\(\int x (a+b \log (c (d+\frac {e}{x^{2/3}})^n)) \, dx\) [510]

Optimal result

Integrand size = 20, antiderivative size = 94 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=-\frac {b e^2 n x^{2/3}}{2 d^2}+\frac {b e n x^{4/3}}{4 d}+\frac {b e^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {b e^3 n \log (x)}{3 d^3} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2504, 2442, 46} \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {b e^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {b e^3 n \log (x)}{3 d^3}-\frac {b e^2 n x^{2/3}}{2 d^2}+\frac {b e n x^{4/3}}{4 d} \]

[In]

Int[x*(a + b*Log[c*(d + e/x^(2/3))^n]),x]

[Out]

-1/2*(b*e^2*n*x^(2/3))/d^2 + (b*e*n*x^(4/3))/(4*d) + (b*e^3*n*Log[d + e/x^(2/3)])/(2*d^3) + (x^2*(a + b*Log[c*
(d + e/x^(2/3))^n]))/2 + (b*e^3*n*Log[x])/(3*d^3)

Rule 46

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2442

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] - Dist[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]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {3}{2} \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^4} \, dx,x,\frac {1}{x^{2/3}}\right )\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \frac {1}{x^3 (d+e x)} \, dx,x,\frac {1}{x^{2/3}}\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^3}-\frac {e}{d^2 x^2}+\frac {e^2}{d^3 x}-\frac {e^3}{d^3 (d+e x)}\right ) \, dx,x,\frac {1}{x^{2/3}}\right ) \\ & = -\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )+\frac {1}{3} b e n \left (-\frac {3 e x^{2/3}}{2 d^2}+\frac {3 x^{4/3}}{4 d}+\frac {3 e^2 \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {e^2 \log (x)}{d^3}\right ) \]

[In]

Integrate[x*(a + b*Log[c*(d + e/x^(2/3))^n]),x]

[Out]

(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*L
og[d + e/x^(2/3)])/(2*d^3) + (e^2*Log[x])/d^3))/3

Maple [F]

\[\int x \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )d x\]

[In]

int(x*(a+b*ln(c*(d+e/x^(2/3))^n)),x)

[Out]

int(x*(a+b*ln(c*(d+e/x^(2/3))^n)),x)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.20 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {2 \, b d^{3} x^{2} \log \left (c\right ) + b d^{2} e n x^{\frac {4}{3}} + 2 \, a d^{3} x^{2} - 4 \, b d^{3} n \log \left (x^{\frac {1}{3}}\right ) - 2 \, b d e^{2} n x^{\frac {2}{3}} + 2 \, {\left (b d^{3} + b e^{3}\right )} n \log \left (d x^{\frac {2}{3}} + e\right ) + 2 \, {\left (b d^{3} n x^{2} - b d^{3} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )}{4 \, d^{3}} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(x*(a+b*ln(c*(d+e/x**(2/3))**n)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.67 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b e n {\left (\frac {2 \, e^{2} \log \left (d x^{\frac {2}{3}} + e\right )}{d^{3}} + \frac {d x^{\frac {4}{3}} - 2 \, e x^{\frac {2}{3}}}{d^{2}}\right )} + \frac {1}{2} \, b x^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + \frac {1}{2} \, a x^{2} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{4} \, {\left (2 \, x^{2} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) + e {\left (\frac {2 \, e^{2} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{d^{3}} + \frac {d x^{\frac {4}{3}} - 2 \, e x^{\frac {2}{3}}}{d^{2}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(2/3))^n)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {x^{4/3}\,\left (\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n}{d^2\,x^{2/3}}\right )}{2}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{2}+\frac {b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{2/3}}+1\right )}{d^3} \]

[In]

int(x*(a + b*log(c*(d + e/x^(2/3))^n)),x)

[Out]

(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