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

Optimal. Leaf size=22 \[ -\log \left (2 x \left (\log (x)-e^{-2-x} \log (x)\right )^2\right ) \]

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Rubi [A]  time = 1.61, antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 9, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.170, Rules used = {6741, 6688, 6742, 2282, 36, 31, 29, 2365, 43} \begin {gather*} 2 x-2 \log \left (1-e^{x+2}\right )-\log (x)-2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2365

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

Rule 6688

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

Rule 6741

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

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

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Mathematica [A]  time = 0.13, size = 25, normalized size = 1.14 \begin {gather*} 2 x-2 \log \left (1-e^{2+x}\right )-\log (x)-2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.93, size = 22, normalized size = 1.00 \begin {gather*} -\log \relax (x) - 2 \, \log \left (e^{\left (-x + \log \left (\log \relax (x)\right ) - 2\right )} - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.49, size = 24, normalized size = 1.09 \begin {gather*} 2 \, x - \log \relax (x) - 2 \, \log \left (-e^{\left (x + 2\right )} + 1\right ) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - log(x) - 2*log(-e^(x + 2) + 1) - 2*log(log(x))

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maple [A]  time = 0.08, size = 24, normalized size = 1.09

method	result	size
risch	\(-\ln \relax (x )-4-2 \ln \left (-\ln \relax (x )+\ln \relax (x ) {\mathrm e}^{-x -2}\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x-1)*ln(x)-2)*exp(ln(ln(x))-x-2)+ln(x)^2+2*ln(x))/(x*ln(x)*exp(ln(ln(x))-x-2)-x*ln(x)^2),x,method=_RE
TURNVERBOSE)

[Out]

-ln(x)-4-2*ln(-ln(x)+ln(x)*exp(-x-2))

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maxima [A]  time = 0.46, size = 25, normalized size = 1.14 \begin {gather*} 2 \, x - 2 \, \log \left ({\left (e^{\left (x + 2\right )} - 1\right )} e^{\left (-2\right )}\right ) - \log \relax (x) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - 2*log((e^(x + 2) - 1)*e^(-2)) - log(x) - 2*log(log(x))

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mupad [B]  time = 0.64, size = 21, normalized size = 0.95 \begin {gather*} -2\,\ln \left (\ln \relax (x)-{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-2}\,\ln \relax (x)\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*log(log(x) - exp(-x)*exp(-2)*log(x)) - log(x)

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sympy [A]  time = 0.28, size = 22, normalized size = 1.00 \begin {gather*} - \log {\relax (x )} - 2 \log {\left (e^{- x - 2} - 1 \right )} - 2 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) - 2*log(exp(-x - 2) - 1) - 2*log(log(x))

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