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

Optimal. Leaf size=143 \[ \frac{1}{4} x^4 \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )+\frac{b e^5 n x^{2/3}}{4 d^5}-\frac{b e^4 n x^{4/3}}{8 d^4}+\frac{b e^3 n x^2}{12 d^3}-\frac{b e^2 n x^{8/3}}{16 d^2}-\frac{b e^6 n \log \left (d+\frac{e}{x^{2/3}}\right )}{4 d^6}-\frac{b e^6 n \log (x)}{6 d^6}+\frac{b e n x^{10/3}}{20 d} \]

[Out]

(b*e^5*n*x^(2/3))/(4*d^5) - (b*e^4*n*x^(4/3))/(8*d^4) + (b*e^3*n*x^2)/(12*d^3) - (b*e^2*n*x^(8/3))/(16*d^2) +
(b*e*n*x^(10/3))/(20*d) - (b*e^6*n*Log[d + e/x^(2/3)])/(4*d^6) + (x^4*(a + b*Log[c*(d + e/x^(2/3))^n]))/4 - (b
*e^6*n*Log[x])/(6*d^6)

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Rubi [A]  time = 0.105398, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2454, 2395, 44} \[ \frac{1}{4} x^4 \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )+\frac{b e^5 n x^{2/3}}{4 d^5}-\frac{b e^4 n x^{4/3}}{8 d^4}+\frac{b e^3 n x^2}{12 d^3}-\frac{b e^2 n x^{8/3}}{16 d^2}-\frac{b e^6 n \log \left (d+\frac{e}{x^{2/3}}\right )}{4 d^6}-\frac{b e^6 n \log (x)}{6 d^6}+\frac{b e n x^{10/3}}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^5*n*x^(2/3))/(4*d^5) - (b*e^4*n*x^(4/3))/(8*d^4) + (b*e^3*n*x^2)/(12*d^3) - (b*e^2*n*x^(8/3))/(16*d^2) +
(b*e*n*x^(10/3))/(20*d) - (b*e^6*n*Log[d + e/x^(2/3)])/(4*d^6) + (x^4*(a + b*Log[c*(d + e/x^(2/3))^n]))/4 - (b
*e^6*n*Log[x])/(6*d^6)

Rule 2454

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])

Rule 2395

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 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^3 \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right ) \, dx &=-\left (\frac{3}{2} \operatorname{Subst}\left (\int \frac{a+b \log \left (c (d+e x)^n\right )}{x^7} \, dx,x,\frac{1}{x^{2/3}}\right )\right )\\ &=\frac{1}{4} x^4 \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )-\frac{1}{4} (b e n) \operatorname{Subst}\left (\int \frac{1}{x^6 (d+e x)} \, dx,x,\frac{1}{x^{2/3}}\right )\\ &=\frac{1}{4} x^4 \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )-\frac{1}{4} (b e n) \operatorname{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}{x^{2/3}}\right )\\ &=\frac{b e^5 n x^{2/3}}{4 d^5}-\frac{b e^4 n x^{4/3}}{8 d^4}+\frac{b e^3 n x^2}{12 d^3}-\frac{b e^2 n x^{8/3}}{16 d^2}+\frac{b e n x^{10/3}}{20 d}-\frac{b e^6 n \log \left (d+\frac{e}{x^{2/3}}\right )}{4 d^6}+\frac{1}{4} x^4 \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )-\frac{b e^6 n \log (x)}{6 d^6}\\ \end{align*}

