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

Optimal. Leaf size=94 \[ -\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}-\frac {b e^3 n \log \left (d+e x^{2/3}\right )}{2 d^3}+\frac {b e^3 n \log (x)}{3 d^3}+\frac {b e^2 n}{2 d^2 x^{2/3}}-\frac {b e n}{4 d x^{4/3}} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 94, 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, 44} \[ -\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}+\frac {b e^2 n}{2 d^2 x^{2/3}}-\frac {b e^3 n \log \left (d+e x^{2/3}\right )}{2 d^3}+\frac {b e^3 n \log (x)}{3 d^3}-\frac {b e n}{4 d x^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*n)/(4*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 44

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] && L
tQ[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+e x^{2/3}\right )^n\right )}{x^3} \, dx &=\frac {3}{2} \operatorname {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) \operatorname {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) \operatorname {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*}

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Mathematica [A]  time = 0.03, size = 91, normalized size = 0.97 \[ -\frac {a}{2 x^2}-\frac {b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{2 x^2}+\frac {1}{2} b e n \left (-\frac {e^2 \log \left (d+e x^{2/3}\right )}{d^3}+\frac {2 e^2 \log (x)}{3 d^3}+\frac {e}{d^2 x^{2/3}}-\frac {1}{2 d x^{4/3}}\right ) \]

Antiderivative was successfully verified.

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

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fricas [A]  time = 0.46, size = 85, normalized size = 0.90 \[ \frac {4 \, b e^{3} n x^{2} \log \left (x^{\frac {1}{3}}\right ) + 2 \, b d e^{2} n x^{\frac {4}{3}} - b d^{2} e n x^{\frac {2}{3}} - 2 \, b d^{3} \log \relax (c) - 2 \, a d^{3} - 2 \, {\left (b e^{3} n x^{2} + b d^{3} n\right )} \log \left (e x^{\frac {2}{3}} + d\right )}{4 \, d^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 95, normalized size = 1.01 \[ \frac {1}{4} \, {\left ({\left (\frac {2 \, \log \left (x^{\frac {2}{3}} e\right )}{d^{3}} - \frac {2 \, \log \left ({\left | x^{\frac {2}{3}} e + d \right |}\right )}{d^{3}} + \frac {{\left (2 \, {\left (x^{\frac {2}{3}} e + d\right )} d - 3 \, d^{2}\right )} e^{\left (-2\right )}}{d^{3} x^{\frac {4}{3}}}\right )} e^{4} - \frac {2 \, e \log \left (x^{\frac {2}{3}} e + d\right )}{x^{2}}\right )} b n e^{\left (-1\right )} - \frac {b \log \relax (c)}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (e \,x^{\frac {2}{3}}+d \right )^{n}\right )+a}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 77, normalized size = 0.82 \[ -\frac {1}{4} \, b e n {\left (\frac {2 \, e^{2} \log \left (e x^{\frac {2}{3}} + d\right )}{d^{3}} - \frac {2 \, e^{2} \log \left (x^{\frac {2}{3}}\right )}{d^{3}} - \frac {2 \, e x^{\frac {2}{3}} - d}{d^{2} x^{\frac {4}{3}}}\right )} - \frac {b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.61, size = 74, normalized size = 0.79 \[ -\frac {\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n\,x^{2/3}}{d^2}}{2\,x^{4/3}}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{2\,x^2}-\frac {b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e\,x^{2/3}}{d}+1\right )}{d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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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**(2/3))**n))/x**3,x)

[Out]

Timed out

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