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

Optimal. Leaf size=17 \[ -13-e^x+2 x-\log (x)+\log (\log (x)) \]

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Rubi [A]  time = 0.30, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6742, 2194, 43, 2302, 29} \begin {gather*} 2 x-e^x-\log (x)+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^x + 2*x - Log[x] + Log[Log[x]]

Rule 29

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

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

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.94 \begin {gather*} -e^x+2 x-\log (x)+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^x + 2*x - Log[x] + Log[Log[x]]

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fricas [A]  time = 0.51, size = 15, normalized size = 0.88 \begin {gather*} 2 \, x - e^{x} - \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^x - log(x) + log(log(x))

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giac [A]  time = 0.19, size = 15, normalized size = 0.88 \begin {gather*} 2 \, x - e^{x} - \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^x - log(x) + log(log(x))

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maple [A]  time = 0.03, size = 16, normalized size = 0.94

method	result	size
default	\(\ln \left (\ln \relax (x )\right )+2 x -\ln \relax (x )-{\mathrm e}^{x}\)	\(16\)
norman	\(\ln \left (\ln \relax (x )\right )+2 x -\ln \relax (x )-{\mathrm e}^{x}\)	\(16\)
risch	\(\ln \left (\ln \relax (x )\right )+2 x -\ln \relax (x )-{\mathrm e}^{x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 15, normalized size = 0.88 \begin {gather*} 2 \, x - e^{x} - \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^x - log(x) + log(log(x))

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mupad [B]  time = 9.26, size = 15, normalized size = 0.88 \begin {gather*} 2\,x+\ln \left (\ln \relax (x)\right )-{\mathrm {e}}^x-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + log(log(x)) - exp(x) - log(x)

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sympy [A]  time = 0.26, size = 14, normalized size = 0.82 \begin {gather*} 2 x - e^{x} - \log {\relax (x )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - exp(x) - log(x) + log(log(x))

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