Mathematica [A]  time = 0.108891, size = 134, normalized size = 0.94 \[ \frac{a x^4}{4}+\frac{1}{4} b x^4 \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )-\frac{1}{4} b e n \left (-\frac{e^4 x^{2/3}}{d^5}+\frac{e^3 x^{4/3}}{2 d^4}-\frac{e^2 x^2}{3 d^3}+\frac{e^5 \log \left (d+\frac{e}{x^{2/3}}\right )}{d^6}+\frac{2 e^5 \log (x)}{3 d^6}+\frac{e x^{8/3}}{4 d^2}-\frac{x^{10/3}}{5 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.6, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b\ln \left ( c \left ( d+{e{x}^{-{\frac{2}{3}}}} \right ) ^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03153, size = 132, normalized size = 0.92 \begin{align*} \frac{1}{4} \, b x^{4} \log \left (c{\left (d + \frac{e}{x^{\frac{2}{3}}}\right )}^{n}\right ) + \frac{1}{4} \, a x^{4} - \frac{1}{240} \, b e n{\left (\frac{60 \, e^{5} \log \left (d x^{\frac{2}{3}} + e\right )}{d^{6}} - \frac{12 \, d^{4} x^{\frac{10}{3}} - 15 \, d^{3} e x^{\frac{8}{3}} + 20 \, d^{2} e^{2} x^{2} - 30 \, d e^{3} x^{\frac{4}{3}} + 60 \, e^{4} x^{\frac{2}{3}}}{d^{5}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4*log(c*(d + e/x^(2/3))^n) + 1/4*a*x^4 - 1/240*b*e*n*(60*e^5*log(d*x^(2/3) + e)/d^6 - (12*d^4*x^(10/3)
- 15*d^3*e*x^(8/3) + 20*d^2*e^2*x^2 - 30*d*e^3*x^(4/3) + 60*e^4*x^(2/3))/d^5)

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Fricas [A]  time = 1.95869, size = 385, normalized size = 2.69 \begin{align*} \frac{60 \, b d^{6} x^{4} \log \left (c\right ) + 60 \, a d^{6} x^{4} + 20 \, b d^{3} e^{3} n x^{2} - 120 \, 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{2}{3}} + e\right ) + 60 \,{\left (b d^{6} n x^{4} - b d^{6} n\right )} \log \left (\frac{d x + e x^{\frac{1}{3}}}{x}\right ) - 15 \,{\left (b d^{4} e^{2} n x^{2} - 4 \, b d e^{5} n\right )} x^{\frac{2}{3}} + 6 \,{\left (2 \, b d^{5} e n x^{3} - 5 \, b d^{2} e^{4} n x\right )} x^{\frac{1}{3}}}{240 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(60*b*d^6*x^4*log(c) + 60*a*d^6*x^4 + 20*b*d^3*e^3*n*x^2 - 120*b*d^6*n*log(x^(1/3)) + 60*(b*d^6 - b*e^6)
*n*log(d*x^(2/3) + e) + 60*(b*d^6*n*x^4 - b*d^6*n)*log((d*x + e*x^(1/3))/x) - 15*(b*d^4*e^2*n*x^2 - 4*b*d*e^5*
n)*x^(2/3) + 6*(2*b*d^5*e*n*x^3 - 5*b*d^2*e^4*n*x)*x^(1/3))/d^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.35651, size = 139, normalized size = 0.97 \begin{align*} \frac{1}{4} \, b x^{4} \log \left (c\right ) + \frac{1}{4} \, a x^{4} + \frac{1}{240} \,{\left (60 \, x^{4} \log \left (d + \frac{e}{x^{\frac{2}{3}}}\right ) +{\left (\frac{12 \, d^{4} x^{\frac{10}{3}} - 15 \, d^{3} x^{\frac{8}{3}} e + 20 \, d^{2} x^{2} e^{2} - 30 \, d x^{\frac{4}{3}} e^{3} + 60 \, x^{\frac{2}{3}} e^{4}}{d^{5}} - \frac{60 \, e^{5} \log \left ({\left | d x^{\frac{2}{3}} + e \right |}\right )}{d^{6}}\right )} e\right )} b n \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4*log(c) + 1/4*a*x^4 + 1/240*(60*x^4*log(d + e/x^(2/3)) + ((12*d^4*x^(10/3) - 15*d^3*x^(8/3)*e + 20*d^
2*x^2*e^2 - 30*d*x^(4/3)*e^3 + 60*x^(2/3)*e^4)/d^5 - 60*e^5*log(abs(d*x^(2/3) + e))/d^6)*e)*b*n