3.6.80 \(\int \frac {e+2 e \log (x)+2 \log ^2(x)}{x^3 \log ^2(x)} \, dx\)

Optimal. Leaf size=13 \[ -\frac {e+\log (x)}{x^2 \log (x)} \]

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Rubi [A]  time = 0.18, antiderivative size = 16, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6742, 2306, 2309, 2178} \begin {gather*} -\frac {1}{x^2}-\frac {e}{x^2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-2) - E/(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]

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

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Mathematica [A]  time = 0.03, size = 16, normalized size = 1.23 \begin {gather*} -\frac {1}{x^2}-\frac {e}{x^2 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-x^(-2) - E/(x^2*Log[x])

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fricas [A]  time = 0.71, size = 14, normalized size = 1.08 \begin {gather*} -\frac {e + \log \relax (x)}{x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+2*exp(1/4)^4*log(x)+exp(1/4)^4)/x^3/log(x)^2,x, algorithm="fricas")

[Out]

-(e + log(x))/(x^2*log(x))

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giac [A]  time = 0.78, size = 14, normalized size = 1.08 \begin {gather*} -\frac {e + \log \relax (x)}{x^{2} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+2*exp(1/4)^4*log(x)+exp(1/4)^4)/x^3/log(x)^2,x, algorithm="giac")

[Out]

-(e + log(x))/(x^2*log(x))

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maple [A]  time = 0.05, size = 18, normalized size = 1.38

method	result	size
risch	\(-\frac {1}{x^{2}}-\frac {{\mathrm e}}{x^{2} \ln \relax (x )}\)	\(18\)
norman	\(\frac {-{\mathrm e}-\ln \relax (x )}{x^{2} \ln \relax (x )}\)	\(20\)
default	\(-\frac {1}{x^{2}}-2 \,{\mathrm e} \expIntegralEi \left (1, 2 \ln \relax (x )\right )+{\mathrm e} \left (-\frac {1}{x^{2} \ln \relax (x )}+2 \expIntegralEi \left (1, 2 \ln \relax (x )\right )\right )\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x)^2+2*exp(1/4)^4*ln(x)+exp(1/4)^4)/x^3/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/x^2-1/x^2*exp(1)/ln(x)

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maxima [C]  time = 0.58, size = 25, normalized size = 1.92 \begin {gather*} 2 \, {\rm Ei}\left (-2 \, \log \relax (x)\right ) e - 2 \, e \Gamma \left (-1, 2 \, \log \relax (x)\right ) - \frac {1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x)^2+2*exp(1/4)^4*log(x)+exp(1/4)^4)/x^3/log(x)^2,x, algorithm="maxima")

[Out]

2*Ei(-2*log(x))*e - 2*e*gamma(-1, 2*log(x)) - 1/x^2

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mupad [B]  time = 0.50, size = 14, normalized size = 1.08 \begin {gather*} -\frac {\mathrm {e}+\ln \relax (x)}{x^2\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1) + 2*log(x)^2 + 2*exp(1)*log(x))/(x^3*log(x)^2),x)

[Out]

-(exp(1) + log(x))/(x^2*log(x))

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sympy [A]  time = 0.09, size = 15, normalized size = 1.15 \begin {gather*} - \frac {1}{x^{2}} - \frac {e}{x^{2} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x)**2+2*exp(1/4)**4*ln(x)+exp(1/4)**4)/x**3/ln(x)**2,x)

[Out]

-1/x**2 - E/(x**2*log(x))

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