\(\int x (a+b \log (c (d+e x^{2/3})^n)) \, dx\) [465]

Optimal result

Integrand size = 20, antiderivative size = 89 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=-\frac {b d^2 n x^{2/3}}{2 e^2}+\frac {b d n x^{4/3}}{4 e}-\frac {1}{6} b n x^2+\frac {b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2504, 2442, 45} \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+\frac {b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}-\frac {b d^2 n x^{2/3}}{2 e^2}+\frac {b d n x^{4/3}}{4 e}-\frac {1}{6} b n x^2 \]

[In]

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

[Out]

-1/2*(b*d^2*n*x^(2/3))/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 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*} \text {integral}& = \frac {3}{2} \text {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) \text {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) \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=-\frac {b d^2 n x^{2/3}}{2 e^2}+\frac {b d n x^{4/3}}{4 e}+\frac {a x^2}{2}-\frac {1}{6} b n x^2+\frac {b d^3 n \log \left (d+e x^{2/3}\right )}{2 e^3}+\frac {1}{2} b x^2 \log \left (c \left (d+e x^{2/3}\right )^n\right ) \]

[In]

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

[Out]

-1/2*(b*d^2*n*x^(2/3))/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

Maple [F]

\[\int x \left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {6 \, b e^{3} x^{2} \log \left (c\right ) + 3 \, b d e^{2} n x^{\frac {4}{3}} - 6 \, b d^{2} e n x^{\frac {2}{3}} - 2 \, {\left (b e^{3} n - 3 \, a e^{3}\right )} x^{2} + 6 \, {\left (b e^{3} n x^{2} + b d^{3} n\right )} \log \left (e x^{\frac {2}{3}} + d\right )}{12 \, e^{3}} \]

[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

Sympy [A] (verification not implemented)

Time = 103.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a x^{2}}{2} + b \left (- \frac {e n \left (- \frac {3 d^{3} \left (\begin {cases} \frac {x^{\frac {2}{3}}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{\frac {2}{3}} \right )}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{3}} + \frac {3 d^{2} x^{\frac {2}{3}}}{2 e^{3}} - \frac {3 d x^{\frac {4}{3}}}{4 e^{2}} + \frac {x^{2}}{2 e}\right )}{3} + \frac {x^{2} \log {\left (c \left (d + e x^{\frac {2}{3}}\right )^{n} \right )}}{2}\right ) \]

[In]

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

[Out]

a*x**2/2 + b*(-e*n*(-3*d**3*Piecewise((x**(2/3)/d, Eq(e, 0)), (log(d + e*x**(2/3))/e, True))/(2*e**3) + 3*d**2
*x**(2/3)/(2*e**3) - 3*d*x**(4/3)/(4*e**2) + x**2/(2*e))/3 + x**2*log(c*(d + e*x**(2/3))**n)/2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{12} \, b e n {\left (\frac {6 \, d^{3} \log \left (e x^{\frac {2}{3}} + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )} + \frac {1}{2} \, b x^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + \frac {1}{2} \, a x^{2} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{12} \, {\left (6 \, x^{2} \log \left (e x^{\frac {2}{3}} + d\right ) + e {\left (\frac {6 \, d^{3} \log \left ({\left | e x^{\frac {2}{3}} + d \right |}\right )}{e^{4}} - \frac {2 \, e^{2} x^{2} - 3 \, d e x^{\frac {4}{3}} + 6 \, d^{2} x^{\frac {2}{3}}}{e^{3}}\right )}\right )} b n + \frac {1}{2} \, a x^{2} \]

[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(e*x^(2/3) + d) + e*(6*d^3*log(abs(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))*b*n + 1/2*a*x^2

Mupad [B] (verification not implemented)

Time = 1.57 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \, dx=\frac {a\,x^2}{2}-\frac {b\,n\,x^2}{6}+\frac {b\,x^2\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2}+\frac {b\,d\,n\,x^{4/3}}{4\,e}+\frac {b\,d^3\,n\,\ln \left (d+e\,x^{2/3}\right )}{2\,e^3}-\frac {b\,d^2\,n\,x^{2/3}}{2\,e^2} \]

[In]

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

[Out]

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