\(\int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx\) [1749]

Optimal result

Integrand size = 28, antiderivative size = 15 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{32} \left (-5 x+\frac {16}{\log (\log (x))}\right ) \]

[Out]

-5/32*x+1/2/ln(ln(x))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6820, 2339, 30} \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{2 \log (\log (x))}-\frac {5 x}{32} \]

[In]

Int[(-16 - 5*x*Log[x]*Log[Log[x]]^2)/(32*x*Log[x]*Log[Log[x]]^2),x]

[Out]

(-5*x)/32 + 1/(2*Log[Log[x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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}& = \frac {1}{32} \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{x \log (x) \log ^2(\log (x))} \, dx \\ & = \frac {1}{32} \int \left (-5-\frac {16}{x \log (x) \log ^2(\log (x))}\right ) \, dx \\ & = -\frac {5 x}{32}-\frac {1}{2} \int \frac {1}{x \log (x) \log ^2(\log (x))} \, dx \\ & = -\frac {5 x}{32}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\log (x)\right ) \\ & = -\frac {5 x}{32}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (\log (x))\right ) \\ & = -\frac {5 x}{32}+\frac {1}{2 \log (\log (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5 x}{32}+\frac {1}{2 \log (\log (x))} \]

[In]

Integrate[(-16 - 5*x*Log[x]*Log[Log[x]]^2)/(32*x*Log[x]*Log[Log[x]]^2),x]

[Out]

(-5*x)/32 + 1/(2*Log[Log[x]])

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \left (x \right )\right )}\)	\(12\)
risch	\(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \left (x \right )\right )}\)	\(12\)
parts	\(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \left (x \right )\right )}\)	\(12\)
norman	\(\frac {\frac {1}{2}-\frac {5 x \ln \left (\ln \left (x \right )\right )}{32}}{\ln \left (\ln \left (x \right )\right )}\)	\(15\)
parallelrisch	\(-\frac {5 x \ln \left (\ln \left (x \right )\right )-16}{32 \ln \left (\ln \left (x \right )\right )}\)	\(16\)

[In]

int(1/32*(-5*x*ln(x)*ln(ln(x))^2-16)/x/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

-5/32*x+1/2/ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5 \, x \log \left (\log \left (x\right )\right ) - 16}{32 \, \log \left (\log \left (x\right )\right )} \]

[In]

integrate(1/32*(-5*x*log(x)*log(log(x))^2-16)/x/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

-1/32*(5*x*log(log(x)) - 16)/log(log(x))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=- \frac {5 x}{32} + \frac {1}{2 \log {\left (\log {\left (x \right )} \right )}} \]

[In]

integrate(1/32*(-5*x*ln(x)*ln(ln(x))**2-16)/x/ln(x)/ln(ln(x))**2,x)

[Out]

-5*x/32 + 1/(2*log(log(x)))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5}{32} \, x + \frac {1}{2 \, \log \left (\log \left (x\right )\right )} \]

[In]

integrate(1/32*(-5*x*log(x)*log(log(x))^2-16)/x/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

-5/32*x + 1/2/log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=-\frac {5}{32} \, x + \frac {1}{2 \, \log \left (\log \left (x\right )\right )} \]

[In]

integrate(1/32*(-5*x*log(x)*log(log(x))^2-16)/x/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

-5/32*x + 1/2/log(log(x))

Mupad [B] (verification not implemented)

Time = 9.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx=\frac {1}{2\,\ln \left (\ln \left (x\right )\right )}-\frac {5\,x}{32} \]

[In]

int(-((5*x*log(log(x))^2*log(x))/32 + 1/2)/(x*log(log(x))^2*log(x)),x)

[Out]

1/(2*log(log(x))) - (5*x)/32