Integrand size = 22, antiderivative size = 87 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {b e n}{2 d x^{2/3}}+\frac {b e^2 n}{d^2 \sqrt [3]{x}}-\frac {b e^3 n \log \left (d+e \sqrt [3]{x}\right )}{d^3}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+\frac {b e^3 n \log (x)}{3 d^3} \] Output:
-1/2*b*e*n/d/x^(2/3)+b*e^2*n/d^2/x^(1/3)-b*e^3*n*ln(d+e*x^(1/3))/d^3-(a+b* ln(c*(d+e*x^(1/3))^n))/x+1/3*b*e^3*n*ln(x)/d^3
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {a}{x}-\frac {b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+\frac {1}{3} b e n \left (-\frac {3}{2 d x^{2/3}}+\frac {3 e}{d^2 \sqrt [3]{x}}-\frac {3 e^2 \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac {e^2 \log (x)}{d^3}\right ) \] Input:
Integrate[(a + b*Log[c*(d + e*x^(1/3))^n])/x^2,x]
Output:
-(a/x) - (b*Log[c*(d + e*x^(1/3))^n])/x + (b*e*n*(-3/(2*d*x^(2/3)) + (3*e) /(d^2*x^(1/3)) - (3*e^2*Log[d + e*x^(1/3)])/d^3 + (e^2*Log[x])/d^3))/3
Time = 0.45 (sec) , antiderivative size = 89, 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, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle 3 \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^{4/3}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle 3 \left (\frac {1}{3} b e n \int \frac {1}{\left (d+e \sqrt [3]{x}\right ) x}d\sqrt [3]{x}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{3 x}\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle 3 \left (\frac {1}{3} b e n \int \left (-\frac {e^3}{d^3 \left (d+e \sqrt [3]{x}\right )}+\frac {e^2}{d^3 \sqrt [3]{x}}-\frac {e}{d^2 x^{2/3}}+\frac {1}{d x}\right )d\sqrt [3]{x}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{3 x}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (\frac {1}{3} b e n \left (-\frac {e^2 \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac {e^2 \log \left (\sqrt [3]{x}\right )}{d^3}+\frac {e}{d^2 \sqrt [3]{x}}-\frac {1}{2 d x^{2/3}}\right )-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{3 x}\right )\) |
Input:
Int[(a + b*Log[c*(d + e*x^(1/3))^n])/x^2,x]
Output:
3*(-1/3*(a + b*Log[c*(d + e*x^(1/3))^n])/x + (b*e*n*(-1/2*1/(d*x^(2/3)) + e/(d^2*x^(1/3)) - (e^2*Log[d + e*x^(1/3)])/d^3 + (e^2*Log[x^(1/3)])/d^3))/ 3)
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])
\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {1}{3}}\right )^{n}\right )}{x^{2}}d x\]
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 = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=\frac {2 \, b e^{3} n x \log \left (x^{\frac {1}{3}}\right ) + 2 \, b d e^{2} n x^{\frac {2}{3}} - b d^{2} e n x^{\frac {1}{3}} - 2 \, b d^{3} \log \left (c\right ) - 2 \, a d^{3} - 2 \, {\left (b e^{3} n x + b d^{3} n\right )} \log \left (e x^{\frac {1}{3}} + d\right )}{2 \, d^{3} x} \] Input:
integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="fricas")
Output:
1/2*(2*b*e^3*n*x*log(x^(1/3)) + 2*b*d*e^2*n*x^(2/3) - b*d^2*e*n*x^(1/3) - 2*b*d^3*log(c) - 2*a*d^3 - 2*(b*e^3*n*x + b*d^3*n)*log(e*x^(1/3) + d))/(d^ 3*x)
Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (82) = 164\).
Time = 82.30 (sec) , antiderivative size = 483, normalized size of antiderivative = 5.55 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=\begin {cases} - \frac {a + b \log {\left (0^{n} c \right )}}{x} & \text {for}\: d = 0 \wedge e = 0 \\- \frac {a}{x} - \frac {b n}{3 x} - \frac {b \log {\left (c \left (e \sqrt [3]{x}\right )^{n} \right )}}{x} & \text {for}\: d = 0 \\- \frac {a + b \log {\left (0^{n} c \right )}}{x} & \text {for}\: d = - e \sqrt [3]{x} \\- \frac {6 a d^{4} x^{\frac {2}{3}}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} - \frac {6 a d^{3} e x}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} - \frac {6 b d^{4} x^{\frac {2}{3}} \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} - \frac {3 b d^{3} e n x}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} - \frac {6 b d^{3} e x \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} + \frac {3 b d^{2} e^{2} n x^{\frac {4}{3}}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} + \frac {2 b d e^{3} n x^{\frac {5}{3}} \log {\left (x \right )}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} + \frac {6 b d e^{3} n x^{\frac {5}{3}}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} - \frac {6 b d e^{3} x^{\frac {5}{3}} \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} + \frac {2 b e^{4} n x^{2} \log {\left (x \right )}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} - \frac {6 b e^{4} x^{2} \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}}{6 d^{4} x^{\frac {5}{3}} + 6 d^{3} e x^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*ln(c*(d+e*x**(1/3))**n))/x**2,x)
