3.24.61 \(\int \frac {-1+\log (x)}{(60-15 e^5+3 \log (5)) \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {x}{3 \left (-5 \left (-4+e^5\right )+\log (5)\right ) \log (x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2360, 2297, 2298} \begin {gather*} \frac {x}{\left (60-15 e^5+\log (125)\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + Log[x])/((60 - 15*E^5 + 3*Log[5])*Log[x]^2),x]

[Out]

x/((60 - 15*E^5 + Log[125])*Log[x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(Log[(c_.)*(x_)^(n_.)]*(e_.) + (d_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*Log[c*x^n])^p*(d + e*Log[c*x^n])^q, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[p
] && IntegerQ[q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-1+\log (x)}{\log ^2(x)} \, dx}{60-15 e^5+\log (125)}\\ &=\frac {\int \left (-\frac {1}{\log ^2(x)}+\frac {1}{\log (x)}\right ) \, dx}{60-15 e^5+\log (125)}\\ &=-\frac {\int \frac {1}{\log ^2(x)} \, dx}{60-15 e^5+\log (125)}+\frac {\int \frac {1}{\log (x)} \, dx}{60-15 e^5+\log (125)}\\ &=\frac {x}{\left (60-15 e^5+\log (125)\right ) \log (x)}+\frac {\text {li}(x)}{60-15 e^5+\log (125)}-\frac {\int \frac {1}{\log (x)} \, dx}{60-15 e^5+\log (125)}\\ &=\frac {x}{\left (60-15 e^5+\log (125)\right ) \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.95 \begin {gather*} \frac {x}{3 \left (20-5 e^5+\log (5)\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Log[x])/((60 - 15*E^5 + 3*Log[5])*Log[x]^2),x]

[Out]

x/(3*(20 - 5*E^5 + Log[5])*Log[x])

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fricas [A]  time = 0.53, size = 19, normalized size = 0.90 \begin {gather*} -\frac {x}{3 \, {\left (5 \, e^{5} - \log \relax (5) - 20\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-1)/(3*log(5)-15*exp(5)+60)/log(x)^2,x, algorithm="fricas")

[Out]

-1/3*x/((5*e^5 - log(5) - 20)*log(x))

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giac [A]  time = 0.32, size = 19, normalized size = 0.90 \begin {gather*} -\frac {x}{3 \, {\left (5 \, e^{5} - \log \relax (5) - 20\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-1)/(3*log(5)-15*exp(5)+60)/log(x)^2,x, algorithm="giac")

[Out]

-1/3*x/((5*e^5 - log(5) - 20)*log(x))

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maple [A]  time = 0.07, size = 18, normalized size = 0.86

method	result	size
default	\(\frac {x}{3 \left (\ln \relax (5)-5 \,{\mathrm e}^{5}+20\right ) \ln \relax (x )}\)	\(18\)
risch	\(\frac {x}{\left (3 \ln \relax (5)-15 \,{\mathrm e}^{5}+60\right ) \ln \relax (x )}\)	\(19\)
norman	\(-\frac {x}{3 \left (-\ln \relax (5)+5 \,{\mathrm e}^{5}-20\right ) \ln \relax (x )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)-1)/(3*ln(5)-15*exp(5)+60)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*x/(ln(5)-5*exp(5)+20)/ln(x)

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maxima [C]  time = 0.67, size = 26, normalized size = 1.24 \begin {gather*} -\frac {{\rm Ei}\left (\log \relax (x)\right ) - \Gamma \left (-1, -\log \relax (x)\right )}{3 \, {\left (5 \, e^{5} - \log \relax (5) - 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)-1)/(3*log(5)-15*exp(5)+60)/log(x)^2,x, algorithm="maxima")

[Out]

-1/3*(Ei(log(x)) - gamma(-1, -log(x)))/(5*e^5 - log(5) - 20)

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mupad [B]  time = 1.36, size = 17, normalized size = 0.81 \begin {gather*} \frac {x}{3\,\ln \relax (x)\,\left (\ln \relax (5)-5\,{\mathrm {e}}^5+20\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) - 1)/(log(x)^2*(3*log(5) - 15*exp(5) + 60)),x)

[Out]

x/(3*log(x)*(log(5) - 5*exp(5) + 20))

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sympy [A]  time = 0.12, size = 17, normalized size = 0.81 \begin {gather*} - \frac {x}{\left (-60 - 3 \log {\relax (5 )} + 15 e^{5}\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)-1)/(3*ln(5)-15*exp(5)+60)/ln(x)**2,x)

[Out]

-x/((-60 - 3*log(5) + 15*exp(5))*log(x))

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