\(\int x^2 (a+b \log (c (d+e x^{2/3})^n)) \, dx\) [464]

Optimal result

Integrand size = 22, antiderivative size = 130 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=-\frac {2 b d^4 n \sqrt [3]{x}}{3 e^4}+\frac {2 b d^3 n x}{9 e^3}-\frac {2 b d^2 n x^{5/3}}{15 e^2}+\frac {2 b d n x^{7/3}}{21 e}-\frac {2}{27} b n x^3+\frac {2 b d^{9/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 e^{9/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]

[Out]

-2/3*b*d^4*n*x^(1/3)/e^4+2/9*b*d^3*n*x/e^3-2/15*b*d^2*n*x^(5/3)/e^2+2/21*b*d*n*x^(7/3)/e-2/27*b*n*x^3+2/3*b*d^
(9/2)*n*arctan(x^(1/3)*e^(1/2)/d^(1/2))/e^(9/2)+1/3*x^3*(a+b*ln(c*(d+e*x^(2/3))^n))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2505, 348, 308, 211} \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+\frac {2 b d^{9/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 e^{9/2}}-\frac {2 b d^4 n \sqrt [3]{x}}{3 e^4}+\frac {2 b d^3 n x}{9 e^3}-\frac {2 b d^2 n x^{5/3}}{15 e^2}+\frac {2 b d n x^{7/3}}{21 e}-\frac {2}{27} b n x^3 \]

[In]

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

[Out]

(-2*b*d^4*n*x^(1/3))/(3*e^4) + (2*b*d^3*n*x)/(9*e^3) - (2*b*d^2*n*x^(5/3))/(15*e^2) + (2*b*d*n*x^(7/3))/(21*e)
- (2*b*n*x^3)/27 + (2*b*d^(9/2)*n*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/(3*e^(9/2)) + (x^3*(a + b*Log[c*(d + e*x
^(2/3))^n]))/3

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 348

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{9} (2 b e n) \int \frac {x^{8/3}}{d+e x^{2/3}} \, dx \\ & = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{3} (2 b e n) \text {Subst}\left (\int \frac {x^{10}}{d+e x^2} \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{3} (2 b e n) \text {Subst}\left (\int \left (\frac {d^4}{e^5}-\frac {d^3 x^2}{e^4}+\frac {d^2 x^4}{e^3}-\frac {d x^6}{e^2}+\frac {x^8}{e}-\frac {d^5}{e^5 \left (d+e x^2\right )}\right ) \, dx,x,\sqrt [3]{x}\right ) \\ & = -\frac {2 b d^4 n \sqrt [3]{x}}{3 e^4}+\frac {2 b d^3 n x}{9 e^3}-\frac {2 b d^2 n x^{5/3}}{15 e^2}+\frac {2 b d n x^{7/3}}{21 e}-\frac {2}{27} b n x^3+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+\frac {\left (2 b d^5 n\right ) \text {Subst}\left (\int \frac {1}{d+e x^2} \, dx,x,\sqrt [3]{x}\right )}{3 e^4} \\ & = -\frac {2 b d^4 n \sqrt [3]{x}}{3 e^4}+\frac {2 b d^3 n x}{9 e^3}-\frac {2 b d^2 n x^{5/3}}{15 e^2}+\frac {2 b d n x^{7/3}}{21 e}-\frac {2}{27} b n x^3+\frac {2 b d^{9/2} n \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 e^{9/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a x^3}{3}-\frac {2}{9} b e n \left (\frac {3 d^4 \sqrt [3]{x}}{e^5}-\frac {d^3 x}{e^4}+\frac {3 d^2 x^{5/3}}{5 e^3}-\frac {3 d x^{7/3}}{7 e^2}+\frac {x^3}{3 e}-\frac {3 d^{9/2} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{e^{11/2}}\right )+\frac {1}{3} b x^3 \log \left (c \left (d+e x^{2/3}\right )^n\right ) \]

[In]

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

[Out]

(a*x^3)/3 - (2*b*e*n*((3*d^4*x^(1/3))/e^5 - (d^3*x)/e^4 + (3*d^2*x^(5/3))/(5*e^3) - (3*d*x^(7/3))/(7*e^2) + x^
3/(3*e) - (3*d^(9/2)*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/e^(11/2)))/9 + (b*x^3*Log[c*(d + e*x^(2/3))^n])/3

