\(\int \frac {a+b \log (c (d+e \sqrt [3]{x})^n)}{x^2} \, dx\) [447]

Optimal result

Integrand size = 22, antiderivative size = 87 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {b e n}{2 d x^{2/3}}+\frac {b e^2 n}{d^2 \sqrt [3]{x}}-\frac {b e^3 n \log \left (d+e \sqrt [3]{x}\right )}{d^3}-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+\frac {b e^3 n \log (x)}{3 d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 87, 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} \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}-\frac {b e^3 n \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac {b e^3 n \log (x)}{3 d^3}+\frac {b e^2 n}{d^2 \sqrt [3]{x}}-\frac {b e n}{2 d x^{2/3}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {a}{x}-\frac {b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+\frac {1}{3} b e n \left (-\frac {3}{2 d x^{2/3}}+\frac {3 e}{d^2 \sqrt [3]{x}}-\frac {3 e^2 \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac {e^2 \log (x)}{d^3}\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {1}{3}}\right )^{n}\right )}{x^{2}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=\frac {2 \, b e^{3} n x \log \left (x^{\frac {1}{3}}\right ) + 2 \, b d e^{2} n x^{\frac {2}{3}} - b d^{2} e n x^{\frac {1}{3}} - 2 \, b d^{3} \log \left (c\right ) - 2 \, a d^{3} - 2 \, {\left (b e^{3} n x + b d^{3} n\right )} \log \left (e x^{\frac {1}{3}} + d\right )}{2 \, d^{3} x} \]

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate((a+b*ln(c*(d+e*x**(1/3))**n))/x**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {1}{6} \, b e n {\left (\frac {6 \, e^{2} \log \left (e x^{\frac {1}{3}} + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac {3 \, {\left (2 \, e x^{\frac {1}{3}} - d\right )}}{d^{2} x^{\frac {2}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right )}{x} - \frac {a}{x} \]

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (75) = 150\).

Time = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.38 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {\frac {2 \, b e^{4} n \log \left (e x^{\frac {1}{3}} + d\right )}{{\left (e x^{\frac {1}{3}} + d\right )}^{3} - 3 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d + 3 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{2} - d^{3}} + \frac {2 \, b e^{4} n \log \left (e x^{\frac {1}{3}} + d\right )}{d^{3}} - \frac {2 \, b e^{4} n \log \left (e x^{\frac {1}{3}}\right )}{d^{3}} - \frac {2 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} b e^{4} n - 5 \, {\left (e x^{\frac {1}{3}} + d\right )} b d e^{4} n + 3 \, b d^{2} e^{4} n - 2 \, b d^{2} e^{4} \log \left (c\right ) - 2 \, a d^{2} e^{4}}{{\left (e x^{\frac {1}{3}} + d\right )}^{3} d^{2} - 3 \, {\left (e x^{\frac {1}{3}} + d\right )}^{2} d^{3} + 3 \, {\left (e x^{\frac {1}{3}} + d\right )} d^{4} - d^{5}}}{2 \, e} \]

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.70 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x^2} \, dx=-\frac {\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n\,x^{1/3}}{d^2}}{x^{2/3}}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}{x}-\frac {2\,b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{1/3}}{d}+1\right )}{d^3} \]

[In]

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

[Out]

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