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

Optimal. Leaf size=89 \[ \frac{1}{2} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac{b d^2 n x^{2/3}}{2 e^2}+\frac{b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac{b d n x^{4/3}}{4 e}-\frac{1}{6} b n x^2 \]

[Out]

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

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Rubi [A]  time = 0.065433, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2454, 2395, 43} \[ \frac{1}{2} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac{b d^2 n x^{2/3}}{2 e^2}+\frac{b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac{b d n x^{4/3}}{4 e}-\frac{1}{6} b n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 43

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])

Rubi steps

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

Mathematica [A]  time = 0.0272043, size = 94, normalized size = 1.06 \[ \frac{a x^2}{2}+\frac{1}{2} b x^2 \log \left (c \left (d+e x^{2/3}\right )^n\right )-\frac{b d^2 n x^{2/3}}{2 e^2}+\frac{b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac{b d n x^{4/3}}{4 e}-\frac{1}{6} b n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.326, size = 0, normalized size = 0. \begin{align*} \int x \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*(a+b*ln(c*(d+e*x^(2/3))^n)),x)

[Out]

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

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Maxima [A]  time = 1.03577, size = 103, normalized size = 1.16 \begin{align*} \frac{1}{12} \, b e n{\left (\frac{6 \, d^{3} \log \left (e x^{\frac{2}{3}} + d\right )}{e^{4}} - \frac{2 \, e^{2} x^{2} - 3 \, d e x^{\frac{4}{3}} + 6 \, d^{2} x^{\frac{2}{3}}}{e^{3}}\right )} + \frac{1}{2} \, b x^{2} \log \left ({\left (e x^{\frac{2}{3}} + d\right )}^{n} c\right ) + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b*e*n*(6*d^3*log(e*x^(2/3) + d)/e^4 - (2*e^2*x^2 - 3*d*e*x^(4/3) + 6*d^2*x^(2/3))/e^3) + 1/2*b*x^2*log((e
*x^(2/3) + d)^n*c) + 1/2*a*x^2

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Fricas [A]  time = 1.82717, size = 203, normalized size = 2.28 \begin{align*} \frac{6 \, b e^{3} x^{2} \log \left (c\right ) + 3 \, b d e^{2} n x^{\frac{4}{3}} - 6 \, b d^{2} e n x^{\frac{2}{3}} - 2 \,{\left (b e^{3} n - 3 \, a e^{3}\right )} x^{2} + 6 \,{\left (b e^{3} n x^{2} + b d^{3} n\right )} \log \left (e x^{\frac{2}{3}} + d\right )}{12 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*b*e^3*x^2*log(c) + 3*b*d*e^2*n*x^(4/3) - 6*b*d^2*e*n*x^(2/3) - 2*(b*e^3*n - 3*a*e^3)*x^2 + 6*(b*e^3*n*
x^2 + b*d^3*n)*log(e*x^(2/3) + d))/e^3

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

[Out]

Timed out

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Giac [A]  time = 1.2815, size = 111, normalized size = 1.25 \begin{align*} \frac{1}{2} \, b x^{2} \log \left (c\right ) + \frac{1}{12} \,{\left (6 \, x^{2} \log \left (x^{\frac{2}{3}} e + d\right ) +{\left (6 \, d^{3} e^{\left (-4\right )} \log \left ({\left | x^{\frac{2}{3}} e + d \right |}\right ) +{\left (3 \, d x^{\frac{4}{3}} e - 2 \, x^{2} e^{2} - 6 \, d^{2} x^{\frac{2}{3}}\right )} e^{\left (-3\right )}\right )} e\right )} b n + \frac{1}{2} \, a x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x^2*log(c) + 1/12*(6*x^2*log(x^(2/3)*e + d) + (6*d^3*e^(-4)*log(abs(x^(2/3)*e + d)) + (3*d*x^(4/3)*e - 2
*x^2*e^2 - 6*d^2*x^(2/3))*e^(-3))*e)*b*n + 1/2*a*x^2