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.59 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\left [\frac {315 \, b e^{4} n x^{3} \log \left (e x^{\frac {2}{3}} + d\right ) + 315 \, b e^{4} x^{3} \log \left (c\right ) - 126 \, b d^{2} e^{2} n x^{\frac {5}{3}} + 315 \, b d^{4} n \sqrt {-\frac {d}{e}} \log \left (\frac {e^{3} x^{2} - 2 \, d e^{2} x \sqrt {-\frac {d}{e}} - d^{3} + 2 \, {\left (e^{3} x \sqrt {-\frac {d}{e}} + d^{2} e\right )} x^{\frac {2}{3}} - 2 \, {\left (d e^{2} x - d^{2} e \sqrt {-\frac {d}{e}}\right )} x^{\frac {1}{3}}}{e^{3} x^{2} + d^{3}}\right ) + 210 \, b d^{3} e n x - 35 \, {\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{3} + 90 \, {\left (b d e^{3} n x^{2} - 7 \, b d^{4} n\right )} x^{\frac {1}{3}}}{945 \, e^{4}}, \frac {315 \, b e^{4} n x^{3} \log \left (e x^{\frac {2}{3}} + d\right ) + 315 \, b e^{4} x^{3} \log \left (c\right ) - 126 \, b d^{2} e^{2} n x^{\frac {5}{3}} + 630 \, b d^{4} n \sqrt {\frac {d}{e}} \arctan \left (\frac {e x^{\frac {1}{3}} \sqrt {\frac {d}{e}}}{d}\right ) + 210 \, b d^{3} e n x - 35 \, {\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{3} + 90 \, {\left (b d e^{3} n x^{2} - 7 \, b d^{4} n\right )} x^{\frac {1}{3}}}{945 \, e^{4}}\right ] \]

[In]

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

[Out]

[1/945*(315*b*e^4*n*x^3*log(e*x^(2/3) + d) + 315*b*e^4*x^3*log(c) - 126*b*d^2*e^2*n*x^(5/3) + 315*b*d^4*n*sqrt
(-d/e)*log((e^3*x^2 - 2*d*e^2*x*sqrt(-d/e) - d^3 + 2*(e^3*x*sqrt(-d/e) + d^2*e)*x^(2/3) - 2*(d*e^2*x - d^2*e*s
qrt(-d/e))*x^(1/3))/(e^3*x^2 + d^3)) + 210*b*d^3*e*n*x - 35*(2*b*e^4*n - 9*a*e^4)*x^3 + 90*(b*d*e^3*n*x^2 - 7*
b*d^4*n)*x^(1/3))/e^4, 1/945*(315*b*e^4*n*x^3*log(e*x^(2/3) + d) + 315*b*e^4*x^3*log(c) - 126*b*d^2*e^2*n*x^(5
/3) + 630*b*d^4*n*sqrt(d/e)*arctan(e*x^(1/3)*sqrt(d/e)/d) + 210*b*d^3*e*n*x - 35*(2*b*e^4*n - 9*a*e^4)*x^3 + 9
0*(b*d*e^3*n*x^2 - 7*b*d^4*n)*x^(1/3))/e^4]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left (c\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{945} \, {\left (315 \, x^{3} \log \left (e x^{\frac {2}{3}} + d\right ) + 2 \, e {\left (\frac {315 \, d^{5} \arctan \left (\frac {e x^{\frac {1}{3}}}{\sqrt {d e}}\right )}{\sqrt {d e} e^{5}} - \frac {35 \, e^{8} x^{3} - 45 \, d e^{7} x^{\frac {7}{3}} + 63 \, d^{2} e^{6} x^{\frac {5}{3}} - 105 \, d^{3} e^{5} x + 315 \, d^{4} e^{4} x^{\frac {1}{3}}}{e^{9}}\right )}\right )} b n \]

[In]

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

[Out]

1/3*b*x^3*log(c) + 1/3*a*x^3 + 1/945*(315*x^3*log(e*x^(2/3) + d) + 2*e*(315*d^5*arctan(e*x^(1/3)/sqrt(d*e))/(s
qrt(d*e)*e^5) - (35*e^8*x^3 - 45*d*e^7*x^(7/3) + 63*d^2*e^6*x^(5/3) - 105*d^3*e^5*x + 315*d^4*e^4*x^(1/3))/e^9
))*b*n

Mupad [F(-1)]

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

[In]

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

[Out]

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