Integrand size = 20, antiderivative size = 89 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=-\frac {b d^2 n x^{2/3}}{2 e^2}+\frac {b d n x^{4/3}}{4 e}-\frac {1}{6} b n x^2+\frac {b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \] Output:
-1/2*b*d^2*n*x^(2/3)/e^2+1/4*b*d*n*x^(4/3)/e-1/6*b*n*x^2+1/2*b*d^3*n*ln(d+ e*x^(2/3))/e^3+1/2*x^2*(a+b*ln(c*(d+e*x^(2/3))^n))
Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=-\frac {b d^2 n x^{2/3}}{2 e^2}+\frac {b d n x^{4/3}}{4 e}+\frac {a x^2}{2}-\frac {1}{6} b n x^2+\frac {b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac {1}{2} b x^2 \log \left (c \left (d+e x^{2/3}\right )^n\right ) \] Input:
Integrate[x*(a + b*Log[c*(d + e*x^(2/3))^n]),x]
Output:
-1/2*(b*d^2*n*x^(2/3))/e^2 + (b*d*n*x^(4/3))/(4*e) + (a*x^2)/2 - (b*n*x^2) /6 + (b*d^3*n*Log[d + e*x^(2/3)])/(2*e^3) + (b*x^2*Log[c*(d + e*x^(2/3))^n ])/2
Time = 0.45 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2904, 2842, 49, 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.
\(\displaystyle \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {3}{2} \int x^{4/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )dx^{2/3}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {3}{2} \left (\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{3} b e n \int \frac {x^2}{d+e x^{2/3}}dx^{2/3}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {3}{2} \left (\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{3} b e n \int \left (-\frac {d^3}{e^3 \left (d+e x^{2/3}\right )}+\frac {d^2}{e^3}-\frac {x^{2/3} d}{e^2}+\frac {x^{4/3}}{e}\right )dx^{2/3}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{2} \left (\frac {1}{3} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{3} b e n \left (-\frac {d^3 \log \left (d+e x^{2/3}\right )}{e^4}+\frac {d^2 x^{2/3}}{e^3}-\frac {d x^{4/3}}{2 e^2}+\frac {x^2}{3 e}\right )\right )\) |
Input:
Int[x*(a + b*Log[c*(d + e*x^(2/3))^n]),x]
Output:
(3*(-1/3*(b*e*n*((d^2*x^(2/3))/e^3 - (d*x^(4/3))/(2*e^2) + x^2/(3*e) - (d^ 3*Log[d + e*x^(2/3)])/e^4)) + (x^2*(a + b*Log[c*(d + e*x^(2/3))^n]))/3))/2
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[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])
\[\int x \left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )d x\]
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 = 0.09 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {6 \, b e^{3} x^{2} \log \left (c\right ) + 3 \, b d e^{2} n x^{\frac {4}{3}} - 6 \, b d^{2} e n x^{\frac {2}{3}} - 2 \, {\left (b e^{3} n - 3 \, a e^{3}\right )} x^{2} + 6 \, {\left (b e^{3} n x^{2} + b d^{3} n\right )} \log \left (e x^{\frac {2}{3}} + d\right )}{12 \, e^{3}} \] Input:
integrate(x*(a+b*log(c*(d+e*x^(2/3))^n)),x, algorithm="fricas")
Output:
1/12*(6*b*e^3*x^2*log(c) + 3*b*d*e^2*n*x^(4/3) - 6*b*d^2*e*n*x^(2/3) - 2*( b*e^3*n - 3*a*e^3)*x^2 + 6*(b*e^3*n*x^2 + b*d^3*n)*log(e*x^(2/3) + d))/e^3
Time = 70.01 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a x^{2}}{2} + b \left (- \frac {e n \left (- \frac {3 d^{3} \left (\begin {cases} \frac {x^{\frac {2}{3}}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{\frac {2}{3}} \right )}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{3}} + \frac {3 d^{2} x^{\frac {2}{3}}}{2 e^{3}} - \frac {3 d x^{\frac {4}{3}}}{4 e^{2}} + \frac {x^{2}}{2 e}\right )}{3} + \frac {x^{2} \log {\left (c \left (d + e x^{\frac {2}{3}}\right )^{n} \right )}}{2}\right ) \] Input:
integrate(x*(a+b*ln(c*(d+e*x**(2/3))**n)),x)
Output:
a*x**2/2 + b*(-e*n*(-3*d**3*Piecewise((x**(2/3)/d, Eq(e, 0)), (log(d + e*x **(2/3))/e, True))/(2*e**3) + 3*d**2*x**(2/3)/(2*e**3) - 3*d*x**(4/3)/(4*e **2) + x**2/(2*e))/3 + x**2*log(c*(d + e*x**(2/3))**n)/2)
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{12} \, b e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )} + \frac {1}{2} \, b x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {1}{2} \, a x^{2} \] Input:
integrate(x*(a+b*log(c*(d+e*x^(2/3))^n)),x, algorithm="maxima")
Output:
1/12*b*e*n*(6*d^3*log(e*x^(2/3) + d)/e^4 - (2*e^2*x^2 - 3*d*e*x^(4/3) + 6* d^2*x^(2/3))/e^3) + 1/2*b*x^2*log((e*x^(2/3) + d)^n*c) + 1/2*a*x^2
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{12} \, {\left (6 \, x^{2} \log \left (e x^{\frac {2}{3}} + d\right ) + e {\left (\frac {6 \, d^{3} \log \left ({\left | e x^{\frac {2}{3}} + d \right |}\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \] Input:
integrate(x*(a+b*log(c*(d+e*x^(2/3))^n)),x, algorithm="giac")
Output:
1/2*b*x^2*log(c) + 1/12*(6*x^2*log(e*x^(2/3) + d) + e*(6*d^3*log(abs(e*x^( 2/3) + d))/e^4 - (2*e^2*x^2 - 3*d*e*x^(4/3) + 6*d^2*x^(2/3))/e^3))*b*n + 1 /2*a*x^2
Time = 14.49 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a\,x^2}{2}-\frac {b\,n\,x^2}{6}+\frac {b\,x^2\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2}+\frac {b\,d\,n\,x^{4/3}}{4\,e}+\frac {b\,d^3\,n\,\ln \left (d+e\,x^{2/3}\right )}{2\,e^3}-\frac {b\,d^2\,n\,x^{2/3}}{2\,e^2} \] Input:
int(x*(a + b*log(c*(d + e*x^(2/3))^n)),x)
Output:
(a*x^2)/2 - (b*n*x^2)/6 + (b*x^2*log(c*(d + e*x^(2/3))^n))/2 + (b*d*n*x^(4 /3))/(4*e) + (b*d^3*n*log(d + e*x^(2/3)))/(2*e^3) - (b*d^2*n*x^(2/3))/(2*e ^2)
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {-6 x^{\frac {2}{3}} b \,d^{2} e n +3 x^{\frac {4}{3}} b d \,e^{2} n +6 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b \,d^{3}+6 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b \,e^{3} x^{2}+6 a \,e^{3} x^{2}-2 b \,e^{3} n \,x^{2}}{12 e^{3}} \] Input:
int(x*(a+b*log(c*(d+e*x^(2/3))^n)),x)
Output:
( - 6*x**(2/3)*b*d**2*e*n + 3*x**(1/3)*b*d*e**2*n*x + 6*log((x**(2/3)*e + d)**n*c)*b*d**3 + 6*log((x**(2/3)*e + d)**n*c)*b*e**3*x**2 + 6*a*e**3*x**2 - 2*b*e**3*n*x**2)/(12*e**3)