Integrand size = 22, antiderivative size = 89 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {b n}{6 x^2}-\frac {b d n}{4 e x^{4/3}}+\frac {b d^2 n}{2 e^2 x^{2/3}}-\frac {b d^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 e^3}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{2 x^2} \] Output:
1/6*b*n/x^2-1/4*b*d*n/e/x^(4/3)+1/2*b*d^2*n/e^2/x^(2/3)-1/2*b*d^3*n*ln(d+e /x^(2/3))/e^3-1/2*(a+b*ln(c*(d+e/x^(2/3))^n))/x^2
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}+\frac {b n}{6 x^2}-\frac {b d n}{4 e x^{4/3}}+\frac {b d^2 n}{2 e^2 x^{2/3}}-\frac {b d^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 e^3}-\frac {b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{2 x^2} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(2/3))^n])/x^3,x]
Output:
-1/2*a/x^2 + (b*n)/(6*x^2) - (b*d*n)/(4*e*x^(4/3)) + (b*d^2*n)/(2*e^2*x^(2 /3)) - (b*d^3*n*Log[d + e/x^(2/3)])/(2*e^3) - (b*Log[c*(d + e/x^(2/3))^n]) /(2*x^2)
Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -\frac {3}{2} \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^{4/3}}d\frac {1}{x^{2/3}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -\frac {3}{2} \left (\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^2}-\frac {1}{3} b e n \int \frac {1}{\left (d+\frac {e}{x^{2/3}}\right ) x^2}d\frac {1}{x^{2/3}}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {3}{2} \left (\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^2}-\frac {1}{3} b e n \int \left (-\frac {d^3}{e^3 \left (d+\frac {e}{x^{2/3}}\right )}+\frac {d^2}{e^3}-\frac {d}{e^2 x^{2/3}}+\frac {1}{e x^{4/3}}\right )d\frac {1}{x^{2/3}}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \left (\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{3 x^2}-\frac {1}{3} b e n \left (-\frac {d^3 \log \left (d+\frac {e}{x^{2/3}}\right )}{e^4}+\frac {d^2}{e^3 x^{2/3}}-\frac {d}{2 e^2 x^{4/3}}+\frac {1}{3 e x^2}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(2/3))^n])/x^3,x]
Output:
(-3*(-1/3*(b*e*n*(1/(3*e*x^2) - d/(2*e^2*x^(4/3)) + d^2/(e^3*x^(2/3)) - (d ^3*Log[d + e/x^(2/3)])/e^4)) + (a + b*Log[c*(d + e/x^(2/3))^n])/(3*x^2)))/ 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])
\[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )}{x^{3}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(2/3))^n))/x^3,x)
Output:
int((a+b*ln(c*(d+e/x^(2/3))^n))/x^3,x)
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {6 \, b d^{2} e n x^{\frac {4}{3}} - 3 \, b d e^{2} n x^{\frac {2}{3}} + 2 \, b e^{3} n - 6 \, b e^{3} \log \left (c\right ) - 6 \, a e^{3} - 6 \, {\left (b d^{3} n x^{2} + b e^{3} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )}{12 \, e^{3} x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x, algorithm="fricas")
Output:
1/12*(6*b*d^2*e*n*x^(4/3) - 3*b*d*e^2*n*x^(2/3) + 2*b*e^3*n - 6*b*e^3*log( c) - 6*a*e^3 - 6*(b*d^3*n*x^2 + b*e^3*n)*log((d*x + e*x^(1/3))/x))/(e^3*x^ 2)
Timed out. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(2/3))**n))/x**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=-\frac {1}{12} \, b e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {2}{3}} + e\right )}{e^{4}} - \frac {6 \, d^{3} \log \left (x^{\frac {2}{3}}\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {4}{3}} - 3 \, d e x^{\frac {2}{3}} + 2 \, e^{2}}{e^{3} x^{2}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x, algorithm="maxima")
Output:
-1/12*b*e*n*(6*d^3*log(d*x^(2/3) + e)/e^4 - 6*d^3*log(x^(2/3))/e^4 - (6*d^ 2*x^(4/3) - 3*d*e*x^(2/3) + 2*e^2)/(e^3*x^2)) - 1/2*b*log(c*(d + e/x^(2/3) )^n)/x^2 - 1/2*a/x^2
Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.18 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {1}{12} \, {\left (e {\left (\frac {12 \, d^{3} \log \left (x^{\frac {1}{3}}\right )}{e^{4}} - \frac {6 \, d^{3} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{e^{4}} - \frac {11 \, d^{3} x^{2} - 6 \, d^{2} e x^{\frac {4}{3}} + 3 \, d e^{2} x^{\frac {2}{3}} - 2 \, e^{3}}{e^{4} x^{2}}\right )} - \frac {6 \, \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right )}{x^{2}}\right )} b n - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x, algorithm="giac")
Output:
1/12*(e*(12*d^3*log(x^(1/3))/e^4 - 6*d^3*log(abs(d*x^(2/3) + e))/e^4 - (11 *d^3*x^2 - 6*d^2*e*x^(4/3) + 3*d*e^2*x^(2/3) - 2*e^3)/(e^4*x^2)) - 6*log(d + e/x^(2/3))/x^2)*b*n - 1/2*b*log(c)/x^2 - 1/2*a/x^2
Time = 25.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {b\,n}{6\,x^2}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{2\,x^2}-\frac {b\,d\,n}{4\,e\,x^{4/3}}-\frac {b\,d^3\,n\,\ln \left (d+\frac {e}{x^{2/3}}\right )}{2\,e^3}+\frac {b\,d^2\,n}{2\,e^2\,x^{2/3}} \] Input:
int((a + b*log(c*(d + e/x^(2/3))^n))/x^3,x)
Output:
(b*n)/(6*x^2) - a/(2*x^2) - (b*log(c*(d + e/x^(2/3))^n))/(2*x^2) - (b*d*n) /(4*e*x^(4/3)) - (b*d^3*n*log(d + e/x^(2/3)))/(2*e^3) + (b*d^2*n)/(2*e^2*x ^(2/3))
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {-3 x^{\frac {2}{3}} b d \,e^{2} n +6 x^{\frac {4}{3}} b \,d^{2} e n -6 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,d^{3} x^{2}-6 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,e^{3}-6 a \,e^{3}+2 b \,e^{3} n}{12 e^{3} x^{2}} \] Input:
int((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x)
Output:
( - 3*x**(2/3)*b*d*e**2*n + 6*x**(1/3)*b*d**2*e*n*x - 6*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b*d**3*x**2 - 6*log(((x**(2/3)*d + e)**n*c)/x**((2 *n)/3))*b*e**3 - 6*a*e**3 + 2*b*e**3*n)/(12*e**3*x**2)