Integrand size = 22, antiderivative size = 143 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {b e^5 n x^{2/3}}{4 d^5}-\frac {b e^4 n x^{4/3}}{8 d^4}+\frac {b e^3 n x^2}{12 d^3}-\frac {b e^2 n x^{8/3}}{16 d^2}+\frac {b e n x^{10/3}}{20 d}-\frac {b e^6 n \log \left (d+\frac {e}{x^{2/3}}\right )}{4 d^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {b e^6 n \log (x)}{6 d^6} \] Output:
1/4*b*e^5*n*x^(2/3)/d^5-1/8*b*e^4*n*x^(4/3)/d^4+1/12*b*e^3*n*x^2/d^3-1/16* b*e^2*n*x^(8/3)/d^2+1/20*b*e*n*x^(10/3)/d-1/4*b*e^6*n*ln(d+e/x^(2/3))/d^6+ 1/4*x^4*(a+b*ln(c*(d+e/x^(2/3))^n))-1/6*b*e^6*n*ln(x)/d^6
Time = 0.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} b x^4 \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )+\frac {1}{6} b e n \left (\frac {3 e^4 x^{2/3}}{2 d^5}-\frac {3 e^3 x^{4/3}}{4 d^4}+\frac {e^2 x^2}{2 d^3}-\frac {3 e x^{8/3}}{8 d^2}+\frac {3 x^{10/3}}{10 d}-\frac {3 e^5 \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^6}-\frac {e^5 \log (x)}{d^6}\right ) \] Input:
Integrate[x^3*(a + b*Log[c*(d + e/x^(2/3))^n]),x]
Output:
(a*x^4)/4 + (b*x^4*Log[c*(d + e/x^(2/3))^n])/4 + (b*e*n*((3*e^4*x^(2/3))/( 2*d^5) - (3*e^3*x^(4/3))/(4*d^4) + (e^2*x^2)/(2*d^3) - (3*e*x^(8/3))/(8*d^ 2) + (3*x^(10/3))/(10*d) - (3*e^5*Log[d + e/x^(2/3)])/(2*d^6) - (e^5*Log[x ])/d^6))/6
Time = 0.53 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.94, 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.
\(\displaystyle \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -\frac {3}{2} \int x^{14/3} \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )d\frac {1}{x^{2/3}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} b e n \int \frac {x^4}{d+\frac {e}{x^{2/3}}}d\frac {1}{x^{2/3}}-\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} b e n \int \left (\frac {e^6}{d^6 \left (d+\frac {e}{x^{2/3}}\right )}-\frac {x^{2/3} e^5}{d^6}+\frac {x^{4/3} e^4}{d^5}-\frac {x^2 e^3}{d^4}+\frac {x^{8/3} e^2}{d^3}-\frac {x^{10/3} e}{d^2}+\frac {x^4}{d}\right )d\frac {1}{x^{2/3}}-\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \left (\frac {1}{6} b e n \left (\frac {e^5 \log \left (d+\frac {e}{x^{2/3}}\right )}{d^6}-\frac {e^5 \log \left (\frac {1}{x^{2/3}}\right )}{d^6}-\frac {e^4 x^{2/3}}{d^5}+\frac {e^3 x^{4/3}}{2 d^4}-\frac {e^2 x^2}{3 d^3}+\frac {e x^{8/3}}{4 d^2}-\frac {x^{10/3}}{5 d}\right )-\frac {1}{6} x^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )\) |
Input:
Int[x^3*(a + b*Log[c*(d + e/x^(2/3))^n]),x]
Output:
(-3*(-1/6*(x^4*(a + b*Log[c*(d + e/x^(2/3))^n])) + (b*e*n*(-((e^4*x^(2/3)) /d^5) + (e^3*x^(4/3))/(2*d^4) - (e^2*x^2)/(3*d^3) + (e*x^(8/3))/(4*d^2) - x^(10/3)/(5*d) + (e^5*Log[d + e/x^(2/3)])/d^6 - (e^5*Log[x^(-2/3)])/d^6))/ 6))/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])
\[\int x^{3} \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )d x\]
Input:
int(x^3*(a+b*ln(c*(d+e/x^(2/3))^n)),x)
Output:
int(x^3*(a+b*ln(c*(d+e/x^(2/3))^n)),x)
Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {60 \, b d^{6} x^{4} \log \left (c\right ) + 60 \, a d^{6} x^{4} + 20 \, b d^{3} e^{3} n x^{2} - 120 \, b d^{6} n \log \left (x^{\frac {1}{3}}\right ) + 60 \, {\left (b d^{6} - b e^{6}\right )} n \log \left (d x^{\frac {2}{3}} + e\right ) + 60 \, {\left (b d^{6} n x^{4} - b d^{6} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) - 15 \, {\left (b d^{4} e^{2} n x^{2} - 4 \, b d e^{5} n\right )} x^{\frac {2}{3}} + 6 \, {\left (2 \, b d^{5} e n x^{3} - 5 \, b d^{2} e^{4} n x\right )} x^{\frac {1}{3}}}{240 \, d^{6}} \] Input:
integrate(x^3*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="fricas")
