Integrand size = 20, antiderivative size = 141 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {b e^5 n \sqrt [3]{x}}{2 d^5}-\frac {b e^4 n x^{2/3}}{4 d^4}+\frac {b e^3 n x}{6 d^3}-\frac {b e^2 n x^{4/3}}{8 d^2}+\frac {b e n x^{5/3}}{10 d}-\frac {b e^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 d^6}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {b e^6 n \log (x)}{6 d^6} \] Output:
1/2*b*e^5*n*x^(1/3)/d^5-1/4*b*e^4*n*x^(2/3)/d^4+1/6*b*e^3*n*x/d^3-1/8*b*e^ 2*n*x^(4/3)/d^2+1/10*b*e*n*x^(5/3)/d-1/2*b*e^6*n*ln(d+e/x^(1/3))/d^6+1/2*x ^2*(a+b*ln(c*(d+e/x^(1/3))^n))-1/6*b*e^6*n*ln(x)/d^6
Time = 0.10 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {1}{6} b e n \left (\frac {3 e^4 \sqrt [3]{x}}{d^5}-\frac {3 e^3 x^{2/3}}{2 d^4}+\frac {e^2 x}{d^3}-\frac {3 e x^{4/3}}{4 d^2}+\frac {3 x^{5/3}}{5 d}-\frac {3 e^5 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^6}-\frac {e^5 \log (x)}{d^6}\right ) \] Input:
Integrate[x*(a + b*Log[c*(d + e/x^(1/3))^n]),x]
Output:
(a*x^2)/2 + (b*x^2*Log[c*(d + e/x^(1/3))^n])/2 + (b*e*n*((3*e^4*x^(1/3))/d ^5 - (3*e^3*x^(2/3))/(2*d^4) + (e^2*x)/d^3 - (3*e*x^(4/3))/(4*d^2) + (3*x^ (5/3))/(5*d) - (3*e^5*Log[d + e/x^(1/3)])/d^6 - (e^5*Log[x])/d^6))/6
Time = 0.50 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93, 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.
\(\displaystyle \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -3 \int x^{7/3} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )d\frac {1}{\sqrt [3]{x}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -3 \left (\frac {1}{6} b e n \int \frac {x^2}{d+\frac {e}{\sqrt [3]{x}}}d\frac {1}{\sqrt [3]{x}}-\frac {1}{6} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -3 \left (\frac {1}{6} b e n \int \left (\frac {e^6}{d^6 \left (d+\frac {e}{\sqrt [3]{x}}\right )}-\frac {\sqrt [3]{x} e^5}{d^6}+\frac {x^{2/3} e^4}{d^5}-\frac {x e^3}{d^4}+\frac {x^{4/3} e^2}{d^3}-\frac {x^{5/3} e}{d^2}+\frac {x^2}{d}\right )d\frac {1}{\sqrt [3]{x}}-\frac {1}{6} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \left (\frac {1}{6} b e n \left (\frac {e^5 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^6}-\frac {e^5 \log \left (\frac {1}{\sqrt [3]{x}}\right )}{d^6}-\frac {e^4 \sqrt [3]{x}}{d^5}+\frac {e^3 x^{2/3}}{2 d^4}-\frac {e^2 x}{3 d^3}+\frac {e x^{4/3}}{4 d^2}-\frac {x^{5/3}}{5 d}\right )-\frac {1}{6} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )\right )\) |
Input:
Int[x*(a + b*Log[c*(d + e/x^(1/3))^n]),x]
Output:
-3*(-1/6*(x^2*(a + b*Log[c*(d + e/x^(1/3))^n])) + (b*e*n*(-((e^4*x^(1/3))/ d^5) + (e^3*x^(2/3))/(2*d^4) - (e^2*x)/(3*d^3) + (e*x^(4/3))/(4*d^2) - x^( 5/3)/(5*d) + (e^5*Log[d + e/x^(1/3)])/d^6 - (e^5*Log[x^(-1/3)])/d^6))/6)
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])
\[\int x \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )\right )d x\]
Input:
int(x*(a+b*ln(c*(d+e/x^(1/3))^n)),x)
Output:
int(x*(a+b*ln(c*(d+e/x^(1/3))^n)),x)
Time = 0.13 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {60 \, b d^{6} x^{2} \log \left (c\right ) + 20 \, b d^{3} e^{3} n x + 60 \, a d^{6} x^{2} - 60 \, b d^{6} n \log \left (x^{\frac {1}{3}}\right ) + 60 \, {\left (b d^{6} - b e^{6}\right )} n \log \left (d x^{\frac {1}{3}} + e\right ) + 60 \, {\left (b d^{6} n x^{2} - b d^{6} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 6 \, {\left (2 \, b d^{5} e n x - 5 \, b d^{2} e^{4} n\right )} x^{\frac {2}{3}} - 15 \, {\left (b d^{4} e^{2} n x - 4 \, b d e^{5} n\right )} x^{\frac {1}{3}}}{120 \, d^{6}} \] Input:
integrate(x*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="fricas")
Output:
1/120*(60*b*d^6*x^2*log(c) + 20*b*d^3*e^3*n*x + 60*a*d^6*x^2 - 60*b*d^6*n* log(x^(1/3)) + 60*(b*d^6 - b*e^6)*n*log(d*x^(1/3) + e) + 60*(b*d^6*n*x^2 - b*d^6*n)*log((d*x + e*x^(2/3))/x) + 6*(2*b*d^5*e*n*x - 5*b*d^2*e^4*n)*x^( 2/3) - 15*(b*d^4*e^2*n*x - 4*b*d*e^5*n)*x^(1/3))/d^6
