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

Optimal. Leaf size=19 \[ 3-2 \left (2+e^{e^x}-\frac {1}{2 \log (x)}\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 12, normalized size of antiderivative = 0.63, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6742, 2282, 2194, 2302, 30} \begin {gather*} \frac {1}{\log (x)}-2 e^{e^x} \end {gather*}

Antiderivative was successfully verified.

[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 2194

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 2282

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 2302

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

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Mathematica [A]  time = 0.01, size = 12, normalized size = 0.63 \begin {gather*} -2 e^{e^x}+\frac {1}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.68, size = 24, normalized size = 1.26 \begin {gather*} -\frac {{\left (2 \, e^{\left (x + e^{x}\right )} \log \relax (x) - e^{x}\right )} e^{\left (-x\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.14, size = 24, normalized size = 1.26 \begin {gather*} -\frac {{\left (2 \, e^{\left (x + e^{x}\right )} \log \relax (x) - e^{x}\right )} e^{\left (-x\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 11, normalized size = 0.58

method	result	size
default	\(\frac {1}{\ln \relax (x )}-2 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(11\)
risch	\(\frac {1}{\ln \relax (x )}-2 \,{\mathrm e}^{{\mathrm e}^{x}}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.36, size = 10, normalized size = 0.53 \begin {gather*} \frac {1}{\log \relax (x)} - 2 \, e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.48, size = 10, normalized size = 0.53 \begin {gather*} \frac {1}{\ln \relax (x)}-2\,{\mathrm {e}}^{{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.29, size = 10, normalized size = 0.53 \begin {gather*} - 2 e^{e^{x}} + \frac {1}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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