Integrand size = 22, antiderivative size = 55 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=\frac {3}{2} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \log \left (-\frac {e x^{2/3}}{d}\right )+\frac {3}{2} b n \operatorname {PolyLog}\left (2,1+\frac {e x^{2/3}}{d}\right ) \] Output:
3/2*(a+b*ln(c*(d+e*x^(2/3))^n))*ln(-e*x^(2/3)/d)+3/2*b*n*polylog(2,1+e*x^( 2/3)/d)
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=a \log (x)+\frac {3}{2} b \left (\log \left (c \left (d+e x^{2/3}\right )^n\right ) \log \left (-\frac {e x^{2/3}}{d}\right )+n \operatorname {PolyLog}\left (2,\frac {d+e x^{2/3}}{d}\right )\right ) \] Input:
Integrate[(a + b*Log[c*(d + e*x^(2/3))^n])/x,x]
Output:
a*Log[x] + (3*b*(Log[c*(d + e*x^(2/3))^n]*Log[-((e*x^(2/3))/d)] + n*PolyLo g[2, (d + e*x^(2/3))/d]))/2
Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2904, 2841, 2752}
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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {3}{2} \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^{2/3}}dx^{2/3}\) |
\(\Big \downarrow \) 2841 |
\(\displaystyle \frac {3}{2} \left (\log \left (-\frac {e x^{2/3}}{d}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-b e n \int \frac {\log \left (-\frac {e x^{2/3}}{d}\right )}{d+e x^{2/3}}dx^{2/3}\right )\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {3}{2} \left (\log \left (-\frac {e x^{2/3}}{d}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+b n \operatorname {PolyLog}\left (2,\frac {x^{2/3} e}{d}+1\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e*x^(2/3))^n])/x,x]
Output:
(3*((a + b*Log[c*(d + e*x^(2/3))^n])*Log[-((e*x^(2/3))/d)] + b*n*PolyLog[2 , 1 + (e*x^(2/3))/d]))/2
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_ )), x_Symbol] :> Simp[Log[e*((f + g*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x )^n])/g), x] - Simp[b*e*(n/g) Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]
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])
\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )}{x}d x\]
Input:
int((a+b*ln(c*(d+e*x^(2/3))^n))/x,x)
Output:
int((a+b*ln(c*(d+e*x^(2/3))^n))/x,x)
\[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=\int { \frac {b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a}{x} \,d x } \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x,x, algorithm="fricas")
Output:
integral((b*log((e*x^(2/3) + d)^n*c) + a)/x, x)
\[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=\int \frac {a + b \log {\left (c \left (d + e x^{\frac {2}{3}}\right )^{n} \right )}}{x}\, dx \] Input:
integrate((a+b*ln(c*(d+e*x**(2/3))**n))/x,x)
Output:
Integral((a + b*log(c*(d + e*x**(2/3))**n))/x, x)
Leaf count of result is larger than twice the leaf count of optimal. 114 vs. \(2 (44) = 88\).
Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.07 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=-\frac {3}{2} \, {\left (2 \, \log \left (\frac {e x^{\frac {2}{3}}}{d} + 1\right ) \log \left (x^{\frac {1}{3}}\right ) + {\rm Li}_2\left (-\frac {e x^{\frac {2}{3}}}{d}\right )\right )} b n + \frac {3 \, {\left (2 \, b e n x^{\frac {2}{3}} \log \left (x^{\frac {1}{3}}\right ) - b e n x^{\frac {2}{3}}\right )}}{2 \, d} + \frac {2 \, b d \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n}\right ) \log \left (x\right ) + 2 \, {\left (b d \log \left (c\right ) + a d\right )} \log \left (x\right ) - \frac {2 \, b e n x \log \left (x\right ) - 3 \, b e n x}{x^{\frac {1}{3}}}}{2 \, d} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x,x, algorithm="maxima")
Output:
-3/2*(2*log(e*x^(2/3)/d + 1)*log(x^(1/3)) + dilog(-e*x^(2/3)/d))*b*n + 3/2 *(2*b*e*n*x^(2/3)*log(x^(1/3)) - b*e*n*x^(2/3))/d + 1/2*(2*b*d*log((e*x^(2 /3) + d)^n)*log(x) + 2*(b*d*log(c) + a*d)*log(x) - (2*b*e*n*x*log(x) - 3*b *e*n*x)/x^(1/3))/d
\[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=\int { \frac {b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a}{x} \,d x } \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x,x, algorithm="giac")
Output:
integrate((b*log((e*x^(2/3) + d)^n*c) + a)/x, x)
Timed out. \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{x} \,d x \] Input:
int((a + b*log(c*(d + e*x^(2/3))^n))/x,x)
Output:
int((a + b*log(c*(d + e*x^(2/3))^n))/x, x)
\[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x} \, dx=\frac {4 \left (\int \frac {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}{x^{\frac {5}{3}} e +d x}d x \right ) b d n +3 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} b +4 \,\mathrm {log}\left (x \right ) a n}{4 n} \] Input:
int((a+b*log(c*(d+e*x^(2/3))^n))/x,x)
Output:
(4*int(log((x**(2/3)*e + d)**n*c)/(x**(2/3)*e*x + d*x),x)*b*d*n + 3*log((x **(2/3)*e + d)**n*c)**2*b + 4*log(x)*a*n)/(4*n)