3.24.55 \(\int \frac {-2 x-\log (\frac {e^x}{6})}{x^2 \log ^3(\frac {e^x}{6})} \, dx\)

Optimal. Leaf size=16 \[ 6+\frac {1}{x \log ^2\left (\frac {e^x}{6}\right )} \]

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Rubi [B]  time = 0.28, antiderivative size = 55, normalized size of antiderivative = 3.44, number of steps used = 14, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6742, 2163, 2160, 2157, 29, 2171} \begin {gather*} \frac {1}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \log ^2\left (\frac {e^x}{6}\right )}-\frac {1}{x \left (x-\log \left (\frac {e^x}{6}\right )\right ) \log \left (\frac {e^x}{6}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x - Log[E^x/6])/(x^2*Log[E^x/6]^3),x]

[Out]

1/((x - Log[E^x/6])*Log[E^x/6]^2) - 1/(x*(x - Log[E^x/6])*Log[E^x/6])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2157

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2160

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Dist[b/(b*u - a*v), Int[1
/v, x], x] - Dist[a/(b*u - a*v), Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 2163

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[(a*(n + 1))/((n + 1)*(b*u - a*v)), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rule 2171

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, -Simp[(u^(m + 1)*v^
(n + 1))/((m + 1)*(b*u - a*v)), x] + Dist[(b*(m + n + 2))/((m + 1)*(b*u - a*v)), Int[u^(m + 1)*v^n, x], x] /;
NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2}{x \log ^3\left (\frac {e^x}{6}\right )}-\frac {1}{x^2 \log ^2\left (\frac {e^x}{6}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x \log ^3\left (\frac {e^x}{6}\right )} \, dx\right )-\int \frac {1}{x^2 \log ^2\left (\frac {e^x}{6}\right )} \, dx\\ &=\frac {1}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \log ^2\left (\frac {e^x}{6}\right )}-\frac {1}{x \left (x-\log \left (\frac {e^x}{6}\right )\right ) \log \left (\frac {e^x}{6}\right )}+\frac {2 \int \frac {1}{x \log ^2\left (\frac {e^x}{6}\right )} \, dx}{x-\log \left (\frac {e^x}{6}\right )}+\frac {2 \int \frac {1}{x \log ^2\left (\frac {e^x}{6}\right )} \, dx}{-x+\log \left (\frac {e^x}{6}\right )}\\ &=\frac {1}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \log ^2\left (\frac {e^x}{6}\right )}-\frac {1}{x \left (x-\log \left (\frac {e^x}{6}\right )\right ) \log \left (\frac {e^x}{6}\right )}+\frac {2 \int \frac {1}{x \log \left (\frac {e^x}{6}\right )} \, dx}{\left (-x+\log \left (\frac {e^x}{6}\right )\right )^2}+\frac {2 \int \frac {1}{x \log \left (\frac {e^x}{6}\right )} \, dx}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \left (-x+\log \left (\frac {e^x}{6}\right )\right )}\\ &=\frac {1}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \log ^2\left (\frac {e^x}{6}\right )}-\frac {1}{x \left (x-\log \left (\frac {e^x}{6}\right )\right ) \log \left (\frac {e^x}{6}\right )}-\frac {2 \int \frac {1}{x} \, dx}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \left (-x+\log \left (\frac {e^x}{6}\right )\right )^2}+\frac {2 \int \frac {1}{\log \left (\frac {e^x}{6}\right )} \, dx}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \left (-x+\log \left (\frac {e^x}{6}\right )\right )^2}-\frac {2 \int \frac {1}{x} \, dx}{\left (x-\log \left (\frac {e^x}{6}\right )\right )^2 \left (-x+\log \left (\frac {e^x}{6}\right )\right )}+\frac {2 \int \frac {1}{\log \left (\frac {e^x}{6}\right )} \, dx}{\left (x-\log \left (\frac {e^x}{6}\right )\right )^2 \left (-x+\log \left (\frac {e^x}{6}\right )\right )}\\ &=\frac {1}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \log ^2\left (\frac {e^x}{6}\right )}-\frac {1}{x \left (x-\log \left (\frac {e^x}{6}\right )\right ) \log \left (\frac {e^x}{6}\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {e^x}{6}\right )\right )}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \left (-x+\log \left (\frac {e^x}{6}\right )\right )^2}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {e^x}{6}\right )\right )}{\left (x-\log \left (\frac {e^x}{6}\right )\right )^2 \left (-x+\log \left (\frac {e^x}{6}\right )\right )}\\ &=\frac {1}{\left (x-\log \left (\frac {e^x}{6}\right )\right ) \log ^2\left (\frac {e^x}{6}\right )}-\frac {1}{x \left (x-\log \left (\frac {e^x}{6}\right )\right ) \log \left (\frac {e^x}{6}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.12, size = 82, normalized size = 5.12 \begin {gather*} \frac {2 x^3+x \log \left (\frac {e^x}{6}\right ) \left (-6 x+5 \log (6)-5 \log \left (e^x\right )\right )+\log ^2\left (\frac {e^x}{6}\right ) \left (11 x+\log (36)-2 \log \left (e^x\right )\right )}{2 x \log ^2\left (\frac {e^x}{6}\right ) \left (x+\log (6)-\log \left (e^x\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x - Log[E^x/6])/(x^2*Log[E^x/6]^3),x]

