\(\int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx\) [2993]

Optimal result

Integrand size = 36, antiderivative size = 22 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log ^2(\log (5)) \log ^2(4-x+5 \log (\log (4)) \log (\log (25))) \]

[Out]

ln(5*ln(2*ln(2))*ln(2*ln(5))-x+4)^2*ln(ln(5))^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2437, 2338} \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log ^2(\log (5)) \log ^2(-x+4+5 \log (\log (4)) \log (\log (25))) \]

[In]

Int[(-2*Log[Log[5]]^2*Log[4 - x + 5*Log[Log[4]]*Log[Log[25]]])/(4 - x + 5*Log[Log[4]]*Log[Log[25]]),x]

[Out]

Log[Log[5]]^2*Log[4 - x + 5*Log[Log[4]]*Log[Log[25]]]^2

Rule 12

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

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (2 \log ^2(\log (5))\right ) \int \frac {\log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx\right ) \\ & = \left (2 \log ^2(\log (5))\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,4-x+5 \log (\log (4)) \log (\log (25))\right ) \\ & = \log ^2(\log (5)) \log ^2(4-x+5 \log (\log (4)) \log (\log (25))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log ^2(\log (5)) \log ^2(4-x+5 \log (\log (4)) \log (\log (25))) \]

[In]

Integrate[(-2*Log[Log[5]]^2*Log[4 - x + 5*Log[Log[4]]*Log[Log[25]]])/(4 - x + 5*Log[Log[4]]*Log[Log[25]]),x]

[Out]

Log[Log[5]]^2*Log[4 - x + 5*Log[Log[4]]*Log[Log[25]]]^2

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\ln \left (5 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2 \ln \left (5\right )\right )-x +4\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}\)	\(27\)
default	\(\ln \left (5 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2 \ln \left (5\right )\right )-x +4\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}\)	\(27\)
norman	\(\ln \left (5 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2 \ln \left (5\right )\right )-x +4\right )^{2} \ln \left (\ln \left (5\right )\right )^{2}\)	\(27\)
risch	\(\ln \left (\ln \left (5\right )\right )^{2} \ln \left (5 \left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) \left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )\right )-x +4\right )^{2}\)	\(29\)

[In]

int(-2*ln(ln(5))^2*ln(5*ln(2*ln(2))*ln(2*ln(5))-x+4)/(5*ln(2*ln(2))*ln(2*ln(5))-x+4),x,method=_RETURNVERBOSE)

[Out]

ln(5*ln(2*ln(2))*ln(2*ln(5))-x+4)^2*ln(ln(5))^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.50 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log \left (5 \, \log \left (2\right ) \log \left (2 \, \log \left (2\right )\right ) + 5 \, \log \left (2 \, \log \left (2\right )\right ) \log \left (\log \left (5\right )\right ) - x + 4\right )^{2} \log \left (\log \left (5\right )\right )^{2} \]

[In]

integrate(-2*log(log(5))^2*log(5*log(2*log(2))*log(2*log(5))-x+4)/(5*log(2*log(2))*log(2*log(5))-x+4),x, algor
ithm="fricas")

[Out]

log(5*log(2)*log(2*log(2)) + 5*log(2*log(2))*log(log(5)) - x + 4)^2*log(log(5))^2

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log {\left (- x + 5 \log {\left (2 \log {\left (2 \right )} \right )} \log {\left (2 \log {\left (5 \right )} \right )} + 4 \right )}^{2} \log {\left (\log {\left (5 \right )} \right )}^{2} \]

[In]

integrate(-2*ln(ln(5))**2*ln(5*ln(2*ln(2))*ln(2*ln(5))-x+4)/(5*ln(2*ln(2))*ln(2*ln(5))-x+4),x)

[Out]

log(-x + 5*log(2*log(2))*log(2*log(5)) + 4)**2*log(log(5))**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log \left (5 \, \log \left (2 \, \log \left (5\right )\right ) \log \left (2 \, \log \left (2\right )\right ) - x + 4\right )^{2} \log \left (\log \left (5\right )\right )^{2} \]

[In]

integrate(-2*log(log(5))^2*log(5*log(2*log(2))*log(2*log(5))-x+4)/(5*log(2*log(2))*log(2*log(5))-x+4),x, algor
ithm="maxima")

[Out]

log(5*log(2*log(5))*log(2*log(2)) - x + 4)^2*log(log(5))^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx=\log \left (5 \, \log \left (2 \, \log \left (5\right )\right ) \log \left (2 \, \log \left (2\right )\right ) - x + 4\right )^{2} \log \left (\log \left (5\right )\right )^{2} \]

[In]

integrate(-2*log(log(5))^2*log(5*log(2*log(2))*log(2*log(5))-x+4)/(5*log(2*log(2))*log(2*log(5))-x+4),x, algor
ithm="giac")

[Out]

log(5*log(2*log(5))*log(2*log(2)) - x + 4)^2*log(log(5))^2

Mupad [B] (verification not implemented)

Time = 11.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int -\frac {2 \log ^2(\log (5)) \log (4-x+5 \log (\log (4)) \log (\log (25)))}{4-x+5 \log (\log (4)) \log (\log (25))} \, dx={\ln \left (\ln \left ({\ln \left (4\right )}^5\right )\,\ln \left (\ln \left (25\right )\right )-x+4\right )}^2\,{\ln \left (\ln \left (5\right )\right )}^2 \]

[In]

int(-(2*log(5*log(2*log(2))*log(2*log(5)) - x + 4)*log(log(5))^2)/(5*log(2*log(2))*log(2*log(5)) - x + 4),x)

[Out]

log(log(log(4)^5)*log(log(25)) - x + 4)^2*log(log(5))^2