3.5.63 \(\int x^3 (a+b \log (c (d+e x^{2/3})^n)) \, dx\) [463]

Optimal. Leaf size=138 \[ \frac {b d^5 n x^{2/3}}{4 e^5}-\frac {b d^4 n x^{4/3}}{8 e^4}+\frac {b d^3 n x^2}{12 e^3}-\frac {b d^2 n x^{8/3}}{16 e^2}+\frac {b d n x^{10/3}}{20 e}-\frac {1}{24} b n x^4-\frac {b d^6 n \log \left (d+e x^{2/3}\right )}{4 e^6}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]

[Out]

1/4*b*d^5*n*x^(2/3)/e^5-1/8*b*d^4*n*x^(4/3)/e^4+1/12*b*d^3*n*x^2/e^3-1/16*b*d^2*n*x^(8/3)/e^2+1/20*b*d*n*x^(10
/3)/e-1/24*b*n*x^4-1/4*b*d^6*n*ln(d+e*x^(2/3))/e^6+1/4*x^4*(a+b*ln(c*(d+e*x^(2/3))^n))

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Rubi [A]
time = 0.07, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 = {2504, 2442, 45} \begin {gather*} \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {b d^6 n \log \left (d+e x^{2/3}\right )}{4 e^6}+\frac {b d^5 n x^{2/3}}{4 e^5}-\frac {b d^4 n x^{4/3}}{8 e^4}+\frac {b d^3 n x^2}{12 e^3}-\frac {b d^2 n x^{8/3}}{16 e^2}+\frac {b d n x^{10/3}}{20 e}-\frac {1}{24} b n x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^5*n*x^(2/3))/(4*e^5) - (b*d^4*n*x^(4/3))/(8*e^4) + (b*d^3*n*x^2)/(12*e^3) - (b*d^2*n*x^(8/3))/(16*e^2) +
(b*d*n*x^(10/3))/(20*e) - (b*n*x^4)/24 - (b*d^6*n*Log[d + e*x^(2/3)])/(4*e^6) + (x^4*(a + b*Log[c*(d + e*x^(2/
3))^n]))/4

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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*} \int x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx &=\frac {3}{2} \text {Subst}\left (\int x^5 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,x^{2/3}\right )\\ &=\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \frac {x^6}{d+e x} \, dx,x,x^{2/3}\right )\\ &=\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \left (-\frac {d^5}{e^6}+\frac {d^4 x}{e^5}-\frac {d^3 x^2}{e^4}+\frac {d^2 x^3}{e^3}-\frac {d x^4}{e^2}+\frac {x^5}{e}+\frac {d^6}{e^6 (d+e x)}\right ) \, dx,x,x^{2/3}\right )\\ &=\frac {b d^5 n x^{2/3}}{4 e^5}-\frac {b d^4 n x^{4/3}}{8 e^4}+\frac {b d^3 n x^2}{12 e^3}-\frac {b d^2 n x^{8/3}}{16 e^2}+\frac {b d n x^{10/3}}{20 e}-\frac {1}{24} b n x^4-\frac {b d^6 n \log \left (d+e x^{2/3}\right )}{4 e^6}+\frac {1}{4} x^4 \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 = 0.09, size = 135, normalized size = 0.98 \begin {gather*} \frac {a x^4}{4}-\frac {1}{4} b e n \left (-\frac {d^5 x^{2/3}}{e^6}+\frac {d^4 x^{4/3}}{2 e^5}-\frac {d^3 x^2}{3 e^4}+\frac {d^2 x^{8/3}}{4 e^3}-\frac {d x^{10/3}}{5 e^2}+\frac {x^4}{6 e}+\frac {d^6 \log \left (d+e x^{2/3}\right )}{e^7}\right )+\frac {1}{4} b x^4 \log \left (c \left (d+e x^{2/3}\right )^n\right ) \end {gather*}

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*e*n*(-((d^5*x^(2/3))/e^6) + (d^4*x^(4/3))/(2*e^5) - (d^3*x^2)/(3*e^4) + (d^2*x^(8/3))/(4*e^3) -
(d*x^(10/3))/(5*e^2) + x^4/(6*e) + (d^6*Log[d + e*x^(2/3)])/e^7))/4 + (b*x^4*Log[c*(d + e*x^(2/3))^n])/4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^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 = 0.29, size = 106, normalized size = 0.77 \begin {gather*} \frac {1}{4} \, b x^{4} \log \left ({\left (x^{\frac {2}{3}} e + d\right )}^{n} c\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{240} \, {\left (60 \, d^{6} e^{\left (-7\right )} \log \left (x^{\frac {2}{3}} e + d\right ) + {\left (30 \, d^{4} x^{\frac {4}{3}} e - 20 \, d^{3} x^{2} e^{2} - 60 \, d^{5} x^{\frac {2}{3}} + 15 \, d^{2} x^{\frac {8}{3}} e^{3} - 12 \, d x^{\frac {10}{3}} e^{4} + 10 \, x^{4} e^{5}\right )} e^{\left (-6\right )}\right )} b n e \end {gather*}

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((x^(2/3)*e + d)^n*c) + 1/4*a*x^4 - 1/240*(60*d^6*e^(-7)*log(x^(2/3)*e + d) + (30*d^4*x^(4/3)*e -
20*d^3*x^2*e^2 - 60*d^5*x^(2/3) + 15*d^2*x^(8/3)*e^3 - 12*d*x^(10/3)*e^4 + 10*x^4*e^5)*e^(-6))*b*n*e

