\(\int x^2 (a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)) \, dx\) [490]

Optimal result

Integrand size = 22, antiderivative size = 190 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=-\frac {b e^8 n \sqrt [3]{x}}{3 d^8}+\frac {b e^7 n x^{2/3}}{6 d^7}-\frac {b e^6 n x}{9 d^6}+\frac {b e^5 n x^{4/3}}{12 d^5}-\frac {b e^4 n x^{5/3}}{15 d^4}+\frac {b e^3 n x^2}{18 d^3}-\frac {b e^2 n x^{7/3}}{21 d^2}+\frac {b e n x^{8/3}}{24 d}+\frac {b e^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 d^9}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )+\frac {b e^9 n \log (x)}{9 d^9} \]

[Out]

-1/3*b*e^8*n*x^(1/3)/d^8+1/6*b*e^7*n*x^(2/3)/d^7-1/9*b*e^6*n*x/d^6+1/12*b*e^5*n*x^(4/3)/d^5-1/15*b*e^4*n*x^(5/
3)/d^4+1/18*b*e^3*n*x^2/d^3-1/21*b*e^2*n*x^(7/3)/d^2+1/24*b*e*n*x^(8/3)/d+1/3*b*e^9*n*ln(d+e/x^(1/3))/d^9+1/3*
x^3*(a+b*ln(c*(d+e/x^(1/3))^n))+1/9*b*e^9*n*ln(x)/d^9

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2504, 2442, 46} \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )+\frac {b e^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 d^9}+\frac {b e^9 n \log (x)}{9 d^9}-\frac {b e^8 n \sqrt [3]{x}}{3 d^8}+\frac {b e^7 n x^{2/3}}{6 d^7}-\frac {b e^6 n x}{9 d^6}+\frac {b e^5 n x^{4/3}}{12 d^5}-\frac {b e^4 n x^{5/3}}{15 d^4}+\frac {b e^3 n x^2}{18 d^3}-\frac {b e^2 n x^{7/3}}{21 d^2}+\frac {b e n x^{8/3}}{24 d} \]

[In]

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

[Out]

-1/3*(b*e^8*n*x^(1/3))/d^8 + (b*e^7*n*x^(2/3))/(6*d^7) - (b*e^6*n*x)/(9*d^6) + (b*e^5*n*x^(4/3))/(12*d^5) - (b
*e^4*n*x^(5/3))/(15*d^4) + (b*e^3*n*x^2)/(18*d^3) - (b*e^2*n*x^(7/3))/(21*d^2) + (b*e*n*x^(8/3))/(24*d) + (b*e
^9*n*Log[d + e/x^(1/3)])/(3*d^9) + (x^3*(a + b*Log[c*(d + e/x^(1/3))^n]))/3 + (b*e^9*n*Log[x])/(9*d^9)

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^{10}} \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\right ) \\ & = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {1}{3} (b e n) \text {Subst}\left (\int \frac {1}{x^9 (d+e x)} \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )-\frac {1}{3} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^9}-\frac {e}{d^2 x^8}+\frac {e^2}{d^3 x^7}-\frac {e^3}{d^4 x^6}+\frac {e^4}{d^5 x^5}-\frac {e^5}{d^6 x^4}+\frac {e^6}{d^7 x^3}-\frac {e^7}{d^8 x^2}+\frac {e^8}{d^9 x}-\frac {e^9}{d^9 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = -\frac {b e^8 n \sqrt [3]{x}}{3 d^8}+\frac {b e^7 n x^{2/3}}{6 d^7}-\frac {b e^6 n x}{9 d^6}+\frac {b e^5 n x^{4/3}}{12 d^5}-\frac {b e^4 n x^{5/3}}{15 d^4}+\frac {b e^3 n x^2}{18 d^3}-\frac {b e^2 n x^{7/3}}{21 d^2}+\frac {b e n x^{8/3}}{24 d}+\frac {b e^9 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{3 d^9}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right )+\frac {b e^9 n \log (x)}{9 d^9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.90 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} b x^3 \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )+\frac {1}{9} b e n \left (-\frac {3 e^7 \sqrt [3]{x}}{d^8}+\frac {3 e^6 x^{2/3}}{2 d^7}-\frac {e^5 x}{d^6}+\frac {3 e^4 x^{4/3}}{4 d^5}-\frac {3 e^3 x^{5/3}}{5 d^4}+\frac {e^2 x^2}{2 d^3}-\frac {3 e x^{7/3}}{7 d^2}+\frac {3 x^{8/3}}{8 d}+\frac {3 e^8 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{d^9}+\frac {e^8 \log (x)}{d^9}\right ) \]

