3.25.96 \(\int \frac {-2-2 x+x \log (x)}{(x+x^2) \log (x)} \, dx\)

Optimal. Leaf size=18 \[ e^2-\log \left (\frac {16 \log ^2(x)}{1+x}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 10, normalized size of antiderivative = 0.56, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1593, 6688, 2302, 29} \begin {gather*} \log (x+1)-2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + x] - 2*Log[Log[x]]

Rule 29

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 10, normalized size = 0.56 \begin {gather*} \log (1+x)-2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[1 + x] - 2*Log[Log[x]]

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fricas [A]  time = 0.95, size = 10, normalized size = 0.56 \begin {gather*} \log \left (x + 1\right ) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

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giac [A]  time = 0.23, size = 10, normalized size = 0.56 \begin {gather*} \log \left (x + 1\right ) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

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maple [A]  time = 0.35, size = 11, normalized size = 0.61

method	result	size
default	\(\ln \left (x +1\right )-2 \ln \left (\ln \relax (x )\right )\)	\(11\)
norman	\(\ln \left (x +1\right )-2 \ln \left (\ln \relax (x )\right )\)	\(11\)
risch	\(\ln \left (x +1\right )-2 \ln \left (\ln \relax (x )\right )\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 10, normalized size = 0.56 \begin {gather*} \log \left (x + 1\right ) - 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

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mupad [B]  time = 1.42, size = 10, normalized size = 0.56 \begin {gather*} \ln \left (x+1\right )-2\,\ln \left (\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

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sympy [A]  time = 0.10, size = 10, normalized size = 0.56 \begin {gather*} \log {\left (x + 1 \right )} - 2 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 1) - 2*log(log(x))

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