3.6.80 \(\int \frac {e+2 e \log (x)+2 \log ^2(x)}{x^3 \log ^2(x)} \, dx\) [580]

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

[Out]

-(exp(1/4)^4+ln(x))/x^2/ln(x)

________________________________________________________________________________________

Rubi [A]
time = 0.11, 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 = {6874, 2343, 2346, 2209} \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 2209

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

Rule 2343

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 2346

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 6874

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) \text {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) \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, 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])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.04, size = 42, normalized size = 3.23

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

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

________________________________________________________________________________________

Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.29, size = 25, normalized size = 1.92 \begin {gather*} 2 \, {\rm Ei}\left (-2 \, \log \left (x\right )\right ) e - 2 \, e \Gamma \left (-1, 2 \, \log \left (x\right )\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

________________________________________________________________________________________

Fricas [A]
time = 0.33, size = 14, normalized size = 1.08 \begin {gather*} -\frac {e + \log \left (x\right )}{x^{2} \log \left (x\right )} \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))

________________________________________________________________________________________

Sympy [A]
time = 0.03, size = 15, normalized size = 1.15 \begin {gather*} - \frac {1}{x^{2}} - \frac {e}{x^{2} \log {\left (x \right )}} \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))

________________________________________________________________________________________

Giac [A]
time = 0.43, size = 14, normalized size = 1.08 \begin {gather*} -\frac {e + \log \left (x\right )}{x^{2} \log \left (x\right )} \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))

________________________________________________________________________________________

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

________________________________________________________________________________________