[In]

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

[Out]

(a*x^3)/3 + (b*x^3*Log[c*(d + e/x^(1/3))^n])/3 + (b*e*n*((-3*e^7*x^(1/3))/d^8 + (3*e^6*x^(2/3))/(2*d^7) - (e^5
*x)/d^6 + (3*e^4*x^(4/3))/(4*d^5) - (3*e^3*x^(5/3))/(5*d^4) + (e^2*x^2)/(2*d^3) - (3*e*x^(7/3))/(7*d^2) + (3*x
^(8/3))/(8*d) + (3*e^8*Log[d + e/x^(1/3)])/d^9 + (e^8*Log[x])/d^9))/9

Maple [F]

\[\int x^{2} \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )\right )d x\]

[In]

int(x^2*(a+b*ln(c*(d+e/x^(1/3))^n)),x)

[Out]

int(x^2*(a+b*ln(c*(d+e/x^(1/3))^n)),x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.01 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {840 \, b d^{9} x^{3} \log \left (c\right ) + 140 \, b d^{6} e^{3} n x^{2} + 840 \, a d^{9} x^{3} - 280 \, b d^{3} e^{6} n x - 840 \, b d^{9} n \log \left (x^{\frac {1}{3}}\right ) + 840 \, {\left (b d^{9} + b e^{9}\right )} n \log \left (d x^{\frac {1}{3}} + e\right ) + 840 \, {\left (b d^{9} n x^{3} - b d^{9} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 21 \, {\left (5 \, b d^{8} e n x^{2} - 8 \, b d^{5} e^{4} n x + 20 \, b d^{2} e^{7} n\right )} x^{\frac {2}{3}} - 30 \, {\left (4 \, b d^{7} e^{2} n x^{2} - 7 \, b d^{4} e^{5} n x + 28 \, b d e^{8} n\right )} x^{\frac {1}{3}}}{2520 \, d^{9}} \]

[In]

integrate(x^2*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="fricas")

[Out]

1/2520*(840*b*d^9*x^3*log(c) + 140*b*d^6*e^3*n*x^2 + 840*a*d^9*x^3 - 280*b*d^3*e^6*n*x - 840*b*d^9*n*log(x^(1/
3)) + 840*(b*d^9 + b*e^9)*n*log(d*x^(1/3) + e) + 840*(b*d^9*n*x^3 - b*d^9*n)*log((d*x + e*x^(2/3))/x) + 21*(5*
b*d^8*e*n*x^2 - 8*b*d^5*e^4*n*x + 20*b*d^2*e^7*n)*x^(2/3) - 30*(4*b*d^7*e^2*n*x^2 - 7*b*d^4*e^5*n*x + 28*b*d*e
^8*n)*x^(1/3))/d^9

Sympy [A] (verification not implemented)

Time = 57.67 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.85 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {a x^{3}}{3} + b \left (\frac {e n \left (\frac {3 x^{\frac {8}{3}}}{8 d} - \frac {3 e x^{\frac {7}{3}}}{7 d^{2}} + \frac {e^{2} x^{2}}{2 d^{3}} - \frac {3 e^{3} x^{\frac {5}{3}}}{5 d^{4}} + \frac {3 e^{4} x^{\frac {4}{3}}}{4 d^{5}} - \frac {e^{5} x}{d^{6}} + \frac {3 e^{6} x^{\frac {2}{3}}}{2 d^{7}} + \frac {3 e^{8} \left (\begin {cases} \frac {\sqrt [3]{x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt [3]{x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{8}} - \frac {3 e^{7} \sqrt [3]{x}}{d^{8}}\right )}{9} + \frac {x^{3} \log {\left (c \left (d + \frac {e}{\sqrt [3]{x}}\right )^{n} \right )}}{3}\right ) \]

[In]

integrate(x**2*(a+b*ln(c*(d+e/x**(1/3))**n)),x)

