\(\int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx\) [7487]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 19 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=3-2 \left (2+e^{e^x}-\frac {1}{2 \log (x)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6874, 2320, 2225, 2339, 30} \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=\frac {1}{\log (x)}-2 e^{e^x} \]

[In]

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

[Out]

-2*E^E^x + Log[x]^(-1)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=-2 e^{e^x}+\frac {1}{\log (x)} \]

[In]

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

[Out]

-2*E^E^x + Log[x]^(-1)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58

method result size
default \(\frac {1}{\ln \left (x \right )}-2 \,{\mathrm e}^{{\mathrm e}^{x}}\) \(11\)
risch \(\frac {1}{\ln \left (x \right )}-2 \,{\mathrm e}^{{\mathrm e}^{x}}\) \(11\)
parts \(\frac {1}{\ln \left (x \right )}-2 \,{\mathrm e}^{{\mathrm e}^{x}}\) \(11\)
parallelrisch \(-\frac {2 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{x}}-1}{\ln \left (x \right )}\) \(16\)

[In]

int((-2*x*exp(x)*ln(x)^2*exp(exp(x))-1)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (11) = 22\).

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=-\frac {{\left (2 \, e^{\left (x + e^{x}\right )} \log \left (x\right ) - e^{x}\right )} e^{\left (-x\right )}}{\log \left (x\right )} \]

[In]

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

[Out]

-(2*e^(x + e^x)*log(x) - e^x)*e^(-x)/log(x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=- 2 e^{e^{x}} + \frac {1}{\log {\left (x \right )}} \]

[In]

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

[Out]

-2*exp(exp(x)) + 1/log(x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=\frac {1}{\log \left (x\right )} - 2 \, e^{\left (e^{x}\right )} \]

[In]

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

[Out]

1/log(x) - 2*e^(e^x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (11) = 22\).

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=-\frac {{\left (2 \, e^{\left (x + e^{x}\right )} \log \left (x\right ) - e^{x}\right )} e^{\left (-x\right )}}{\log \left (x\right )} \]

[In]

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

[Out]

-(2*e^(x + e^x)*log(x) - e^x)*e^(-x)/log(x)

Mupad [B] (verification not implemented)

Time = 12.95 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.53 \[ \int \frac {-1-2 e^{e^x+x} x \log ^2(x)}{x \log ^2(x)} \, dx=\frac {1}{\ln \left (x\right )}-2\,{\mathrm {e}}^{{\mathrm {e}}^x} \]

[In]

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

[Out]

1/log(x) - 2*exp(exp(x))