3.615 \(\int \frac {1}{x^3 \log ^4(x)} \, dx\)

Optimal. Leaf size=43 \[ -\frac {4}{3} \text {Ei}(-2 \log (x))-\frac {1}{3 x^2 \log ^3(x)}+\frac {1}{3 x^2 \log ^2(x)}-\frac {2}{3 x^2 \log (x)} \]

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Rubi [A]  time = 0.06, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2306, 2309, 2178} \[ -\frac {4}{3} \text {ExpIntegralEi}(-2 \log (x))+\frac {1}{3 x^2 \log ^2(x)}-\frac {1}{3 x^2 \log ^3(x)}-\frac {2}{3 x^2 \log (x)} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Log[x]^4),x]

[Out]

(-4*ExpIntegralEi[-2*Log[x]])/3 - 1/(3*x^2*Log[x]^3) + 1/(3*x^2*Log[x]^2) - 2/(3*x^2*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^3 \log ^4(x)} \, dx &=-\frac {1}{3 x^2 \log ^3(x)}-\frac {2}{3} \int \frac {1}{x^3 \log ^3(x)} \, dx\\ &=-\frac {1}{3 x^2 \log ^3(x)}+\frac {1}{3 x^2 \log ^2(x)}+\frac {2}{3} \int \frac {1}{x^3 \log ^2(x)} \, dx\\ &=-\frac {1}{3 x^2 \log ^3(x)}+\frac {1}{3 x^2 \log ^2(x)}-\frac {2}{3 x^2 \log (x)}-\frac {4}{3} \int \frac {1}{x^3 \log (x)} \, dx\\ &=-\frac {1}{3 x^2 \log ^3(x)}+\frac {1}{3 x^2 \log ^2(x)}-\frac {2}{3 x^2 \log (x)}-\frac {4}{3} \operatorname {Subst}\left (\int \frac {e^{-2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {4}{3} \text {Ei}(-2 \log (x))-\frac {1}{3 x^2 \log ^3(x)}+\frac {1}{3 x^2 \log ^2(x)}-\frac {2}{3 x^2 \log (x)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 1.00 \[ -\frac {4}{3} \text {Ei}(-2 \log (x))-\frac {1}{3 x^2 \log ^3(x)}+\frac {1}{3 x^2 \log ^2(x)}-\frac {2}{3 x^2 \log (x)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Log[x]^4),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^3 \log ^4(x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^3*Log[x]^4),x]

[Out]

Could not integrate

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fricas [A]  time = 0.96, size = 34, normalized size = 0.79 \[ -\frac {4 \, x^{2} \log \relax (x)^{3} \operatorname {log\_integral}\left (\frac {1}{x^{2}}\right ) + 2 \, \log \relax (x)^{2} - \log \relax (x) + 1}{3 \, x^{2} \log \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/log(x)^4,x, algorithm="fricas")

[Out]

-1/3*(4*x^2*log(x)^3*log_integral(x^(-2)) + 2*log(x)^2 - log(x) + 1)/(x^2*log(x)^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \log \relax (x)^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/log(x)^4,x, algorithm="giac")

[Out]

integrate(1/(x^3*log(x)^4), x)

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maple [A]  time = 0.03, size = 31, normalized size = 0.72




method result size



risch \(-\frac {2 \ln \relax (x )^{2}-\ln \relax (x )+1}{3 x^{2} \ln \relax (x )^{3}}+\frac {4 \expIntegralEi \left (1, 2 \ln \relax (x )\right )}{3}\) \(31\)
default \(-\frac {1}{3 x^{2} \ln \relax (x )^{3}}+\frac {1}{3 x^{2} \ln \relax (x )^{2}}-\frac {2}{3 x^{2} \ln \relax (x )}+\frac {4 \expIntegralEi \left (1, 2 \ln \relax (x )\right )}{3}\) \(37\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/ln(x)^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*(2*ln(x)^2-ln(x)+1)/x^2/ln(x)^3+4/3*Ei(1,2*ln(x))

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maxima [A]  time = 0.54, size = 8, normalized size = 0.19 \[ -8 \, \Gamma \left (-3, 2 \, \log \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/log(x)^4,x, algorithm="maxima")

[Out]

-8*gamma(-3, 2*log(x))

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mupad [B]  time = 0.28, size = 29, normalized size = 0.67 \[ -\frac {4\,\mathrm {ei}\left (-2\,\ln \relax (x)\right )}{3}-\frac {\frac {2\,{\ln \relax (x)}^2}{3}-\frac {\ln \relax (x)}{3}+\frac {1}{3}}{x^2\,{\ln \relax (x)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*log(x)^4),x)

[Out]

- (4*ei(-2*log(x)))/3 - ((2*log(x)^2)/3 - log(x)/3 + 1/3)/(x^2*log(x)^3)

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sympy [A]  time = 0.72, size = 32, normalized size = 0.74 \[ - \frac {4 \operatorname {Ei}{\left (- 2 \log {\relax (x )} \right )}}{3} + \frac {- 2 \log {\relax (x )}^{2} + \log {\relax (x )} - 1}{3 x^{2} \log {\relax (x )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/ln(x)**4,x)

[Out]

-4*Ei(-2*log(x))/3 + (-2*log(x)**2 + log(x) - 1)/(3*x**2*log(x)**3)

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