Integrand size = 20, antiderivative size = 94 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=-\frac {b e^2 n x^{2/3}}{2 d^2}+\frac {b e n x^{4/3}}{4 d}+\frac {b e^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {b e^3 n \log (x)}{3 d^3} \] Output:
-1/2*b*e^2*n*x^(2/3)/d^2+1/4*b*e*n*x^(4/3)/d+1/2*b*e^3*n*ln(d+e/x^(2/3))/d ^3+1/2*x^2*(a+b*ln(c*(d+e/x^(2/3))^n))+1/3*b*e^3*n*ln(x)/d^3
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )+\frac {1}{3} b e n \left (-\frac {3 e x^{2/3}}{2 d^2}+\frac {3 x^{4/3}}{4 d}+\frac {3 e^2 \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {e^2 \log (x)}{d^3}\right ) \] Input:
Integrate[x*(a + b*Log[c*(d + e/x^(2/3))^n]),x]
Output:
(a*x^2)/2 + (b*x^2*Log[c*(d + e/x^(2/3))^n])/2 + (b*e*n*((-3*e*x^(2/3))/(2 *d^2) + (3*x^(4/3))/(4*d) + (3*e^2*Log[d + e/x^(2/3)])/(2*d^3) + (e^2*Log[ x])/d^3))/3
Time = 0.44 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -\frac {3}{2} \int x^{8/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )d\frac {1}{x^{2/3}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{3} b e n \int \frac {x^2}{d+\frac {e}{x^{2/3}}}d\frac {1}{x^{2/3}}-\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{3} b e n \int \left (-\frac {e^3}{d^3 \left (d+\frac {e}{x^{2/3}}\right )}+\frac {x^{2/3} e^2}{d^3}-\frac {x^{4/3} e}{d^2}+\frac {x^2}{d}\right )d\frac {1}{x^{2/3}}-\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{3} b e n \left (-\frac {e^2 \log \left (d+\frac {e}{x^{2/3}}\right )}{d^3}+\frac {e^2 \log \left (\frac {1}{x^{2/3}}\right )}{d^3}+\frac {e x^{2/3}}{d^2}-\frac {x^{4/3}}{2 d}\right )-\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )\) |
Input:
Int[x*(a + b*Log[c*(d + e/x^(2/3))^n]),x]
Output:
(-3*(-1/3*(x^2*(a + b*Log[c*(d + e/x^(2/3))^n])) + (b*e*n*((e*x^(2/3))/d^2 - x^(4/3)/(2*d) - (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])
\[\int x \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )d x\]
Input:
int(x*(a+b*ln(c*(d+e/x^(2/3))^n)),x)
Output:
int(x*(a+b*ln(c*(d+e/x^(2/3))^n)),x)
Time = 0.10 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.20 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {2 \, b d^{3} x^{2} \log \left (c\right ) + b d^{2} e n x^{\frac {4}{3}} + 2 \, a d^{3} x^{2} - 4 \, b d^{3} n \log \left (x^{\frac {1}{3}}\right ) - 2 \, b d e^{2} n x^{\frac {2}{3}} + 2 \, {\left (b d^{3} + b e^{3}\right )} n \log \left (d x^{\frac {2}{3}} + e\right ) + 2 \, {\left (b d^{3} n x^{2} - b d^{3} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )}{4 \, d^{3}} \] Input:
integrate(x*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="fricas")
Output:
1/4*(2*b*d^3*x^2*log(c) + b*d^2*e*n*x^(4/3) + 2*a*d^3*x^2 - 4*b*d^3*n*log( x^(1/3)) - 2*b*d*e^2*n*x^(2/3) + 2*(b*d^3 + b*e^3)*n*log(d*x^(2/3) + e) + 2*(b*d^3*n*x^2 - b*d^3*n)*log((d*x + e*x^(1/3))/x))/d^3
Timed out. \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x*(a+b*ln(c*(d+e/x**(2/3))**n)),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b e n {\left (\frac {2 \, e^{2} \log \left (d x^{\frac {2}{3}} + e\right )}{d^{3}} + \frac {d x^{\frac {4}{3}} - 2 \, e x^{\frac {2}{3}}}{d^{2}}\right )} + \frac {1}{2} \, b x^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + \frac {1}{2} \, a x^{2} \] Input:
integrate(x*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="maxima")
Output:
1/4*b*e*n*(2*e^2*log(d*x^(2/3) + e)/d^3 + (d*x^(4/3) - 2*e*x^(2/3))/d^2) + 1/2*b*x^2*log(c*(d + e/x^(2/3))^n) + 1/2*a*x^2
Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{4} \, {\left (2 \, x^{2} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) + e {\left (\frac {2 \, e^{2} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{d^{3}} + \frac {d x^{\frac {4}{3}} - 2 \, e x^{\frac {2}{3}}}{d^{2}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \] Input:
integrate(x*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="giac")
Output:
1/2*b*x^2*log(c) + 1/4*(2*x^2*log(d + e/x^(2/3)) + e*(2*e^2*log(abs(d*x^(2 /3) + e))/d^3 + (d*x^(4/3) - 2*e*x^(2/3))/d^2))*b*n + 1/2*a*x^2
Time = 25.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {x^{4/3}\,\left (\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n}{d^2\,x^{2/3}}\right )}{2}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{2}+\frac {b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{2/3}}+1\right )}{d^3} \] Input:
int(x*(a + b*log(c*(d + e/x^(2/3))^n)),x)
Output:
(x^(4/3)*((b*e*n)/(2*d) - (b*e^2*n)/(d^2*x^(2/3))))/2 + (a*x^2)/2 + (b*x^2 *log(c*(d + e/x^(2/3))^n))/2 + (b*e^3*n*atanh((2*e)/(d*x^(2/3)) + 1))/d^3
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {-2 x^{\frac {2}{3}} b d \,e^{2} n +x^{\frac {4}{3}} b \,d^{2} e n +4 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{3} n +2 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,d^{3} x^{2}+2 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,e^{3}+2 a \,d^{3} x^{2}}{4 d^{3}} \] Input:
int(x*(a+b*log(c*(d+e/x^(2/3))^n)),x)
Output:
( - 2*x**(2/3)*b*d*e**2*n + x**(1/3)*b*d**2*e*n*x + 4*log(x**(1/3))*b*e**3 *n + 2*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b*d**3*x**2 + 2*log(((x** (2/3)*d + e)**n*c)/x**((2*n)/3))*b*e**3 + 2*a*d**3*x**2)/(4*d**3)