Time = 9.53 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^{2}}{2} + b \left (\frac {e n \left (\frac {3 x^{\frac {5}{3}}}{5 d} - \frac {3 e x^{\frac {4}{3}}}{4 d^{2}} + \frac {e^{2} x}{d^{3}} - \frac {3 e^{3} x^{\frac {2}{3}}}{2 d^{4}} - \frac {3 e^{5} \left (\begin {cases} \frac {\sqrt [3]{x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt [3]{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{5}} + \frac {3 e^{4} \sqrt [3]{x}}{d^{5}}\right )}{6} + \frac {x^{2} \log {\left (c \left (d + \frac {e}{\sqrt [3]{x}}\right )^{n} \right )}}{2}\right ) \] Input:
integrate(x*(a+b*ln(c*(d+e/x**(1/3))**n)),x)
Output:
a*x**2/2 + b*(e*n*(3*x**(5/3)/(5*d) - 3*e*x**(4/3)/(4*d**2) + e**2*x/d**3 - 3*e**3*x**(2/3)/(2*d**4) - 3*e**5*Piecewise((x**(1/3)/e, Eq(d, 0)), (log (d*x**(1/3) + e)/d, True))/d**5 + 3*e**4*x**(1/3)/d**5)/6 + x**2*log(c*(d + e/x**(1/3))**n)/2)
Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.68 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=-\frac {1}{120} \, b e n {\left (\frac {60 \, e^{5} \log \left (d x^{\frac {1}{3}} + e\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {5}{3}} - 15 \, d^{3} e x^{\frac {4}{3}} + 20 \, d^{2} e^{2} x - 30 \, d e^{3} x^{\frac {2}{3}} + 60 \, e^{4} x^{\frac {1}{3}}}{d^{5}}\right )} + \frac {1}{2} \, b x^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right ) + \frac {1}{2} \, a x^{2} \] Input:
integrate(x*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="maxima")
Output:
-1/120*b*e*n*(60*e^5*log(d*x^(1/3) + e)/d^6 - (12*d^4*x^(5/3) - 15*d^3*e*x ^(4/3) + 20*d^2*e^2*x - 30*d*e^3*x^(2/3) + 60*e^4*x^(1/3))/d^5) + 1/2*b*x^ 2*log(c*(d + e/x^(1/3))^n) + 1/2*a*x^2
Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.73 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{120} \, {\left (60 \, x^{2} \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right ) - e {\left (\frac {60 \, e^{5} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {5}{3}} - 15 \, d^{3} e x^{\frac {4}{3}} + 20 \, d^{2} e^{2} x - 30 \, d e^{3} x^{\frac {2}{3}} + 60 \, e^{4} x^{\frac {1}{3}}}{d^{5}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \] Input:
integrate(x*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="giac")
Output:
1/2*b*x^2*log(c) + 1/120*(60*x^2*log(d + e/x^(1/3)) - e*(60*e^5*log(abs(d* x^(1/3) + e))/d^6 - (12*d^4*x^(5/3) - 15*d^3*e*x^(4/3) + 20*d^2*e^2*x - 30 *d*e^3*x^(2/3) + 60*e^4*x^(1/3))/d^5))*b*n + 1/2*a*x^2
Time = 26.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {x^{5/3}\,\left (\frac {b\,e\,n}{5\,d}-\frac {b\,e^2\,n}{4\,d^2\,x^{1/3}}-\frac {b\,e^4\,n}{2\,d^4\,x}+\frac {b\,e^3\,n}{3\,d^3\,x^{2/3}}+\frac {b\,e^5\,n}{d^5\,x^{4/3}}\right )}{2}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{2}-\frac {b\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{1/3}}+1\right )}{d^6} \] Input:
int(x*(a + b*log(c*(d + e/x^(1/3))^n)),x)
Output:
(x^(5/3)*((b*e*n)/(5*d) - (b*e^2*n)/(4*d^2*x^(1/3)) - (b*e^4*n)/(2*d^4*x) + (b*e^3*n)/(3*d^3*x^(2/3)) + (b*e^5*n)/(d^5*x^(4/3))))/2 + (a*x^2)/2 + (b *x^2*log(c*(d + e/x^(1/3))^n))/2 - (b*e^6*n*atanh((2*e)/(d*x^(1/3)) + 1))/ d^6
Time = 0.17 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.98 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {12 x^{\frac {5}{3}} b \,d^{5} e n -30 x^{\frac {2}{3}} b \,d^{2} e^{4} n -15 x^{\frac {4}{3}} b \,d^{4} e^{2} n +60 x^{\frac {1}{3}} b d \,e^{5} n -60 \,\mathrm {log}\left (x^{\frac {1}{3}}\right ) b \,e^{6} n +60 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,d^{6} x^{2}-60 \,\mathrm {log}\left (\frac {\left (x^{\frac {1}{3}} d +e \right )^{n} c}{x^{\frac {n}{3}}}\right ) b \,e^{6}+60 a \,d^{6} x^{2}+20 b \,d^{3} e^{3} n x}{120 d^{6}} \] Input:
int(x*(a+b*log(c*(d+e/x^(1/3))^n)),x)
Output:
(12*x**(2/3)*b*d**5*e*n*x - 30*x**(2/3)*b*d**2*e**4*n - 15*x**(1/3)*b*d**4 *e**2*n*x + 60*x**(1/3)*b*d*e**5*n - 60*log(x**(1/3))*b*e**6*n + 60*log((( x**(1/3)*d + e)**n*c)/x**(n/3))*b*d**6*x**2 - 60*log(((x**(1/3)*d + e)**n* c)/x**(n/3))*b*e**6 + 60*a*d**6*x**2 + 20*b*d**3*e**3*n*x)/(120*d**6)