[Out]

(2*x^3 + x*Log[E^x/6]*(-6*x + 5*Log[6] - 5*Log[E^x]) + Log[E^x/6]^2*(11*x + Log[36] - 2*Log[E^x]))/(2*x*Log[E^
x/6]^2*(x + Log[6] - Log[E^x])^3)

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fricas [A]  time = 0.63, size = 19, normalized size = 1.19 \begin {gather*} \frac {1}{x^{3} - 2 \, x^{2} \log \relax (6) + x \log \relax (6)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(1/6*exp(x))-2*x)/x^2/log(1/6*exp(x))^3,x, algorithm="fricas")

[Out]

1/(x^3 - 2*x^2*log(6) + x*log(6)^2)

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giac [B]  time = 0.29, size = 29, normalized size = 1.81 \begin {gather*} -\frac {x - 2 \, \log \relax (6)}{{\left (x - \log \relax (6)\right )}^{2} \log \relax (6)^{2}} + \frac {1}{x \log \relax (6)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(1/6*exp(x))-2*x)/x^2/log(1/6*exp(x))^3,x, algorithm="giac")

[Out]

-(x - 2*log(6))/((x - log(6))^2*log(6)^2) + 1/(x*log(6)^2)

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maple [C]  time = 0.09, size = 25, normalized size = 1.56

method	result	size
risch	\(-\frac {4}{x \left (2 i \ln \relax (2)+2 i \ln \relax (3)-2 i \ln \left ({\mathrm e}^{x}\right )\right )^{2}}\)	\(25\)
default	\(-\frac {1}{\left (\ln \left (\frac {{\mathrm e}^{x}}{6}\right )-x \right )^{2} \ln \left (\frac {{\mathrm e}^{x}}{6}\right )}+\frac {1}{\left (\ln \left (\frac {{\mathrm e}^{x}}{6}\right )-x \right )^{2} x}-\frac {1}{\left (\ln \left (\frac {{\mathrm e}^{x}}{6}\right )-x \right ) \ln \left (\frac {{\mathrm e}^{x}}{6}\right )^{2}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(1/6*exp(x))-2*x)/x^2/ln(1/6*exp(x))^3,x,method=_RETURNVERBOSE)

[Out]

-4/x/(2*I*ln(2)+2*I*ln(3)-2*I*ln(exp(x)))^2

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maxima [B]  time = 0.82, size = 162, normalized size = 10.12 \begin {gather*} \frac {2 \, x - \log \relax (3) - \log \relax (2)}{{\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x^{2} - {\left (\log \relax (3)^{3} + 3 \, \log \relax (3)^{2} \log \relax (2) + 3 \, \log \relax (3) \log \relax (2)^{2} + \log \relax (2)^{3}\right )} x} - \frac {2 \, x - 3 \, \log \relax (3) - 3 \, \log \relax (2)}{\log \relax (3)^{4} + 4 \, \log \relax (3)^{3} \log \relax (2) + 6 \, \log \relax (3)^{2} \log \relax (2)^{2} + 4 \, \log \relax (3) \log \relax (2)^{3} + \log \relax (2)^{4} + {\left (\log \relax (3)^{2} + 2 \, \log \relax (3) \log \relax (2) + \log \relax (2)^{2}\right )} x^{2} - 2 \, {\left (\log \relax (3)^{3} + 3 \, \log \relax (3)^{2} \log \relax (2) + 3 \, \log \relax (3) \log \relax (2)^{2} + \log \relax (2)^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(1/6*exp(x))-2*x)/x^2/log(1/6*exp(x))^3,x, algorithm="maxima")

[Out]

(2*x - log(3) - log(2))/((log(3)^2 + 2*log(3)*log(2) + log(2)^2)*x^2 - (log(3)^3 + 3*log(3)^2*log(2) + 3*log(3
)*log(2)^2 + log(2)^3)*x) - (2*x - 3*log(3) - 3*log(2))/(log(3)^4 + 4*log(3)^3*log(2) + 6*log(3)^2*log(2)^2 +
4*log(3)*log(2)^3 + log(2)^4 + (log(3)^2 + 2*log(3)*log(2) + log(2)^2)*x^2 - 2*(log(3)^3 + 3*log(3)^2*log(2) +
3*log(3)*log(2)^2 + log(2)^3)*x)

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mupad [B]  time = 1.79, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{x\,{\left (x-\ln \relax (6)\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(exp(x)/6))/(x^2*log(exp(x)/6)^3),x)

[Out]

1/(x*(x - log(6))^2)

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sympy [A]  time = 0.22, size = 19, normalized size = 1.19 \begin {gather*} \frac {1}{x^{3} - 2 x^{2} \log {\relax (6 )} + x \log {\relax (6 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(1/6*exp(x))-2*x)/x**2/ln(1/6*exp(x))**3,x)

[Out]

1/(x**3 - 2*x**2*log(6) + x*log(6)**2)

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