Output:
1/240*(60*b*d^6*x^4*log(c) + 60*a*d^6*x^4 + 20*b*d^3*e^3*n*x^2 - 120*b*d^6 *n*log(x^(1/3)) + 60*(b*d^6 - b*e^6)*n*log(d*x^(2/3) + e) + 60*(b*d^6*n*x^ 4 - b*d^6*n)*log((d*x + e*x^(1/3))/x) - 15*(b*d^4*e^2*n*x^2 - 4*b*d*e^5*n) *x^(2/3) + 6*(2*b*d^5*e*n*x^3 - 5*b*d^2*e^4*n*x)*x^(1/3))/d^6
Timed out. \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\text {Timed out} \] Input:
integrate(x**3*(a+b*ln(c*(d+e/x**(2/3))**n)),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.69 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{240} \, b e n {\left (\frac {60 \, e^{5} \log \left (d x^{\frac {2}{3}} + e\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {10}{3}} - 15 \, d^{3} e x^{\frac {8}{3}} + 20 \, d^{2} e^{2} x^{2} - 30 \, d e^{3} x^{\frac {4}{3}} + 60 \, e^{4} x^{\frac {2}{3}}}{d^{5}}\right )} \] Input:
integrate(x^3*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="maxima")
Output:
1/4*b*x^4*log(c*(d + e/x^(2/3))^n) + 1/4*a*x^4 - 1/240*b*e*n*(60*e^5*log(d *x^(2/3) + e)/d^6 - (12*d^4*x^(10/3) - 15*d^3*e*x^(8/3) + 20*d^2*e^2*x^2 - 30*d*e^3*x^(4/3) + 60*e^4*x^(2/3))/d^5)
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.73 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} + \frac {1}{240} \, {\left (60 \, x^{4} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) - e {\left (\frac {60 \, e^{5} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {10}{3}} - 15 \, d^{3} e x^{\frac {8}{3}} + 20 \, d^{2} e^{2} x^{2} - 30 \, d e^{3} x^{\frac {4}{3}} + 60 \, e^{4} x^{\frac {2}{3}}}{d^{5}}\right )}\right )} b n \] Input:
integrate(x^3*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="giac")
Output:
1/4*b*x^4*log(c) + 1/4*a*x^4 + 1/240*(60*x^4*log(d + e/x^(2/3)) - e*(60*e^ 5*log(abs(d*x^(2/3) + e))/d^6 - (12*d^4*x^(10/3) - 15*d^3*e*x^(8/3) + 20*d ^2*e^2*x^2 - 30*d*e^3*x^(4/3) + 60*e^4*x^(2/3))/d^5))*b*n
Time = 25.59 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.78 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {x^{10/3}\,\left (\frac {b\,e\,n}{5\,d}-\frac {b\,e^2\,n}{4\,d^2\,x^{2/3}}-\frac {b\,e^4\,n}{2\,d^4\,x^2}+\frac {b\,e^3\,n}{3\,d^3\,x^{4/3}}+\frac {b\,e^5\,n}{d^5\,x^{8/3}}\right )}{4}+\frac {a\,x^4}{4}+\frac {b\,x^4\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{4}-\frac {b\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{2/3}}+1\right )}{2\,d^6} \] Input:
int(x^3*(a + b*log(c*(d + e/x^(2/3))^n)),x)
Output:
(x^(10/3)*((b*e*n)/(5*d) - (b*e^2*n)/(4*d^2*x^(2/3)) - (b*e^4*n)/(2*d^4*x^ 2) + (b*e^3*n)/(3*d^3*x^(4/3)) + (b*e^5*n)/(d^5*x^(8/3))))/4 + (a*x^4)/4 + (b*x^4*log(c*(d + e/x^(2/3))^n))/4 - (b*e^6*n*atanh((2*e)/(d*x^(2/3)) + 1 ))/(2*d^6)
Time = 0.15 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.98 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {-15 x^{\frac {8}{3}} b \,d^{4} e^{2} n +60 x^{\frac {2}{3}} b d \,e^{5} n +12 x^{\frac {10}{3}} b \,d^{5} e n -30 x^{\frac {4}{3}} b \,d^{2} e^{4} n -120 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{6} n +60 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,d^{6} x^{4}-60 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,e^{6}+60 a \,d^{6} x^{4}+20 b \,d^{3} e^{3} n \,x^{2}}{240 d^{6}} \] Input:
int(x^3*(a+b*log(c*(d+e/x^(2/3))^n)),x)
Output:
( - 15*x**(2/3)*b*d**4*e**2*n*x**2 + 60*x**(2/3)*b*d*e**5*n + 12*x**(1/3)* b*d**5*e*n*x**3 - 30*x**(1/3)*b*d**2*e**4*n*x - 120*log(x**(1/3))*b*e**6*n + 60*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b*d**6*x**4 - 60*log(((x** (2/3)*d + e)**n*c)/x**((2*n)/3))*b*e**6 + 60*a*d**6*x**4 + 20*b*d**3*e**3* n*x**2)/(240*d**6)