\(\int \frac {a+b \log (c (d+\frac {e}{x^{2/3}})^n)}{x^3} \, dx\) [514]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {b n}{6 x^2}-\frac {b d n}{4 e x^{4/3}}+\frac {b d^2 n}{2 e^2 x^{2/3}}-\frac {b d^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 e^3}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2504, 2442, 45} \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{2 x^2}-\frac {b d^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 e^3}+\frac {b d^2 n}{2 e^2 x^{2/3}}-\frac {b d n}{4 e x^{4/3}}+\frac {b n}{6 x^2} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}+\frac {b n}{6 x^2}-\frac {b d n}{4 e x^{4/3}}+\frac {b d^2 n}{2 e^2 x^{2/3}}-\frac {b d^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 e^3}-\frac {b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{2 x^2} \]

[In]

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

[Out]

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

Maple [F]

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

[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 = 0.37 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {6 \, b d^{2} e n x^{\frac {4}{3}} - 3 \, b d e^{2} n x^{\frac {2}{3}} + 2 \, b e^{3} n - 6 \, b e^{3} \log \left (c\right ) - 6 \, a e^{3} - 6 \, {\left (b d^{3} n x^{2} + b e^{3} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )}{12 \, e^{3} x^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[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 = 0.20 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=-\frac {1}{12} \, b e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {2}{3}} + e\right )}{e^{4}} - \frac {6 \, d^{3} \log \left (x^{\frac {2}{3}}\right )}{e^{4}} - \frac {6 \, d^{2} x^{\frac {4}{3}} - 3 \, d e x^{\frac {2}{3}} + 2 \, e^{2}}{e^{3} x^{2}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.18 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {1}{12} \, {\left (e {\left (\frac {12 \, d^{3} \log \left (x^{\frac {1}{3}}\right )}{e^{4}} - \frac {6 \, d^{3} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{e^{4}} - \frac {11 \, d^{3} x^{2} - 6 \, d^{2} e x^{\frac {4}{3}} + 3 \, d e^{2} x^{\frac {2}{3}} - 2 \, e^{3}}{e^{4} x^{2}}\right )} - \frac {6 \, \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right )}{x^{2}}\right )} b n - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[In]

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

[Out]

1/12*(e*(12*d^3*log(x^(1/3))/e^4 - 6*d^3*log(abs(d*x^(2/3) + e))/e^4 - (11*d^3*x^2 - 6*d^2*e*x^(4/3) + 3*d*e^2
*x^(2/3) - 2*e^3)/(e^4*x^2)) - 6*log(d + e/x^(2/3))/x^2)*b*n - 1/2*b*log(c)/x^2 - 1/2*a/x^2

Mupad [B] (verification not implemented)

Time = 1.68 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^3} \, dx=\frac {b\,n}{6\,x^2}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{2\,x^2}-\frac {b\,d\,n}{4\,e\,x^{4/3}}-\frac {b\,d^3\,n\,\ln \left (d+\frac {e}{x^{2/3}}\right )}{2\,e^3}+\frac {b\,d^2\,n}{2\,e^2\,x^{2/3}} \]

[In]

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

[Out]

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