^6}-\frac {e^5 \log (x)}{d^6}\right ) \] Input:
Integrate[(a + b*Log[c*(d + e*x^(1/3))^n])/x^3,x]
Output:
-1/2*a/x^2 - (b*Log[c*(d + e*x^(1/3))^n])/(2*x^2) + (b*e*n*(-3/(5*d*x^(5/3 )) + (3*e)/(4*d^2*x^(4/3)) - e^2/(d^3*x) + (3*e^3)/(2*d^4*x^(2/3)) - (3*e^ 4)/(d^5*x^(1/3)) + (3*e^5*Log[d + e*x^(1/3)])/d^6 - (e^5*Log[x])/d^6))/6
Time = 0.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.93, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle 3 \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^{7/3}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle 3 \left (\frac {1}{6} b e n \int \frac {1}{\left (d+e \sqrt [3]{x}\right ) x^2}d\sqrt [3]{x}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{6 x^2}\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle 3 \left (\frac {1}{6} b e n \int \left (\frac {e^6}{d^6 \left (d+e \sqrt [3]{x}\right )}-\frac {e^5}{d^6 \sqrt [3]{x}}+\frac {e^4}{d^5 x^{2/3}}-\frac {e^3}{d^4 x}+\frac {e^2}{d^3 x^{4/3}}-\frac {e}{d^2 x^{5/3}}+\frac {1}{d x^2}\right )d\sqrt [3]{x}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{6 x^2}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (\frac {1}{6} b e n \left (\frac {e^5 \log \left (d+e \sqrt [3]{x}\right )}{d^6}-\frac {e^5 \log \left (\sqrt [3]{x}\right )}{d^6}-\frac {e^4}{d^5 \sqrt [3]{x}}+\frac {e^3}{2 d^4 x^{2/3}}-\frac {e^2}{3 d^3 x}+\frac {e}{4 d^2 x^{4/3}}-\frac {1}{5 d x^{5/3}}\right )-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{6 x^2}\right )\) |
Input:
Int[(a + b*Log[c*(d + e*x^(1/3))^n])/x^3,x]
Output:
3*(-1/6*(a + b*Log[c*(d + e*x^(1/3))^n])/x^2 + (b*e*n*(-1/5*1/(d*x^(5/3)) + e/(4*d^2*x^(4/3)) - e^2/(3*d^3*x) + e^3/(2*d^4*x^(2/3)) - e^4/(d^5*x^(1/ 3)) + (e^5*Log[d + e*x^(1/3)])/d^6 - (e^5*Log[x^(1/3)])/d^6))/6)
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])
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])
\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {1}{3}}\right )^{n}\right )}{x^{3}}d x\]
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 = 0.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=-\frac {60 \, b e^{6} n x^{2} \log \left (x^{\frac {1}{3}}\right ) + 20 \, b d^{3} e^{3} n x + 60 \, b d^{6} \log \left (c\right ) + 60 \, a d^{6} - 60 \, {\left (b e^{6} n x^{2} - b d^{6} n\right )} \log \left (e x^{\frac {1}{3}} + d\right ) + 15 \, {\left (4 \, b d e^{5} n x - b d^{4} e^{2} n\right )} x^{\frac {2}{3}} - 6 \, {\left (5 \, b d^{2} e^{4} n x - 2 \, b d^{5} e n\right )} x^{\frac {1}{3}}}{120 \, d^{6} x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^3,x, algorithm="fricas")
Output:
-1/120*(60*b*e^6*n*x^2*log(x^(1/3)) + 20*b*d^3*e^3*n*x + 60*b*d^6*log(c) + 60*a*d^6 - 60*(b*e^6*n*x^2 - b*d^6*n)*log(e*x^(1/3) + d) + 15*(4*b*d*e^5* n*x - b*d^4*e^2*n)*x^(2/3) - 6*(5*b*d^2*e^4*n*x - 2*b*d^5*e*n)*x^(1/3))/(d ^6*x^2)
Timed out. \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e*x**(1/3))**n))/x**3,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.74 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=\frac {1}{120} \, b e n {\left (\frac {60 \, e^{5} \log \left (e x^{\frac {1}{3}} + d\right )}{d^{6}} - \frac {20 \, e^{5} \log \left (x\right )}{d^{6}} - \frac {60 \, e^{4} x^{\frac {4}{3}} - 30 \, d e^{3} x + 20 \, d^{2} e^{2} x^{\frac {2}{3}} - 15 \, d^{3} e x^{\frac {1}{3}} + 12 \, d^{4}}{d^{5} x^{\frac {5}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^3,x, algorithm="maxima")
Output:
1/120*b*e*n*(60*e^5*log(e*x^(1/3) + d)/d^6 - 20*e^5*log(x)/d^6 - (60*e^4*x ^(4/3) - 30*d*e^3*x + 20*d^2*e^2*x^(2/3) - 15*d^3*e*x^(1/3) + 12*d^4)/(d^5 *x^(5/3))) - 1/2*b*log((e*x^(1/3) + d)^n*c)/x^2 - 1/2*a/x^2
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (115) = 230\).
