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

Optimal. Leaf size=130 \[ -\frac {2 b d^4 n \sqrt [3]{x}}{3 e^4}+\frac {2 b d^3 n x}{9 e^3}-\frac {2 b d^2 n x^{5/3}}{15 e^2}+\frac {2 b d n x^{7/3}}{21 e}-\frac {2}{27} b n x^3+\frac {2 b d^{9/2} n \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 e^{9/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]

[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))

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Rubi [A]
time = 0.06, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2505, 348, 308, 211} \begin {gather*} \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 \text {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 \end {gather*}

Antiderivative was successfully verified.

[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*} \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 {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*}

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Mathematica [A]
time = 0.07, size = 135, normalized size = 1.04 \begin {gather*} -\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 {a x^3}{3}-\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} b x^3 \log \left (c \left (d+e x^{2/3}\right )^n\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.51, size = 95, normalized size = 0.73 \begin {gather*} \frac {1}{3} \, b x^{3} \log \left ({\left (x^{\frac {2}{3}} e + d\right )}^{n} c\right ) + \frac {1}{3} \, a x^{3} + \frac {2}{945} \, {\left (315 \, d^{\frac {9}{2}} \arctan \left (\frac {x^{\frac {1}{3}} e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {11}{2}\right )} + {\left (105 \, d^{3} x e - 63 \, d^{2} x^{\frac {5}{3}} e^{2} - 315 \, d^{4} x^{\frac {1}{3}} + 45 \, d x^{\frac {7}{3}} e^{3} - 35 \, x^{3} e^{4}\right )} e^{\left (-5\right )}\right )} b n e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3*log((x^(2/3)*e + d)^n*c) + 1/3*a*x^3 + 2/945*(315*d^(9/2)*arctan(x^(1/3)*e^(1/2)/sqrt(d))*e^(-11/2)
+ (105*d^3*x*e - 63*d^2*x^(5/3)*e^2 - 315*d^4*x^(1/3) + 45*d*x^(7/3)*e^3 - 35*x^3*e^4)*e^(-5))*b*n*e

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Fricas [A]
time = 0.42, size = 307, normalized size = 2.36 \begin {gather*} \left [\frac {1}{945} \, {\left (315 \, \sqrt {-d e^{\left (-1\right )}} b d^{4} n \log \left (-\frac {2 \, \sqrt {-d e^{\left (-1\right )}} d x e^{2} + d^{3} - x^{2} e^{3} - 2 \, {\left (d^{2} e + \sqrt {-d e^{\left (-1\right )}} x e^{3}\right )} x^{\frac {2}{3}} - 2 \, {\left (\sqrt {-d e^{\left (-1\right )}} d^{2} e - d x e^{2}\right )} x^{\frac {1}{3}}}{d^{3} + x^{2} e^{3}}\right ) + 210 \, b d^{3} n x e + 315 \, b n x^{3} e^{4} \log \left (x^{\frac {2}{3}} e + d\right ) - 126 \, b d^{2} n x^{\frac {5}{3}} e^{2} + 315 \, b x^{3} e^{4} \log \left (c\right ) - 35 \, {\left (2 \, b n - 9 \, a\right )} x^{3} e^{4} - 90 \, {\left (7 \, b d^{4} n - b d n x^{2} e^{3}\right )} x^{\frac {1}{3}}\right )} e^{\left (-4\right )}, \frac {1}{945} \, {\left (630 \, b d^{\frac {9}{2}} n \arctan \left (\frac {x^{\frac {1}{3}} e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} + 210 \, b d^{3} n x e + 315 \, b n x^{3} e^{4} \log \left (x^{\frac {2}{3}} e + d\right ) - 126 \, b d^{2} n x^{\frac {5}{3}} e^{2} + 315 \, b x^{3} e^{4} \log \left (c\right ) - 35 \, {\left (2 \, b n - 9 \, a\right )} x^{3} e^{4} - 90 \, {\left (7 \, b d^{4} n - b d n x^{2} e^{3}\right )} x^{\frac {1}{3}}\right )} e^{\left (-4\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 3.09, size = 104, normalized size = 0.80 \begin {gather*} \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 (x^{\frac {2}{3}} e + d\right ) + 2 \, {\left (315 \, d^{\frac {9}{2}} \arctan \left (\frac {x^{\frac {1}{3}} e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {11}{2}\right )} - {\left (315 \, d^{4} x^{\frac {1}{3}} e^{4} - 105 \, d^{3} x e^{5} + 63 \, d^{2} x^{\frac {5}{3}} e^{6} - 45 \, d x^{\frac {7}{3}} e^{7} + 35 \, x^{3} e^{8}\right )} e^{\left (-9\right )}\right )} e\right )} b n \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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(x^(2/3)*e + d) + 2*(315*d^(9/2)*arctan(x^(1/3)*e^(1/2)/sqrt(
d))*e^(-11/2) - (315*d^4*x^(1/3)*e^4 - 105*d^3*x*e^5 + 63*d^2*x^(5/3)*e^6 - 45*d*x^(7/3)*e^7 + 35*x^3*e^8)*e^(
-9))*e)*b*n

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\left (a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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