3.63 \(\int \frac {1}{x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ -\text {Ei}(-\log (x))-\frac {1}{x \log (x)} \]

[Out]

-Ei(-ln(x))-1/x/ln(x)

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2306, 2309, 2178} \[ -\text {ExpIntegralEi}(-\log (x))-\frac {1}{x \log (x)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Log[x]^2),x]

[Out]

-ExpIntegralEi[-Log[x]] - 1/(x*Log[x])

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \log ^2(x)} \, dx &=-\frac {1}{x \log (x)}-\int \frac {1}{x^2 \log (x)} \, dx\\ &=-\frac {1}{x \log (x)}-\operatorname {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right )\\ &=-\text {Ei}(-\log (x))-\frac {1}{x \log (x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 1.00 \[ -\text {Ei}(-\log (x))-\frac {1}{x \log (x)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Log[x]^2),x]

[Out]

-ExpIntegralEi[-Log[x]] - 1/(x*Log[x])

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fricas [A]  time = 0.40, size = 19, normalized size = 1.12 \[ -\frac {x \log \relax (x) \operatorname {log\_integral}\left (\frac {1}{x}\right ) + 1}{x \log \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/log(x)^2,x, algorithm="fricas")

[Out]

-(x*log(x)*log_integral(1/x) + 1)/(x*log(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{2} \log \relax (x)^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/log(x)^2,x, algorithm="giac")

[Out]

integrate(1/(x^2*log(x)^2), x)

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maple [A]  time = 0.01, size = 15, normalized size = 0.88 \[ \Ei \left (1, \ln \relax (x )\right )-\frac {1}{x \ln \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/ln(x)^2,x)

[Out]

-1/x/ln(x)+Ei(1,ln(x))

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maxima [A]  time = 0.52, size = 6, normalized size = 0.35 \[ -\Gamma \left (-1, \log \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/log(x)^2,x, algorithm="maxima")

[Out]

-gamma(-1, log(x))

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mupad [B]  time = 0.03, size = 17, normalized size = 1.00 \[ -\mathrm {ei}\left (-\ln \relax (x)\right )-\frac {1}{x\,\ln \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*log(x)^2),x)

[Out]

- ei(-log(x)) - 1/(x*log(x))

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sympy [A]  time = 0.65, size = 14, normalized size = 0.82 \[ - \operatorname {Ei}{\left (- \log {\relax (x )} \right )} - \frac {1}{x \log {\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/ln(x)**2,x)

[Out]

-Ei(-log(x)) - 1/(x*log(x))

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