Output:
Piecewise((-(a + b*log(0**n*c))/x, Eq(d, 0) & Eq(e, 0)), (-a/x - b*n/(3*x) - b*log(c*(e*x**(1/3))**n)/x, Eq(d, 0)), (-(a + b*log(0**n*c))/x, Eq(d, - e*x**(1/3))), (-6*a*d**4*x**(2/3)/(6*d**4*x**(5/3) + 6*d**3*e*x**2) - 6*a* d**3*e*x/(6*d**4*x**(5/3) + 6*d**3*e*x**2) - 6*b*d**4*x**(2/3)*log(c*(d + e*x**(1/3))**n)/(6*d**4*x**(5/3) + 6*d**3*e*x**2) - 3*b*d**3*e*n*x/(6*d**4 *x**(5/3) + 6*d**3*e*x**2) - 6*b*d**3*e*x*log(c*(d + e*x**(1/3))**n)/(6*d* *4*x**(5/3) + 6*d**3*e*x**2) + 3*b*d**2*e**2*n*x**(4/3)/(6*d**4*x**(5/3) + 6*d**3*e*x**2) + 2*b*d*e**3*n*x**(5/3)*log(x)/(6*d**4*x**(5/3) + 6*d**3*e *x**2) + 6*b*d*e**3*n*x**(5/3)/(6*d**4*x**(5/3) + 6*d**3*e*x**2) - 6*b*d*e **3*x**(5/3)*log(c*(d + e*x**(1/3))**n)/(6*d**4*x**(5/3) + 6*d**3*e*x**2) + 2*b*e**4*n*x**2*log(x)/(6*d**4*x**(5/3) + 6*d**3*e*x**2) - 6*b*e**4*x**2 *log(c*(d + e*x**(1/3))**n)/(6*d**4*x**(5/3) + 6*d**3*e*x**2), True))
Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {1}{6} \, b e n {\left (\frac {6 \, e^{2} \log \left (e x^{\frac {1}{3}} + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {3 \, {\left (2 \, e x^{\frac {1}{3}} - d\right )}}{d^{2} x^{\frac {2}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )}{x} - \frac {a}{x} \] Input:
integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="maxima")
Output:
-1/6*b*e*n*(6*e^2*log(e*x^(1/3) + d)/d^3 - 2*e^2*log(x)/d^3 - 3*(2*e*x^(1/ 3) - d)/(d^2*x^(2/3))) - b*log((e*x^(1/3) + d)^n*c)/x - a/x
Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (75) = 150\).
Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.38 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {\frac {2 \, b e^{4} n \log \left (e x^{\frac {1}{3}} + d\right )}{{\left (e x^{\frac {1}{3}} + d\right )}^{3} - 3 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d + 3 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{2} - d^{3}} + \frac {2 \, b e^{4} n \log \left (e x^{\frac {1}{3}} + d\right )}{d^{3}} - \frac {2 \, b e^{4} n \log \left (e x^{\frac {1}{3}}\right )}{d^{3}} - \frac {2 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} b e^{4} n - 5 \, {\left (e x^{\frac {1}{3}} + d\right )} b d e^{4} n + 3 \, b d^{2} e^{4} n - 2 \, b d^{2} e^{4} \log \left (c\right ) - 2 \, a d^{2} e^{4}}{{\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{2} - 3 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{3} + 3 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{4} - d^{5}}}{2 \, e} \] Input:
integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="giac")
Output:
-1/2*(2*b*e^4*n*log(e*x^(1/3) + d)/((e*x^(1/3) + d)^3 - 3*(e*x^(1/3) + d)^ 2*d + 3*(e*x^(1/3) + d)*d^2 - d^3) + 2*b*e^4*n*log(e*x^(1/3) + d)/d^3 - 2* b*e^4*n*log(e*x^(1/3))/d^3 - (2*(e*x^(1/3) + d)^2*b*e^4*n - 5*(e*x^(1/3) + d)*b*d*e^4*n + 3*b*d^2*e^4*n - 2*b*d^2*e^4*log(c) - 2*a*d^2*e^4)/((e*x^(1 /3) + d)^3*d^2 - 3*(e*x^(1/3) + d)^2*d^3 + 3*(e*x^(1/3) + d)*d^4 - d^5))/e
Time = 14.69 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n\,x^{1/3}}{d^2}}{x^{2/3}}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}{x}-\frac {2\,b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{1/3}}{d}+1\right )}{d^3} \] Input:
int((a + b*log(c*(d + e*x^(1/3))^n))/x^2,x)
Output:
- ((b*e*n)/(2*d) - (b*e^2*n*x^(1/3))/d^2)/x^(2/3) - a/x - (b*log(c*(d + e* x^(1/3))^n))/x - (2*b*e^3*n*atanh((2*e*x^(1/3))/d + 1))/d^3
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=\frac {2 x^{\frac {2}{3}} b d \,e^{2} n -x^{\frac {1}{3}} b \,d^{2} e n +2 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{3} n x -2 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,d^{3}-2 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,e^{3} x -2 a \,d^{3}}{2 d^{3} x} \] Input:
int((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x)
Output:
(2*x**(2/3)*b*d*e**2*n - x**(1/3)*b*d**2*e*n + 2*log(x**(1/3))*b*e**3*n*x - 2*log((x**(1/3)*e + d)**n*c)*b*d**3 - 2*log((x**(1/3)*e + d)**n*c)*b*e** 3*x - 2*a*d**3)/(2*d**3*x)