\(\int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx\) [5283]

Optimal result

Integrand size = 34, antiderivative size = 12 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log (\log (4 (-5+\log (5-20 x)))) \]

[Out]

ln(ln(4*ln(-20*x+5)-20))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6816} \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log (\log (4 \log (5-20 x)-20)) \]

[In]

Int[4/((5 - 20*x + (-1 + 4*x)*Log[5 - 20*x])*Log[-20 + 4*Log[5 - 20*x]]),x]

[Out]

Log[Log[-20 + 4*Log[5 - 20*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 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps \begin{align*} \text {integral}& = 4 \int \frac {1}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx \\ & = \log (\log (-20+4 \log (5-20 x))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log (\log (4 (-5+\log (5-20 x)))) \]

[In]

Integrate[4/((5 - 20*x + (-1 + 4*x)*Log[5 - 20*x])*Log[-20 + 4*Log[5 - 20*x]]),x]

[Out]

Log[Log[4*(-5 + Log[5 - 20*x])]]

Maple [A] (verified)

Time = 12.62 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
norman	\(\ln \left (\ln \left (4 \ln \left (-20 x +5\right )-20\right )\right )\)	\(13\)
risch	\(\ln \left (\ln \left (4 \ln \left (-20 x +5\right )-20\right )\right )\)	\(13\)
parallelrisch	\(\ln \left (\ln \left (4 \ln \left (-20 x +5\right )-20\right )\right )\)	\(13\)

[In]

int(4/((-1+4*x)*ln(-20*x+5)-20*x+5)/ln(4*ln(-20*x+5)-20),x,method=_RETURNVERBOSE)

[Out]

ln(ln(4*ln(-20*x+5)-20))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log \left (\log \left (4 \, \log \left (-20 \, x + 5\right ) - 20\right )\right ) \]

[In]

integrate(4/((-1+4*x)*log(-20*x+5)-20*x+5)/log(4*log(-20*x+5)-20),x, algorithm="fricas")

[Out]

log(log(4*log(-20*x + 5) - 20))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log {\left (\log {\left (4 \log {\left (5 - 20 x \right )} - 20 \right )} \right )} \]

[In]

integrate(4/((-1+4*x)*ln(-20*x+5)-20*x+5)/ln(4*ln(-20*x+5)-20),x)

[Out]

log(log(4*log(5 - 20*x) - 20))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.67 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log \left (2 \, \log \left (2\right ) + \log \left (i \, \pi + \log \left (5\right ) + \log \left (4 \, x - 1\right ) - 5\right )\right ) \]

[In]

integrate(4/((-1+4*x)*log(-20*x+5)-20*x+5)/log(4*log(-20*x+5)-20),x, algorithm="maxima")

[Out]

log(2*log(2) + log(I*pi + log(5) + log(4*x - 1) - 5))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\log \left (\log \left (4 \, \log \left (-20 \, x + 5\right ) - 20\right )\right ) \]

[In]

integrate(4/((-1+4*x)*log(-20*x+5)-20*x+5)/log(4*log(-20*x+5)-20),x, algorithm="giac")

[Out]

log(log(4*log(-20*x + 5) - 20))

Mupad [B] (verification not implemented)

Time = 15.43 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {4}{(5-20 x+(-1+4 x) \log (5-20 x)) \log (-20+4 \log (5-20 x))} \, dx=\ln \left (\ln \left (4\,\ln \left (5-20\,x\right )-20\right )\right ) \]

[In]

int(4/(log(4*log(5 - 20*x) - 20)*(log(5 - 20*x)*(4*x - 1) - 20*x + 5)),x)

[Out]

log(log(4*log(5 - 20*x) - 20))