\(\int \frac {a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)}{x^3} \, dx\) [495]

Optimal result

Integrand size = 22, antiderivative size = 138 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {b n}{12 x^2}-\frac {b d n}{10 e x^{5/3}}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 138, 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} \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d n}{10 e x^{5/3}}+\frac {b n}{12 x^2} \]

[In]

Int[(a + b*Log[c*(d + e/x^(1/3))^n])/x^3,x]

[Out]

(b*n)/(12*x^2) - (b*d*n)/(10*e*x^(5/3)) + (b*d^2*n)/(8*e^2*x^(4/3)) - (b*d^3*n)/(6*e^3*x) + (b*d^4*n)/(4*e^4*x
^(2/3)) - (b*d^5*n)/(2*e^5*x^(1/3)) + (b*d^6*n*Log[d + e/x^(1/3)])/(2*e^6) - (a + b*Log[c*(d + e/x^(1/3))^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 (3 \text {Subst}\left (\int x^5 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\right ) \\ & = -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}+\frac {1}{2} (b e n) \text {Subst}\left (\int \frac {x^6}{d+e x} \, dx,x,\frac {1}{\sqrt [3]{x}}\right ) \\ & = -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}+\frac {1}{2} (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,\frac {1}{\sqrt [3]{x}}\right ) \\ & = \frac {b n}{12 x^2}-\frac {b d n}{10 e x^{5/3}}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {1}{6} b e n \left (-\frac {1}{2 e x^2}+\frac {3 d}{5 e^2 x^{5/3}}-\frac {3 d^2}{4 e^3 x^{4/3}}+\frac {d^3}{e^4 x}-\frac {3 d^4}{2 e^5 x^{2/3}}+\frac {3 d^5}{e^6 \sqrt [3]{x}}-\frac {3 d^6 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^7}\right )-\frac {b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \]

[In]

Integrate[(a + b*Log[c*(d + e/x^(1/3))^n])/x^3,x]

[Out]

-1/2*a/x^2 - (b*e*n*(-1/2*1/(e*x^2) + (3*d)/(5*e^2*x^(5/3)) - (3*d^2)/(4*e^3*x^(4/3)) + d^3/(e^4*x) - (3*d^4)/
(2*e^5*x^(2/3)) + (3*d^5)/(e^6*x^(1/3)) - (3*d^6*Log[d + e/x^(1/3)])/e^7))/6 - (b*Log[c*(d + e/x^(1/3))^n])/(2
*x^2)

Maple [F]

\[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )}{x^{3}}d x\]

[In]

int((a+b*ln(c*(d+e/x^(1/3))^n))/x^3,x)

[Out]

int((a+b*ln(c*(d+e/x^(1/3))^n))/x^3,x)

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=-\frac {20 \, b d^{3} e^{3} n x - 10 \, b e^{6} n + 60 \, a e^{6} - 10 \, {\left (6 \, a e^{6} + {\left (2 \, b d^{3} e^{3} - b e^{6}\right )} n\right )} x^{2} - 60 \, {\left (b e^{6} x^{2} - b e^{6}\right )} \log \left (c\right ) - 60 \, {\left (b d^{6} n x^{2} - b e^{6} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right ) + 15 \, {\left (4 \, b d^{5} e n x - b d^{2} e^{4} n\right )} x^{\frac {2}{3}} - 6 \, {\left (5 \, b d^{4} e^{2} n x - 2 \, b d e^{5} n\right )} x^{\frac {1}{3}}}{120 \, e^{6} x^{2}} \]

[In]

integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x, algorithm="fricas")

[Out]

-1/120*(20*b*d^3*e^3*n*x - 10*b*e^6*n + 60*a*e^6 - 10*(6*a*e^6 + (2*b*d^3*e^3 - b*e^6)*n)*x^2 - 60*(b*e^6*x^2
- b*e^6)*log(c) - 60*(b*d^6*n*x^2 - b*e^6*n)*log((d*x + e*x^(2/3))/x) + 15*(4*b*d^5*e*n*x - b*d^2*e^4*n)*x^(2/
3) - 6*(5*b*d^4*e^2*n*x - 2*b*d*e^5*n)*x^(1/3))/(e^6*x^2)

Sympy [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((a+b*ln(c*(d+e/x**(1/3))**n))/x**3,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {1}{120} \, b e n {\left (\frac {60 \, d^{6} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{7}} - \frac {20 \, d^{6} \log \left (x\right )}{e^{7}} - \frac {60 \, d^{5} x^{\frac {5}{3}} - 30 \, d^{4} e x^{\frac {4}{3}} + 20 \, d^{3} e^{2} x - 15 \, d^{2} e^{3} x^{\frac {2}{3}} + 12 \, d e^{4} x^{\frac {1}{3}} - 10 \, e^{5}}{e^{6} x^{2}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[In]

integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x, algorithm="maxima")

[Out]

1/120*b*e*n*(60*d^6*log(d*x^(1/3) + e)/e^7 - 20*d^6*log(x)/e^7 - (60*d^5*x^(5/3) - 30*d^4*e*x^(4/3) + 20*d^3*e
^2*x - 15*d^2*e^3*x^(2/3) + 12*d*e^4*x^(1/3) - 10*e^5)/(e^6*x^2)) - 1/2*b*log(c*(d + e/x^(1/3))^n)/x^2 - 1/2*a
/x^2

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {1}{120} \, {\left (e {\left (\frac {60 \, d^{6} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right )}{e^{7}} - \frac {20 \, d^{6} \log \left ({\left | x \right |}\right )}{e^{7}} - \frac {60 \, d^{5} e x^{\frac {5}{3}} - 30 \, d^{4} e^{2} x^{\frac {4}{3}} + 20 \, d^{3} e^{3} x - 15 \, d^{2} e^{4} x^{\frac {2}{3}} + 12 \, d e^{5} x^{\frac {1}{3}} - 10 \, e^{6}}{e^{7} x^{2}}\right )} - \frac {60 \, \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )}{x^{2}}\right )} b n - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[In]

integrate((a+b*log(c*(d+e/x^(1/3))^n))/x^3,x, algorithm="giac")

[Out]

1/120*(e*(60*d^6*log(abs(d*x^(1/3) + e))/e^7 - 20*d^6*log(abs(x))/e^7 - (60*d^5*e*x^(5/3) - 30*d^4*e^2*x^(4/3)
+ 20*d^3*e^3*x - 15*d^2*e^4*x^(2/3) + 12*d*e^5*x^(1/3) - 10*e^6)/(e^7*x^2)) - 60*log(d + e/x^(1/3))/x^2)*b*n
- 1/2*b*log(c)/x^2 - 1/2*a/x^2

Mupad [B] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx=\frac {b\,n}{12\,x^2}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{2\,x^2}-\frac {b\,d\,n}{10\,e\,x^{5/3}}+\frac {b\,d^6\,n\,\ln \left (d+\frac {e}{x^{1/3}}\right )}{2\,e^6}-\frac {b\,d^3\,n}{6\,e^3\,x}+\frac {b\,d^2\,n}{8\,e^2\,x^{4/3}}+\frac {b\,d^4\,n}{4\,e^4\,x^{2/3}}-\frac {b\,d^5\,n}{2\,e^5\,x^{1/3}} \]

[In]

int((a + b*log(c*(d + e/x^(1/3))^n))/x^3,x)

[Out]

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