\(\int \frac {a+b \log (c (d+e x^{2/3})^n)}{x^3} \, dx\) [469]

Optimal result

Integrand size = 22, antiderivative size = 94 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {b e n}{4 d x^{4/3}}+\frac {b e^2 n}{2 d^2 x^{2/3}}-\frac {b e^3 n \log \left (d+e x^{2/3}\right )}{2 d^3}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}+\frac {b e^3 n \log (x)}{3 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2504, 2442, 46} \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}-\frac {b e^3 n \log \left (d+e x^{2/3}\right )}{2 d^3}+\frac {b e^3 n \log (x)}{3 d^3}+\frac {b e^2 n}{2 d^2 x^{2/3}}-\frac {b e n}{4 d x^{4/3}} \]

[In]

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

[Out]

-1/4*(b*e*n)/(d*x^(4/3)) + (b*e^2*n)/(2*d^2*x^(2/3)) - (b*e^3*n*Log[d + e*x^(2/3)])/(2*d^3) - (a + b*Log[c*(d
+ e*x^(2/3))^n])/(2*x^2) + (b*e^3*n*Log[x])/(3*d^3)

Rule 46

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}+\frac {1}{3} b e n \left (-\frac {3}{4 d x^{4/3}}+\frac {3 e}{2 d^2 x^{2/3}}-\frac {3 e^2 \log \left (d+e x^{2/3}\right )}{2 d^3}+\frac {e^2 \log (x)}{d^3}\right ) \]

[In]

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

[Out]

-1/2*a/x^2 - (b*Log[c*(d + e*x^(2/3))^n])/(2*x^2) + (b*e*n*(-3/(4*d*x^(4/3)) + (3*e)/(2*d^2*x^(2/3)) - (3*e^2*
Log[d + e*x^(2/3)])/(2*d^3) + (e^2*Log[x])/d^3))/3

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=\frac {4 \, b e^{3} n x^{2} \log \left (x^{\frac {1}{3}}\right ) + 2 \, b d e^{2} n x^{\frac {4}{3}} - b d^{2} e n x^{\frac {2}{3}} - 2 \, b d^{3} \log \left (c\right ) - 2 \, a d^{3} - 2 \, {\left (b e^{3} n x^{2} + b d^{3} n\right )} \log \left (e x^{\frac {2}{3}} + d\right )}{4 \, d^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {1}{4} \, b e n {\left (\frac {2 \, e^{2} \log \left (e x^{\frac {2}{3}} + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x^{\frac {2}{3}}\right )}{d^{3}} - \frac {2 \, e x^{\frac {2}{3}} - d}{d^{2} x^{\frac {4}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {{\left (e^{4} {\left (\frac {2 \, \log \left ({\left | e x^{\frac {2}{3}} + d \right |}\right )}{d^{3}} - \frac {2 \, \log \left ({\left | e x^{\frac {2}{3}} \right |}\right )}{d^{3}} - \frac {2 \, {\left (e x^{\frac {2}{3}} + d\right )} d - 3 \, d^{2}}{d^{3} e^{2} x^{\frac {4}{3}}}\right )} + \frac {2 \, e \log \left (e x^{\frac {2}{3}} + d\right )}{x^{2}}\right )} b n}{4 \, e} - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.83 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^3} \, dx=-\frac {\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n\,x^{2/3}}{d^2}}{2\,x^{4/3}}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2\,x^2}-\frac {b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{2/3}}{d}+1\right )}{d^3} \]

[In]

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

[Out]

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