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

Optimal. Leaf size=82 \[ -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x}-\frac {b d^3 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^3}+\frac {b d^2 n}{e^2 \sqrt [3]{x}}-\frac {b d n}{2 e x^{2/3}}+\frac {b n}{3 x} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2454, 2395, 43} \[ -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x}+\frac {b d^2 n}{e^2 \sqrt [3]{x}}-\frac {b d^3 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^3}-\frac {b d n}{2 e x^{2/3}}+\frac {b n}{3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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 2395

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 2454

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

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Mathematica [A]  time = 0.04, size = 85, normalized size = 1.04 \[ -\frac {a}{x}-\frac {b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x}-\frac {b d^3 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^3}+\frac {b d^2 n}{e^2 \sqrt [3]{x}}-\frac {b d n}{2 e x^{2/3}}+\frac {b n}{3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 107, normalized size = 1.30 \[ \frac {6 \, b d^{2} e n x^{\frac {2}{3}} - 3 \, b d e^{2} n x^{\frac {1}{3}} + 2 \, b e^{3} n - 6 \, a e^{3} - 2 \, {\left (b e^{3} n - 3 \, a e^{3}\right )} x + 6 \, {\left (b e^{3} x - b e^{3}\right )} \log \relax (c) - 6 \, {\left (b d^{3} n x + b e^{3} n\right )} \log \left (\frac {d x + e x^{\frac {2}{3}}}{x}\right )}{6 \, e^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 95, normalized size = 1.16 \[ -\frac {1}{6} \, {\left ({\left (6 \, d^{3} e^{\left (-4\right )} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right ) - 2 \, d^{3} e^{\left (-4\right )} \log \left ({\left | x \right |}\right ) - \frac {{\left (6 \, d^{2} x^{\frac {2}{3}} e - 3 \, d x^{\frac {1}{3}} e^{2} + 2 \, e^{3}\right )} e^{\left (-4\right )}}{x}\right )} e + \frac {6 \, \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )}{x}\right )} b n - \frac {b \log \relax (c)}{x} - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )+a}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 86, normalized size = 1.05 \[ -\frac {1}{6} \, b e n {\left (\frac {6 \, d^{3} \log \left (d x^{\frac {1}{3}} + e\right )}{e^{4}} - \frac {2 \, d^{3} \log \relax (x)}{e^{4}} - \frac {6 \, d^{2} x^{\frac {2}{3}} - 3 \, d e x^{\frac {1}{3}} + 2 \, e^{2}}{e^{3} x}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{x} - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.43, size = 73, normalized size = 0.89 \[ \frac {b\,n}{3\,x}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{x}-\frac {b\,d\,n}{2\,e\,x^{2/3}}-\frac {b\,d^3\,n\,\ln \left (d+\frac {e}{x^{1/3}}\right )}{e^3}+\frac {b\,d^2\,n}{e^2\,x^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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