Integrand size = 22, antiderivative size = 94 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {b e n}{4 d x^{4/3}}+\frac {b e^2 n}{2 d^2 x^{2/3}}-\frac {b e^3 n \log \left (d+e x^{2/3}\right )}{2 d^3}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}+\frac {b e^3 n \log (x)}{3 d^3} \] Output:
-1/4*b*e*n/d/x^(4/3)+1/2*b*e^2*n/d^2/x^(2/3)-1/2*b*e^3*n*ln(d+e*x^(2/3))/d ^3-1/2*(a+b*ln(c*(d+e*x^(2/3))^n))/x^2+1/3*b*e^3*n*ln(x)/d^3
Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}+\frac {1}{3} b e n \left (-\frac {3}{4 d x^{4/3}}+\frac {3 e}{2 d^2 x^{2/3}}-\frac {3 e^2 \log \left (d+e x^{2/3}\right )}{2 d^3}+\frac {e^2 \log (x)}{d^3}\right ) \] Input:
Integrate[(a + b*Log[c*(d + e*x^(2/3))^n])/x^3,x]
Output:
-1/2*a/x^2 - (b*Log[c*(d + e*x^(2/3))^n])/(2*x^2) + (b*e*n*(-3/(4*d*x^(4/3 )) + (3*e)/(2*d^2*x^(2/3)) - (3*e^2*Log[d + e*x^(2/3)])/(2*d^3) + (e^2*Log [x])/d^3))/3
Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {3}{2} \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^{8/3}}dx^{2/3}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {3}{2} \left (\frac {1}{3} b e n \int \frac {1}{\left (d+e x^{2/3}\right ) x^2}dx^{2/3}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^2}\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {3}{2} \left (\frac {1}{3} b e n \int \left (-\frac {e^3}{d^3 \left (d+e x^{2/3}\right )}+\frac {e^2}{d^3 x^{2/3}}-\frac {e}{d^2 x^{4/3}}+\frac {1}{d x^2}\right )dx^{2/3}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^2}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{2} \left (\frac {1}{3} b e n \left (-\frac {e^2 \log \left (d+e x^{2/3}\right )}{d^3}+\frac {e^2 \log \left (x^{2/3}\right )}{d^3}+\frac {e}{d^2 x^{2/3}}-\frac {1}{2 d x^{4/3}}\right )-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^2}\right )\) |
Input:
Int[(a + b*Log[c*(d + e*x^(2/3))^n])/x^3,x]
Output:
(3*(-1/3*(a + b*Log[c*(d + e*x^(2/3))^n])/x^2 + (b*e*n*(-1/2*1/(d*x^(4/3)) + e/(d^2*x^(2/3)) - (e^2*Log[d + e*x^(2/3)])/d^3 + (e^2*Log[x^(2/3)])/d^3 ))/3))/2
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])
\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )}{x^{3}}d x\]
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 = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=\frac {4 \, b e^{3} n x^{2} \log \left (x^{\frac {1}{3}}\right ) + 2 \, b d e^{2} n x^{\frac {4}{3}} - b d^{2} e n x^{\frac {2}{3}} - 2 \, b d^{3} \log \left (c\right ) - 2 \, a d^{3} - 2 \, {\left (b e^{3} n x^{2} + b d^{3} n\right )} \log \left (e x^{\frac {2}{3}} + d\right )}{4 \, d^{3} x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x^3,x, algorithm="fricas")
Output:
1/4*(4*b*e^3*n*x^2*log(x^(1/3)) + 2*b*d*e^2*n*x^(4/3) - b*d^2*e*n*x^(2/3) - 2*b*d^3*log(c) - 2*a*d^3 - 2*(b*e^3*n*x^2 + b*d^3*n)*log(e*x^(2/3) + d)) /(d^3*x^2)
Timed out. \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e*x**(2/3))**n))/x**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {1}{4} \, b e n {\left (\frac {2 \, e^{2} \log \left (e x^{\frac {2}{3}} + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x^{\frac {2}{3}}\right )}{d^{3}} - \frac {2 \, e x^{\frac {2}{3}} - d}{d^{2} x^{\frac {4}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x^3,x, algorithm="maxima")
Output:
-1/4*b*e*n*(2*e^2*log(e*x^(2/3) + d)/d^3 - 2*e^2*log(x^(2/3))/d^3 - (2*e*x ^(2/3) - d)/(d^2*x^(4/3))) - 1/2*b*log((e*x^(2/3) + d)^n*c)/x^2 - 1/2*a/x^ 2
Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {{\left (e^{4} {\left (\frac {2 \, \log \left ({\left | e x^{\frac {2}{3}} + d \right |}\right )}{d^{3}} - \frac {2 \, \log \left ({\left | e x^{\frac {2}{3}} \right |}\right )}{d^{3}} - \frac {2 \, {\left (e x^{\frac {2}{3}} + d\right )} d - 3 \, d^{2}}{d^{3} e^{2} x^{\frac {4}{3}}}\right )} + \frac {2 \, e \log \left (e x^{\frac {2}{3}} + d\right )}{x^{2}}\right )} b n}{4 \, e} - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x^3,x, algorithm="giac")
Output:
-1/4*(e^4*(2*log(abs(e*x^(2/3) + d))/d^3 - 2*log(abs(e*x^(2/3)))/d^3 - (2* (e*x^(2/3) + d)*d - 3*d^2)/(d^3*e^2*x^(4/3))) + 2*e*log(e*x^(2/3) + d)/x^2 )*b*n/e - 1/2*b*log(c)/x^2 - 1/2*a/x^2
Time = 14.91 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n\,x^{2/3}}{d^2}}{2\,x^{4/3}}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2\,x^2}-\frac {b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{2/3}}{d}+1\right )}{d^3} \] Input:
int((a + b*log(c*(d + e*x^(2/3))^n))/x^3,x)
Output:
- ((b*e*n)/(2*d) - (b*e^2*n*x^(2/3))/d^2)/(2*x^(4/3)) - a/(2*x^2) - (b*log (c*(d + e*x^(2/3))^n))/(2*x^2) - (b*e^3*n*atanh((2*e*x^(2/3))/d + 1))/d^3
Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=\frac {-x^{\frac {2}{3}} b \,d^{2} e n +2 x^{\frac {4}{3}} b d \,e^{2} n +4 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{3} n \,x^{2}-2 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b \,d^{3}-2 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b \,e^{3} x^{2}-2 a \,d^{3}}{4 d^{3} x^{2}} \] Input:
int((a+b*log(c*(d+e*x^(2/3))^n))/x^3,x)
Output:
( - x**(2/3)*b*d**2*e*n + 2*x**(1/3)*b*d*e**2*n*x + 4*log(x**(1/3))*b*e**3 *n*x**2 - 2*log((x**(2/3)*e + d)**n*c)*b*d**3 - 2*log((x**(2/3)*e + d)**n* c)*b*e**3*x**2 - 2*a*d**3)/(4*d**3*x**2)