3.6.9 \(\int x^2 (a+b \log (c (d+\frac {e}{x^{2/3}})^n)) \, dx\) [509]

Optimal. Leaf size=121 \[ -\frac {2 b e^4 n \sqrt [3]{x}}{3 d^4}+\frac {2 b e^3 n x}{9 d^3}-\frac {2 b e^2 n x^{5/3}}{15 d^2}+\frac {2 b e n x^{7/3}}{21 d}+\frac {2 b e^{9/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 d^{9/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2505, 269, 348, 308, 211} \begin {gather*} \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {2 b e^{9/2} n \text {ArcTan}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 d^{9/2}}-\frac {2 b e^4 n \sqrt [3]{x}}{3 d^4}+\frac {2 b e^3 n x}{9 d^3}-\frac {2 b e^2 n x^{5/3}}{15 d^2}+\frac {2 b e n x^{7/3}}{21 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*e^4*n*x^(1/3))/(3*d^4) + (2*b*e^3*n*x)/(9*d^3) - (2*b*e^2*n*x^(5/3))/(15*d^2) + (2*b*e*n*x^(7/3))/(21*d)
+ (2*b*e^(9/2)*n*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]])/(3*d^(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 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 65, normalized size = 0.54 \begin {gather*} \frac {a x^3}{3}+\frac {2 b e n x^{7/3} \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};-\frac {e}{d x^{2/3}}\right )}{21 d}+\frac {1}{3} b x^3 \log \left (c \left (d+\frac {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]

(a*x^3)/3 + (2*b*e*n*x^(7/3)*Hypergeometric2F1[-7/2, 1, -5/2, -(e/(d*x^(2/3)))])/(21*d) + (b*x^3*Log[c*(d + e/
x^(2/3))^n])/3

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )\, dx\]

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 = 0.52, size = 86, normalized size = 0.71 \begin {gather*} \frac {1}{3} \, b x^{3} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + \frac {1}{3} \, a x^{3} + \frac {2}{315} \, b n {\left (\frac {105 \, \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {7}{2}}}{d^{\frac {9}{2}}} + \frac {15 \, d^{3} x^{\frac {7}{3}} - 21 \, d^{2} x^{\frac {5}{3}} e + 35 \, d x e^{2} - 105 \, x^{\frac {1}{3}} e^{3}}{d^{4}}\right )} 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(c*(d + e/x^(2/3))^n) + 1/3*a*x^3 + 2/315*b*n*(105*arctan(sqrt(d)*x^(1/3)*e^(-1/2))*e^(7/2)/d^(9/
2) + (15*d^3*x^(7/3) - 21*d^2*x^(5/3)*e + 35*d*x*e^2 - 105*x^(1/3)*e^3)/d^4)*e

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Fricas [A]
time = 0.41, size = 389, normalized size = 3.21 \begin {gather*} \left [\frac {105 \, b d^{4} x^{3} \log \left (c\right ) + 105 \, a d^{4} x^{3} + 105 \, b d^{4} n \log \left (d x^{\frac {2}{3}} + e\right ) - 210 \, b d^{4} n \log \left (x^{\frac {1}{3}}\right ) - 42 \, b d^{2} n x^{\frac {5}{3}} e^{2} + 70 \, b d n x e^{3} + 105 \, b n \sqrt {-\frac {e}{d}} e^{4} \log \left (\frac {d^{3} x^{2} - 2 \, d^{2} x \sqrt {-\frac {e}{d}} e + 2 \, {\left (d^{3} x \sqrt {-\frac {e}{d}} + d e^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{2} x e - d \sqrt {-\frac {e}{d}} e^{2}\right )} x^{\frac {1}{3}} - e^{3}}{d^{3} x^{2} + e^{3}}\right ) + 105 \, {\left (b d^{4} n x^{3} - b d^{4} n\right )} \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right ) + 30 \, {\left (b d^{3} n x^{2} e - 7 \, b n e^{4}\right )} x^{\frac {1}{3}}}{315 \, d^{4}}, \frac {105 \, b d^{4} x^{3} \log \left (c\right ) + 105 \, a d^{4} x^{3} + 105 \, b d^{4} n \log \left (d x^{\frac {2}{3}} + e\right ) - 210 \, b d^{4} n \log \left (x^{\frac {1}{3}}\right ) - 42 \, b d^{2} n x^{\frac {5}{3}} e^{2} + 70 \, b d n x e^{3} + \frac {210 \, b n \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {9}{2}}}{\sqrt {d}} + 105 \, {\left (b d^{4} n x^{3} - b d^{4} n\right )} \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right ) + 30 \, {\left (b d^{3} n x^{2} e - 7 \, b n e^{4}\right )} x^{\frac {1}{3}}}{315 \, d^{4}}\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]

[1/315*(105*b*d^4*x^3*log(c) + 105*a*d^4*x^3 + 105*b*d^4*n*log(d*x^(2/3) + e) - 210*b*d^4*n*log(x^(1/3)) - 42*
b*d^2*n*x^(5/3)*e^2 + 70*b*d*n*x*e^3 + 105*b*n*sqrt(-e/d)*e^4*log((d^3*x^2 - 2*d^2*x*sqrt(-e/d)*e + 2*(d^3*x*s
qrt(-e/d) + d*e^2)*x^(2/3) - 2*(d^2*x*e - d*sqrt(-e/d)*e^2)*x^(1/3) - e^3)/(d^3*x^2 + e^3)) + 105*(b*d^4*n*x^3
- b*d^4*n)*log((d*x + x^(1/3)*e)/x) + 30*(b*d^3*n*x^2*e - 7*b*n*e^4)*x^(1/3))/d^4, 1/315*(105*b*d^4*x^3*log(c
) + 105*a*d^4*x^3 + 105*b*d^4*n*log(d*x^(2/3) + e) - 210*b*d^4*n*log(x^(1/3)) - 42*b*d^2*n*x^(5/3)*e^2 + 70*b*
d*n*x*e^3 + 210*b*n*arctan(sqrt(d)*x^(1/3)*e^(-1/2))*e^(9/2)/sqrt(d) + 105*(b*d^4*n*x^3 - b*d^4*n)*log((d*x +
x^(1/3)*e)/x) + 30*(b*d^3*n*x^2*e - 7*b*n*e^4)*x^(1/3))/d^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 = 3.56, size = 97, normalized size = 0.80 \begin {gather*} \frac {1}{3} \, b x^{3} \log \left (c\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{315} \, {\left (105 \, x^{3} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) + 2 \, {\left (\frac {105 \, \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {7}{2}}}{d^{\frac {9}{2}}} + \frac {15 \, d^{6} x^{\frac {7}{3}} - 21 \, d^{5} x^{\frac {5}{3}} e + 35 \, d^{4} x e^{2} - 105 \, d^{3} x^{\frac {1}{3}} e^{3}}{d^{7}}\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 + 1/315*(105*x^3*log(d + e/x^(2/3)) + 2*(105*arctan(sqrt(d)*x^(1/3)*e^(-1/2))*e^(
7/2)/d^(9/2) + (15*d^6*x^(7/3) - 21*d^5*x^(5/3)*e + 35*d^4*x*e^2 - 105*d^3*x^(1/3)*e^3)/d^7)*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+\frac {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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