Integrand size = 59, antiderivative size = 27 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \log (5) \left (1+\log \left (\frac {1}{2} \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right )\right )\right ) \] Output:
ln(5)*(1/4*ln(1/2+1/2*ln(-ln(3/4*x)+x+1))+1/4)
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \log (5) \log \left (1+\log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )\right ) \] Input:
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 - Log[(3*x)/4]]),x]
Output:
(Log[5]*Log[1 + Log[1 + x - Log[(3*x)/4]]])/4
Time = 0.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {27, 27, 7292, 7235}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(1-x) \log (5)}{-4 x^2+\left (-4 x^2-4 x+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (x-\log \left (\frac {3 x}{4}\right )+1\right )-4 x+4 x \log \left (\frac {3 x}{4}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (5) \int -\frac {1-x}{4 \left (x^2-\log \left (\frac {3 x}{4}\right ) x+x+\left (x^2-\log \left (\frac {3 x}{4}\right ) x+x\right ) \log \left (x-\log \left (\frac {3 x}{4}\right )+1\right )\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} \log (5) \int \frac {1-x}{x^2-\log \left (\frac {3 x}{4}\right ) x+x+\left (x^2-\log \left (\frac {3 x}{4}\right ) x+x\right ) \log \left (x-\log \left (\frac {3 x}{4}\right )+1\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{4} \log (5) \int \frac {1-x}{x \left (x-\log \left (\frac {3 x}{4}\right )+1\right ) \left (\log \left (x-\log \left (\frac {3 x}{4}\right )+1\right )+1\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \frac {1}{4} \log (5) \log \left (\log \left (x-\log \left (\frac {3 x}{4}\right )+1\right )+1\right )\) |
Input:
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]
Output:
(Log[5]*Log[1 + Log[1 + x - Log[(3*x)/4]]])/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 7.51 (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 | \(\frac {\ln \left (5\right ) \ln \left (\ln \left (-\ln \left (\frac {3 x}{4}\right )+x +1\right )+1\right )}{4}\) | \(18\) |
Input:
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)
Output:
1/4*ln(5)*ln(ln(-ln(3/4*x)+x+1)+1)
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \, \log \left (5\right ) \log \left (\log \left (x - \log \left (\frac {3}{4} \, x\right ) + 1\right ) + 1\right ) \] Input:
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")
Output:
1/4*log(5)*log(log(x - log(3/4*x) + 1) + 1)
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {\log {\left (5 \right )} \log {\left (\log {\left (x - \log {\left (\frac {3 x}{4} \right )} + 1 \right )} + 1 \right )}}{4} \] Input:
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*l n(3/4*x)-4*x**2-4*x),x)
Output:
log(5)*log(log(x - log(3*x/4) + 1) + 1)/4
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {1}{4} \, \log \left (5\right ) \log \left (\log \left (x - \log \left (3\right ) + 2 \, \log \left (2\right ) - \log \left (x\right ) + 1\right ) + 1\right ) \] Input:
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")
Output:
1/4*log(5)*log(log(x - log(3) + 2*log(2) - log(x) + 1) + 1)
\[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\int { \frac {{\left (x - 1\right )} \log \left (5\right )}{4 \, {\left (x^{2} - x \log \left (\frac {3}{4} \, x\right ) + {\left (x^{2} - x \log \left (\frac {3}{4} \, x\right ) + x\right )} \log \left (x - \log \left (\frac {3}{4} \, x\right ) + 1\right ) + x\right )}} \,d x } \] Input:
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")
Output:
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)
Time = 3.73 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {\ln \left (5\right )\,\ln \left (\ln \left (x-\ln \left (\frac {3\,x}{4}\right )+1\right )+1\right )}{4} \] Input:
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)
Output:
(log(5)*log(log(x - log((3*x)/4) + 1) + 1))/4
Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(1-x) \log (5)}{-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )+\left (-4 x-4 x^2+4 x \log \left (\frac {3 x}{4}\right )\right ) \log \left (1+x-\log \left (\frac {3 x}{4}\right )\right )} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (-\mathrm {log}\left (\frac {3 x}{4}\right )+x +1\right )+1\right ) \mathrm {log}\left (5\right )}{4} \] Input:
int((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)
Output:
(log(log( - log((3*x)/4) + x + 1) + 1)*log(5))/4