Time = 0.13 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.40 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=-\frac {\frac {60 \, b e^{7} n \log \left (e x^{\frac {1}{3}} + d\right )}{{\left (e x^{\frac {1}{3}} + d\right )}^{6} - 6 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} d + 15 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} d^{2} - 20 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{3} + 15 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{4} - 6 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{5} + d^{6}} - \frac {60 \, b e^{7} n \log \left (e x^{\frac {1}{3}} + d\right )}{d^{6}} + \frac {60 \, b e^{7} n \log \left (e x^{\frac {1}{3}}\right )}{d^{6}} + \frac {60 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} b e^{7} n - 330 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} b d e^{7} n + 740 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} b d^{2} e^{7} n - 855 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} b d^{3} e^{7} n + 522 \, {\left (e x^{\frac {1}{3}} + d\right )} b d^{4} e^{7} n - 137 \, b d^{5} e^{7} n + 60 \, b d^{5} e^{7} \log \left (c\right ) + 60 \, a d^{5} e^{7}}{{\left (e x^{\frac {1}{3}} + d\right )}^{6} d^{5} - 6 \, {\left (e x^{\frac {1}{3}} + d\right )}^{5} d^{6} + 15 \, {\left (e x^{\frac {1}{3}} + d\right )}^{4} d^{7} - 20 \, {\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{8} + 15 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{9} - 6 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{10} + d^{11}}}{120 \, e} \] Input:
integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^3,x, algorithm="giac")
Output:
-1/120*(60*b*e^7*n*log(e*x^(1/3) + d)/((e*x^(1/3) + d)^6 - 6*(e*x^(1/3) + d)^5*d + 15*(e*x^(1/3) + d)^4*d^2 - 20*(e*x^(1/3) + d)^3*d^3 + 15*(e*x^(1/ 3) + d)^2*d^4 - 6*(e*x^(1/3) + d)*d^5 + d^6) - 60*b*e^7*n*log(e*x^(1/3) + d)/d^6 + 60*b*e^7*n*log(e*x^(1/3))/d^6 + (60*(e*x^(1/3) + d)^5*b*e^7*n - 3 30*(e*x^(1/3) + d)^4*b*d*e^7*n + 740*(e*x^(1/3) + d)^3*b*d^2*e^7*n - 855*( e*x^(1/3) + d)^2*b*d^3*e^7*n + 522*(e*x^(1/3) + d)*b*d^4*e^7*n - 137*b*d^5 *e^7*n + 60*b*d^5*e^7*log(c) + 60*a*d^5*e^7)/((e*x^(1/3) + d)^6*d^5 - 6*(e *x^(1/3) + d)^5*d^6 + 15*(e*x^(1/3) + d)^4*d^7 - 20*(e*x^(1/3) + d)^3*d^8 + 15*(e*x^(1/3) + d)^2*d^9 - 6*(e*x^(1/3) + d)*d^10 + d^11))/e
Time = 14.78 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.76 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=\frac {b\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{1/3}}{d}+1\right )}{d^6}-\frac {\frac {b\,e\,n}{5\,d}-\frac {b\,e^4\,n\,x}{2\,d^4}-\frac {b\,e^2\,n\,x^{1/3}}{4\,d^2}+\frac {b\,e^3\,n\,x^{2/3}}{3\,d^3}+\frac {b\,e^5\,n\,x^{4/3}}{d^5}}{2\,x^{5/3}}-\frac {b\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}{2\,x^2}-\frac {a}{2\,x^2} \] Input:
int((a + b*log(c*(d + e*x^(1/3))^n))/x^3,x)
Output:
(b*e^6*n*atanh((2*e*x^(1/3))/d + 1))/d^6 - ((b*e*n)/(5*d) - (b*e^4*n*x)/(2 *d^4) - (b*e^2*n*x^(1/3))/(4*d^2) + (b*e^3*n*x^(2/3))/(3*d^3) + (b*e^5*n*x ^(4/3))/d^5)/(2*x^(5/3)) - (b*log(c*(d + e*x^(1/3))^n))/(2*x^2) - a/(2*x^2 )
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^3} \, dx=\frac {15 x^{\frac {2}{3}} b \,d^{4} e^{2} n -60 x^{\frac {5}{3}} b d \,e^{5} n -12 x^{\frac {1}{3}} b \,d^{5} e n +30 x^{\frac {4}{3}} b \,d^{2} e^{4} n -60 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{6} n \,x^{2}-60 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,d^{6}+60 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,e^{6} x^{2}-60 a \,d^{6}-20 b \,d^{3} e^{3} n x}{120 d^{6} x^{2}} \] Input:
int((a+b*log(c*(d+e*x^(1/3))^n))/x^3,x)
Output:
(15*x**(2/3)*b*d**4*e**2*n - 60*x**(2/3)*b*d*e**5*n*x - 12*x**(1/3)*b*d**5 *e*n + 30*x**(1/3)*b*d**2*e**4*n*x - 60*log(x**(1/3))*b*e**6*n*x**2 - 60*l og((x**(1/3)*e + d)**n*c)*b*d**6 + 60*log((x**(1/3)*e + d)**n*c)*b*e**6*x* *2 - 60*a*d**6 - 20*b*d**3*e**3*n*x)/(120*d**6*x**2)