Integrand size = 22, antiderivative size = 82 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=\frac {b n}{3 x}-\frac {b d n}{2 e x^{2/3}}+\frac {b d^2 n}{e^2 \sqrt [3]{x}}-\frac {b d^3 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^3}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x} \] Output:
1/3*b*n/x-1/2*b*d*n/e/x^(2/3)+b*d^2*n/e^2/x^(1/3)-b*d^3*n*ln(d+e/x^(1/3))/ e^3-(a+b*ln(c*(d+e/x^(1/3))^n))/x
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=-\frac {a}{x}+\frac {b n}{3 x}-\frac {b d n}{2 e x^{2/3}}+\frac {b d^2 n}{e^2 \sqrt [3]{x}}-\frac {b d^3 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^3}-\frac {b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(1/3))^n])/x^2,x]
Output:
-(a/x) + (b*n)/(3*x) - (b*d*n)/(2*e*x^(2/3)) + (b*d^2*n)/(e^2*x^(1/3)) - ( b*d^3*n*Log[d + e/x^(1/3)])/e^3 - (b*Log[c*(d + e/x^(1/3))^n])/x
Time = 0.46 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -3 \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^{2/3}}d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -3 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{3 x}-\frac {1}{3} b e n \int \frac {1}{\left (d+\frac {e}{\sqrt [3]{x}}\right ) x}d\frac {1}{\sqrt [3]{x}}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -3 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{3 x}-\frac {1}{3} b e n \int \left (-\frac {d^3}{e^3 \left (d+\frac {e}{\sqrt [3]{x}}\right )}+\frac {d^2}{e^3}-\frac {d}{e^2 \sqrt [3]{x}}+\frac {1}{e x^{2/3}}\right )d\frac {1}{\sqrt [3]{x}}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{3 x}-\frac {1}{3} b e n \left (-\frac {d^3 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^4}+\frac {d^2}{e^3 \sqrt [3]{x}}-\frac {d}{2 e^2 x^{2/3}}+\frac {1}{3 e x}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(1/3))^n])/x^2,x]
Output:
-3*(-1/3*(b*e*n*(1/(3*e*x) - d/(2*e^2*x^(2/3)) + d^2/(e^3*x^(1/3)) - (d^3* Log[d + e/x^(1/3)])/e^4)) + (a + b*Log[c*(d + e/x^(1/3))^n])/(3*x))
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])
\[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )}{x^{2}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(1/3))^n))/x^2,x)
Output:
int((a+b*ln(c*(d+e/x^(1/3))^n))/x^2,x)
Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.30 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=\frac {6 \, b d^{2} e n x^{\frac {2}{3}} - 3 \, b d e^{2} n x^{\frac {1}{3}} + 2 \, b e^{3} n - 6 \, a e^{3} - 2 \, {\left (b e^{3} n - 3 \, a e^{3}\right )} x + 6 \, {\left (b e^{3} x - b e^{3}\right )} \log \left (c\right ) - 6 \, {\left (b d^{3} n x + b e^{3} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right )}{6 \, e^{3} x} \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^2,x, algorithm="fricas")
Output:
1/6*(6*b*d^2*e*n*x^(2/3) - 3*b*d*e^2*n*x^(1/3) + 2*b*e^3*n - 6*a*e^3 - 2*( b*e^3*n - 3*a*e^3)*x + 6*(b*e^3*x - b*e^3)*log(c) - 6*(b*d^3*n*x + b*e^3*n )*log((d*x + e*x^(2/3))/x))/(e^3*x)
Timed out. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(1/3))**n))/x**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=-\frac {1}{6} \, b e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{4}} - \frac {2 \, d^{3} \log \left (x\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {2}{3}} - 3 \, d e x^{\frac {1}{3}} + 2 \, e^{2}}{e^{3} x}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{x} - \frac {a}{x} \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^2,x, algorithm="maxima")
Output:
-1/6*b*e*n*(6*d^3*log(d*x^(1/3) + e)/e^4 - 2*d^3*log(x)/e^4 - (6*d^2*x^(2/ 3) - 3*d*e*x^(1/3) + 2*e^2)/(e^3*x)) - b*log(c*(d + e/x^(1/3))^n)/x - a/x
Time = 0.17 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.17 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=-\frac {1}{6} \, {\left (e {\left (\frac {6 \, d^{3} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{e^{4}} - \frac {2 \, d^{3} \log \left ({\left | x \right |}\right )}{e^{4}} - \frac {6 \, d^{2} e x^{\frac {2}{3}} - 3 \, d e^{2} x^{\frac {1}{3}} + 2 \, e^{3}}{e^{4} x}\right )} + \frac {6 \, \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )}{x}\right )} b n - \frac {b \log \left (c\right )}{x} - \frac {a}{x} \] Input:
integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^2,x, algorithm="giac")
Output:
-1/6*(e*(6*d^3*log(abs(d*x^(1/3) + e))/e^4 - 2*d^3*log(abs(x))/e^4 - (6*d^ 2*e*x^(2/3) - 3*d*e^2*x^(1/3) + 2*e^3)/(e^4*x)) + 6*log(d + e/x^(1/3))/x)* b*n - b*log(c)/x - a/x
Time = 25.92 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=\frac {b\,n}{3\,x}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{x}-\frac {b\,d\,n}{2\,e\,x^{2/3}}-\frac {b\,d^3\,n\,\ln \left (d+\frac {e}{x^{1/3}}\right )}{e^3}+\frac {b\,d^2\,n}{e^2\,x^{1/3}} \] Input:
int((a + b*log(c*(d + e/x^(1/3))^n))/x^2,x)
Output:
(b*n)/(3*x) - a/x - (b*log(c*(d + e/x^(1/3))^n))/x - (b*d*n)/(2*e*x^(2/3)) - (b*d^3*n*log(d + e/x^(1/3)))/e^3 + (b*d^2*n)/(e^2*x^(1/3))
Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^2} \, dx=\frac {6 x^{\frac {2}{3}} b \,d^{2} e n -3 x^{\frac {1}{3}} b d \,e^{2} n -6 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,d^{3} x -6 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,e^{3}-6 a \,e^{3}+2 b \,e^{3} n}{6 e^{3} x} \] Input:
int((a+b*log(c*(d+e/x^(1/3))^n))/x^2,x)
Output:
(6*x**(2/3)*b*d**2*e*n - 3*x**(1/3)*b*d*e**2*n - 6*log(((x**(1/3)*d + e)** n*c)/x**(n/3))*b*d**3*x - 6*log(((x**(1/3)*d + e)**n*c)/x**(n/3))*b*e**3 - 6*a*e**3 + 2*b*e**3*n)/(6*e**3*x)