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Fricas [A]
time = 0.38, size = 121, normalized size = 0.88 \begin {gather*} \frac {1}{240} \, {\left (20 \, b d^{3} n x^{2} e^{3} + 60 \, b x^{4} e^{6} \log \left (c\right ) - 10 \, {\left (b n - 6 \, a\right )} x^{4} e^{6} - 60 \, {\left (b d^{6} n - b n x^{4} e^{6}\right )} \log \left (x^{\frac {2}{3}} e + d\right ) + 15 \, {\left (4 \, b d^{5} n e - b d^{2} n x^{2} e^{4}\right )} x^{\frac {2}{3}} - 6 \, {\left (5 \, b d^{4} n x e^{2} - 2 \, b d n x^{3} e^{5}\right )} x^{\frac {1}{3}}\right )} e^{\left (-6\right )} \end {gather*}

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

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (106) = 212\).
time = 4.41, size = 261, normalized size = 1.89 \begin {gather*} \frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} + \frac {1}{240} \, {\left (60 \, {\left (x^{\frac {2}{3}} e + d\right )}^{6} e^{\left (-6\right )} \log \left (x^{\frac {2}{3}} e + d\right ) - 360 \, {\left (x^{\frac {2}{3}} e + d\right )}^{5} d e^{\left (-6\right )} \log \left (x^{\frac {2}{3}} e + d\right ) + 900 \, {\left (x^{\frac {2}{3}} e + d\right )}^{4} d^{2} e^{\left (-6\right )} \log \left (x^{\frac {2}{3}} e + d\right ) - 1200 \, {\left (x^{\frac {2}{3}} e + d\right )}^{3} d^{3} e^{\left (-6\right )} \log \left (x^{\frac {2}{3}} e + d\right ) + 900 \, {\left (x^{\frac {2}{3}} e + d\right )}^{2} d^{4} e^{\left (-6\right )} \log \left (x^{\frac {2}{3}} e + d\right ) - 10 \, {\left (x^{\frac {2}{3}} e + d\right )}^{6} e^{\left (-6\right )} + 72 \, {\left (x^{\frac {2}{3}} e + d\right )}^{5} d e^{\left (-6\right )} - 225 \, {\left (x^{\frac {2}{3}} e + d\right )}^{4} d^{2} e^{\left (-6\right )} + 400 \, {\left (x^{\frac {2}{3}} e + d\right )}^{3} d^{3} e^{\left (-6\right )} - 450 \, {\left (x^{\frac {2}{3}} e + d\right )}^{2} d^{4} e^{\left (-6\right )} - 360 \, {\left ({\left (x^{\frac {2}{3}} e + d\right )} \log \left (x^{\frac {2}{3}} e + d\right ) - x^{\frac {2}{3}} e - d\right )} d^{5} e^{\left (-6\right )}\right )} b n \end {gather*}

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^(2/3)*e + d)^6*e^(-6)*log(x^(2/3)*e + d) - 360*(x^(2/3)*e + d)^5*d
*e^(-6)*log(x^(2/3)*e + d) + 900*(x^(2/3)*e + d)^4*d^2*e^(-6)*log(x^(2/3)*e + d) - 1200*(x^(2/3)*e + d)^3*d^3*
e^(-6)*log(x^(2/3)*e + d) + 900*(x^(2/3)*e + d)^2*d^4*e^(-6)*log(x^(2/3)*e + d) - 10*(x^(2/3)*e + d)^6*e^(-6)
+ 72*(x^(2/3)*e + d)^5*d*e^(-6) - 225*(x^(2/3)*e + d)^4*d^2*e^(-6) + 400*(x^(2/3)*e + d)^3*d^3*e^(-6) - 450*(x
^(2/3)*e + d)^2*d^4*e^(-6) - 360*((x^(2/3)*e + d)*log(x^(2/3)*e + d) - x^(2/3)*e - d)*d^5*e^(-6))*b*n

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Mupad [B]
time = 0.45, size = 113, normalized size = 0.82 \begin {gather*} \frac {a\,x^4}{4}-\frac {b\,n\,x^4}{24}+\frac {b\,x^4\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{4}+\frac {b\,d\,n\,x^{10/3}}{20\,e}-\frac {b\,d^6\,n\,\ln \left (d+e\,x^{2/3}\right )}{4\,e^6}+\frac {b\,d^3\,n\,x^2}{12\,e^3}-\frac {b\,d^2\,n\,x^{8/3}}{16\,e^2}-\frac {b\,d^4\,n\,x^{4/3}}{8\,e^4}+\frac {b\,d^5\,n\,x^{2/3}}{4\,e^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x^4)/4 - (b*n*x^4)/24 + (b*x^4*log(c*(d + e*x^(2/3))^n))/4 + (b*d*n*x^(10/3))/(20*e) - (b*d^6*n*log(d + e*x
^(2/3)))/(4*e^6) + (b*d^3*n*x^2)/(12*e^3) - (b*d^2*n*x^(8/3))/(16*e^2) - (b*d^4*n*x^(4/3))/(8*e^4) + (b*d^5*n*
x^(2/3))/(4*e^5)

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