[Out]

a*x**3/3 + b*(e*n*(3*x**(8/3)/(8*d) - 3*e*x**(7/3)/(7*d**2) + e**2*x**2/(2*d**3) - 3*e**3*x**(5/3)/(5*d**4) +
3*e**4*x**(4/3)/(4*d**5) - e**5*x/d**6 + 3*e**6*x**(2/3)/(2*d**7) + 3*e**8*Piecewise((x**(1/3)/e, Eq(d, 0)), (
log(d*x**(1/3) + e)/d, True))/d**8 - 3*e**7*x**(1/3)/d**8)/9 + x**3*log(c*(d + e/x**(1/3))**n)/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.67 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{2520} \, b e n {\left (\frac {840 \, e^{8} \log \left (d x^{\frac {1}{3}} + e\right )}{d^{9}} + \frac {105 \, d^{7} x^{\frac {8}{3}} - 120 \, d^{6} e x^{\frac {7}{3}} + 140 \, d^{5} e^{2} x^{2} - 168 \, d^{4} e^{3} x^{\frac {5}{3}} + 210 \, d^{3} e^{4} x^{\frac {4}{3}} - 280 \, d^{2} e^{5} x + 420 \, d e^{6} x^{\frac {2}{3}} - 840 \, e^{7} x^{\frac {1}{3}}}{d^{8}}\right )} \]

[In]

integrate(x^2*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="maxima")

[Out]

1/3*b*x^3*log(c*(d + e/x^(1/3))^n) + 1/3*a*x^3 + 1/2520*b*e*n*(840*e^8*log(d*x^(1/3) + e)/d^9 + (105*d^7*x^(8/
3) - 120*d^6*e*x^(7/3) + 140*d^5*e^2*x^2 - 168*d^4*e^3*x^(5/3) + 210*d^3*e^4*x^(4/3) - 280*d^2*e^5*x + 420*d*e
^6*x^(2/3) - 840*e^7*x^(1/3))/d^8)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.71 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left (c\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{2520} \, {\left (840 \, x^{3} \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right ) + e {\left (\frac {840 \, e^{8} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{d^{9}} + \frac {105 \, d^{7} x^{\frac {8}{3}} - 120 \, d^{6} e x^{\frac {7}{3}} + 140 \, d^{5} e^{2} x^{2} - 168 \, d^{4} e^{3} x^{\frac {5}{3}} + 210 \, d^{3} e^{4} x^{\frac {4}{3}} - 280 \, d^{2} e^{5} x + 420 \, d e^{6} x^{\frac {2}{3}} - 840 \, e^{7} x^{\frac {1}{3}}}{d^{8}}\right )}\right )} b n \]

[In]

integrate(x^2*(a+b*log(c*(d+e/x^(1/3))^n)),x, algorithm="giac")

[Out]

1/3*b*x^3*log(c) + 1/3*a*x^3 + 1/2520*(840*x^3*log(d + e/x^(1/3)) + e*(840*e^8*log(abs(d*x^(1/3) + e))/d^9 + (
105*d^7*x^(8/3) - 120*d^6*e*x^(7/3) + 140*d^5*e^2*x^2 - 168*d^4*e^3*x^(5/3) + 210*d^3*e^4*x^(4/3) - 280*d^2*e^
5*x + 420*d*e^6*x^(2/3) - 840*e^7*x^(1/3))/d^8))*b*n

Mupad [B] (verification not implemented)

Time = 1.91 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.81 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )\right ) \, dx=\frac {840\,a\,d^9\,x^3+1680\,b\,e^9\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{1/3}}+1\right )+840\,b\,d^9\,x^3\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )-280\,b\,d^3\,e^6\,n\,x-840\,b\,d\,e^8\,n\,x^{1/3}+105\,b\,d^8\,e\,n\,x^{8/3}+140\,b\,d^6\,e^3\,n\,x^2+420\,b\,d^2\,e^7\,n\,x^{2/3}+210\,b\,d^4\,e^5\,n\,x^{4/3}-168\,b\,d^5\,e^4\,n\,x^{5/3}-120\,b\,d^7\,e^2\,n\,x^{7/3}}{2520\,d^9} \]

[In]

int(x^2*(a + b*log(c*(d + e/x^(1/3))^n)),x)

[Out]

(840*a*d^9*x^3 + 1680*b*e^9*n*atanh((2*e)/(d*x^(1/3)) + 1) + 840*b*d^9*x^3*log(c*(d + e/x^(1/3))^n) - 280*b*d^
3*e^6*n*x - 840*b*d*e^8*n*x^(1/3) + 105*b*d^8*e*n*x^(8/3) + 140*b*d^6*e^3*n*x^2 + 420*b*d^2*e^7*n*x^(2/3) + 21
0*b*d^4*e^5*n*x^(4/3) - 168*b*d^5*e^4*n*x^(5/3) - 120*b*d^7*e^2*n*x^(7/3))/(2520*d^9)