\(\int x (a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)) \, dx\) [491]

Optimal result

Integrand size = 20, antiderivative size = 141 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {b e^5 n \sqrt [3]{x}}{2 d^5}-\frac {b e^4 n x^{2/3}}{4 d^4}+\frac {b e^3 n x}{6 d^3}-\frac {b e^2 n x^{4/3}}{8 d^2}+\frac {b e n x^{5/3}}{10 d}-\frac {b e^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 d^6}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {b e^6 n \log (x)}{6 d^6} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 141, 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} \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {b e^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 d^6}-\frac {b e^6 n \log (x)}{6 d^6}+\frac {b e^5 n \sqrt [3]{x}}{2 d^5}-\frac {b e^4 n x^{2/3}}{4 d^4}+\frac {b e^3 n x}{6 d^3}-\frac {b e^2 n x^{4/3}}{8 d^2}+\frac {b e n x^{5/3}}{10 d} \]

[In]

Int[x*(a + b*Log[c*(d + e/x^(1/3))^n]),x]

[Out]

(b*e^5*n*x^(1/3))/(2*d^5) - (b*e^4*n*x^(2/3))/(4*d^4) + (b*e^3*n*x)/(6*d^3) - (b*e^2*n*x^(4/3))/(8*d^2) + (b*e
*n*x^(5/3))/(10*d) - (b*e^6*n*Log[d + e/x^(1/3)])/(2*d^6) + (x^2*(a + b*Log[c*(d + e/x^(1/3))^n]))/2 - (b*e^6*
n*Log[x])/(6*d^6)

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 (3 \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^7} \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \frac {1}{x^6 (d+e x)} \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^6}-\frac {e}{d^2 x^5}+\frac {e^2}{d^3 x^4}-\frac {e^3}{d^4 x^3}+\frac {e^4}{d^5 x^2}-\frac {e^5}{d^6 x}+\frac {e^6}{d^6 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} b x^2 \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {1}{6} b e n \left (\frac {3 e^4 \sqrt [3]{x}}{d^5}-\frac {3 e^3 x^{2/3}}{2 d^4}+\frac {e^2 x}{d^3}-\frac {3 e x^{4/3}}{4 d^2}+\frac {3 x^{5/3}}{5 d}-\frac {3 e^5 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^6}-\frac {e^5 \log (x)}{d^6}\right ) \]

[In]

Integrate[x*(a + b*Log[c*(d + e/x^(1/3))^n]),x]

[Out]

(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

Maple [F]

\[\int x \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )\right )d x\]

[In]

int(x*(a+b*ln(c*(d+e/x^(1/3))^n)),x)

[Out]

int(x*(a+b*ln(c*(d+e/x^(1/3))^n)),x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {60 \, b d^{6} x^{2} \log \left (c\right ) + 20 \, b d^{3} e^{3} n x + 60 \, a d^{6} x^{2} - 60 \, b d^{6} n \log \left (x^{\frac {1}{3}}\right ) + 60 \, {\left (b d^{6} - b e^{6}\right )} n \log \left (d x^{\frac {1}{3}} + e\right ) + 60 \, {\left (b d^{6} n x^{2} - b d^{6} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 6 \, {\left (2 \, b d^{5} e n x - 5 \, b d^{2} e^{4} n\right )} x^{\frac {2}{3}} - 15 \, {\left (b d^{4} e^{2} n x - 4 \, b d e^{5} n\right )} x^{\frac {1}{3}}}{120 \, d^{6}} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 11.43 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^{2}}{2} + b \left (\frac {e n \left (\frac {3 x^{\frac {5}{3}}}{5 d} - \frac {3 e x^{\frac {4}{3}}}{4 d^{2}} + \frac {e^{2} x}{d^{3}} - \frac {3 e^{3} x^{\frac {2}{3}}}{2 d^{4}} - \frac {3 e^{5} \left (\begin {cases} \frac {\sqrt [3]{x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt [3]{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{5}} + \frac {3 e^{4} \sqrt [3]{x}}{d^{5}}\right )}{6} + \frac {x^{2} \log {\left (c \left (d + \frac {e}{\sqrt [3]{x}}\right )^{n} \right )}}{2}\right ) \]

[In]

integrate(x*(a+b*ln(c*(d+e/x**(1/3))**n)),x)

[Out]

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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.68 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=-\frac {1}{120} \, b e n {\left (\frac {60 \, e^{5} \log \left (d x^{\frac {1}{3}} + e\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {5}{3}} - 15 \, d^{3} e x^{\frac {4}{3}} + 20 \, d^{2} e^{2} x - 30 \, d e^{3} x^{\frac {2}{3}} + 60 \, e^{4} x^{\frac {1}{3}}}{d^{5}}\right )} + \frac {1}{2} \, b x^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right ) + \frac {1}{2} \, a x^{2} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="maxima")

[Out]

-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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.73 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{120} \, {\left (60 \, x^{2} \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right ) - e {\left (\frac {60 \, e^{5} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {5}{3}} - 15 \, d^{3} e x^{\frac {4}{3}} + 20 \, d^{2} e^{2} x - 30 \, d e^{3} x^{\frac {2}{3}} + 60 \, e^{4} x^{\frac {1}{3}}}{d^{5}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \]

[In]

integrate(x*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.96 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int x \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {x^{5/3}\,\left (\frac {b\,e\,n}{5\,d}-\frac {b\,e^2\,n}{4\,d^2\,x^{1/3}}-\frac {b\,e^4\,n}{2\,d^4\,x}+\frac {b\,e^3\,n}{3\,d^3\,x^{2/3}}+\frac {b\,e^5\,n}{d^5\,x^{4/3}}\right )}{2}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{2}-\frac {b\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{1/3}}+1\right )}{d^6} \]

[In]

int(x*(a + b*log(c*(d + e/x^(1/3))^n)),x)

[Out]

(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