\(\int \frac {1+(-3 x-2 x^2) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx\) [7812]

Optimal result

Integrand size = 30, antiderivative size = 21 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=2+e^3-(-1-x)^2-x+\log (\log (\log (x))) \]

[Out]

2+ln(ln(ln(x)))+exp(3)-x-(-1-x)^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6820, 2339, 29} \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-x^2-3 x+\log (\log (\log (x))) \]

[In]

Int[(1 + (-3*x - 2*x^2)*Log[x]*Log[Log[x]])/(x*Log[x]*Log[Log[x]]),x]

[Out]

-3*x - x^2 + Log[Log[Log[x]]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-3-2 x+\frac {1}{x \log (x) \log (\log (x))}\right ) \, dx \\ & = -3 x-x^2+\int \frac {1}{x \log (x) \log (\log (x))} \, dx \\ & = -3 x-x^2+\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,\log (x)\right ) \\ & = -3 x-x^2+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (\log (x))\right ) \\ & = -3 x-x^2+\log (\log (\log (x))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-3 x-x^2+\log (\log (\log (x))) \]

[In]

Integrate[(1 + (-3*x - 2*x^2)*Log[x]*Log[Log[x]])/(x*Log[x]*Log[Log[x]]),x]

[Out]

-3*x - x^2 + Log[Log[Log[x]]]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
default	\(-3 x +\ln \left (\ln \left (\ln \left (x \right )\right )\right )-x^{2}\)	\(14\)
norman	\(-3 x +\ln \left (\ln \left (\ln \left (x \right )\right )\right )-x^{2}\)	\(14\)
risch	\(-3 x +\ln \left (\ln \left (\ln \left (x \right )\right )\right )-x^{2}\)	\(14\)
parallelrisch	\(-3 x +\ln \left (\ln \left (\ln \left (x \right )\right )\right )-x^{2}\)	\(14\)
parts	\(-3 x +\ln \left (\ln \left (\ln \left (x \right )\right )\right )-x^{2}\)	\(14\)

[In]

int(((-2*x^2-3*x)*ln(x)*ln(ln(x))+1)/x/ln(x)/ln(ln(x)),x,method=_RETURNVERBOSE)

[Out]

-3*x+ln(ln(ln(x)))-x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-x^{2} - 3 \, x + \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

integrate(((-2*x^2-3*x)*log(x)*log(log(x))+1)/x/log(x)/log(log(x)),x, algorithm="fricas")

[Out]

-x^2 - 3*x + log(log(log(x)))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=- x^{2} - 3 x + \log {\left (\log {\left (\log {\left (x \right )} \right )} \right )} \]

[In]

integrate(((-2*x**2-3*x)*ln(x)*ln(ln(x))+1)/x/ln(x)/ln(ln(x)),x)

[Out]

-x**2 - 3*x + log(log(log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-x^{2} - 3 \, x + \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

integrate(((-2*x^2-3*x)*log(x)*log(log(x))+1)/x/log(x)/log(log(x)),x, algorithm="maxima")

[Out]

-x^2 - 3*x + log(log(log(x)))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=-x^{2} - 3 \, x + \log \left (\log \left (\log \left (x\right )\right )\right ) \]

[In]

integrate(((-2*x^2-3*x)*log(x)*log(log(x))+1)/x/log(x)/log(log(x)),x, algorithm="giac")

[Out]

-x^2 - 3*x + log(log(log(x)))

Mupad [B] (verification not implemented)

Time = 14.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+\left (-3 x-2 x^2\right ) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx=\ln \left (\ln \left (\ln \left (x\right )\right )\right )-3\,x-x^2 \]

[In]

int(-(log(log(x))*log(x)*(3*x + 2*x^2) - 1)/(x*log(log(x))*log(x)),x)

[Out]

log(log(log(x))) - 3*x - x^2