\(\int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log (\frac {3 x}{4})+(-4 x-4 x^2+4 x \log (\frac {3 x}{4})) \log (1+x-\log (\frac {3 x}{4}))} \, dx\) [4244]

Optimal result

Integrand size = 59, antiderivative size = 27 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \log (5) \left (1+\log \left (\frac {1}{2} \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right )\right )\right ) \]

[Out]

ln(5)*(1/4*ln(1/2+1/2*ln(-ln(3/4*x)+x+1))+1/4)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 6820, 6816} \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \log (5) \log \left (\log \left (x-\log \left (\frac {3 x}{4}\right )+1\right )+1\right ) \]

[In]

Int[((1 - x)*Log[5])/(-4*x - 4*x^2 + 4*x*Log[(3*x)/4] + (-4*x - 4*x^2 + 4*x*Log[(3*x)/4])*Log[1 + x - Log[(3*x
)/4]]),x]

[Out]

(Log[5]*Log[1 + Log[1 + x - Log[(3*x)/4]]])/4

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]]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \log (5) \int \frac {1-x}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx \\ & = \log (5) \int \frac {-1+x}{4 x \left (1+x-\log \left (\frac {3 x}{4}\right )\right ) \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right )} \, dx \\ & = \frac {1}{4} \log (5) \int \frac {-1+x}{x \left (1+x-\log \left (\frac {3 x}{4}\right )\right ) \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right )} \, dx \\ & = \frac {1}{4} \log (5) \log \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \log (5) \log \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right ) \]

[In]

Integrate[((1 - x)*Log[5])/(-4*x - 4*x^2 + 4*x*Log[(3*x)/4] + (-4*x - 4*x^2 + 4*x*Log[(3*x)/4])*Log[1 + x - Lo
g[(3*x)/4]]),x]

[Out]

(Log[5]*Log[1 + Log[1 + x - Log[(3*x)/4]]])/4

Maple [A] (verified)

Time = 19.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67

method	result	size
norman	\(\frac {\ln \left (5\right ) \ln \left (\ln \left (-\ln \left (\frac {3 x}{4}\right )+x +1\right )+1\right )}{4}\)	\(18\)
risch	\(\frac {\ln \left (5\right ) \ln \left (\ln \left (-\ln \left (\frac {3 x}{4}\right )+x +1\right )+1\right )}{4}\)	\(18\)
parallelrisch	\(\frac {\ln \left (5\right ) \ln \left (\ln \left (-\ln \left (\frac {3 x}{4}\right )+x +1\right )+1\right )}{4}\)	\(18\)

[In]

int((1-x)*ln(5)/((4*x*ln(3/4*x)-4*x^2-4*x)*ln(-ln(3/4*x)+x+1)+4*x*ln(3/4*x)-4*x^2-4*x),x,method=_RETURNVERBOSE
)

[Out]

1/4*ln(5)*ln(ln(-ln(3/4*x)+x+1)+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \, \log \left (5\right ) \log \left (\log \left (x - \log \left (\frac {3}{4} \, x\right ) + 1\right ) + 1\right ) \]

[In]

integrate((1-x)*log(5)/((4*x*log(3/4*x)-4*x^2-4*x)*log(-log(3/4*x)+x+1)+4*x*log(3/4*x)-4*x^2-4*x),x, algorithm
="fricas")

[Out]

1/4*log(5)*log(log(x - log(3/4*x) + 1) + 1)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {\log {\left (5 \right )} \log {\left (\log {\left (x - \log {\left (\frac {3 x}{4} \right )} + 1 \right )} + 1 \right )}}{4} \]

[In]

integrate((1-x)*ln(5)/((4*x*ln(3/4*x)-4*x**2-4*x)*ln(-ln(3/4*x)+x+1)+4*x*ln(3/4*x)-4*x**2-4*x),x)

[Out]

log(5)*log(log(x - log(3*x/4) + 1) + 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \, \log \left (5\right ) \log \left (\log \left (x - \log \left (3\right ) + 2 \, \log \left (2\right ) - \log \left (x\right ) + 1\right ) + 1\right ) \]

[In]

integrate((1-x)*log(5)/((4*x*log(3/4*x)-4*x^2-4*x)*log(-log(3/4*x)+x+1)+4*x*log(3/4*x)-4*x^2-4*x),x, algorithm
="maxima")

[Out]

1/4*log(5)*log(log(x - log(3) + 2*log(2) - log(x) + 1) + 1)

Giac [F]

\[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\int { \frac {{\left (x - 1\right )} \log \left (5\right )}{4 \, {\left (x^{2} - x \log \left (\frac {3}{4} \, x\right ) + {\left (x^{2} - x \log \left (\frac {3}{4} \, x\right ) + x\right )} \log \left (x - \log \left (\frac {3}{4} \, x\right ) + 1\right ) + x\right )}} \,d x } \]

[In]

integrate((1-x)*log(5)/((4*x*log(3/4*x)-4*x^2-4*x)*log(-log(3/4*x)+x+1)+4*x*log(3/4*x)-4*x^2-4*x),x, algorithm
="giac")

[Out]

integrate(1/4*(x - 1)*log(5)/(x^2 - x*log(3/4*x) + (x^2 - x*log(3/4*x) + x)*log(x - log(3/4*x) + 1) + x), x)

Mupad [B] (verification not implemented)

Time = 9.47 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {\ln \left (5\right )\,\ln \left (\ln \left (x-\ln \left (\frac {3\,x}{4}\right )+1\right )+1\right )}{4} \]

[In]

int((log(5)*(x - 1))/(4*x - 4*x*log((3*x)/4) + log(x - log((3*x)/4) + 1)*(4*x - 4*x*log((3*x)/4) + 4*x^2) + 4*
x^2),x)

[Out]

(log(5)*log(log(x - log((3*x)/4) + 1) + 1))/4