Integrand size = 71, antiderivative size = 33 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=(1+x) \left (5-\frac {4+5 (-2 x+\log (5))}{x}+\log \left (\frac {4 x}{x-\log (2)}\right )\right ) \]
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=-\frac {4}{x}+15 x-\frac {5 \log (5)}{x}+\log (x)+x \log \left (\frac {4 x}{x-\log (2)}\right )-\log (x-\log (2)) \]
Integrate[(-4*x - 15*x^3 + (4 + x + 16*x^2)*Log[2] + (-5*x + 5*Log[2])*Log [5] + (-x^3 + x^2*Log[2])*Log[(-4*x)/(-x + Log[2])])/(-x^3 + x^2*Log[2]),x ]
Time = 0.57 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2026, 7293, 2009}
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.
\(\displaystyle \int \frac {-15 x^3+\left (16 x^2+x+4\right ) \log (2)+\left (x^2 \log (2)-x^3\right ) \log \left (-\frac {4 x}{\log (2)-x}\right )-4 x+\log (5) (5 \log (2)-5 x)}{x^2 \log (2)-x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-15 x^3+\left (16 x^2+x+4\right ) \log (2)+\left (x^2 \log (2)-x^3\right ) \log \left (-\frac {4 x}{\log (2)-x}\right )-4 x+\log (5) (5 \log (2)-5 x)}{x^2 (\log (2)-x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {15 x^3-16 x^2 \log (2)+x \left (4+\log \left (\frac {3125}{2}\right )\right )-\log (2) (4+5 \log (5))}{x^2 (x-\log (2))}+\log \left (\frac {4 x}{x-\log (2)}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 15 x+x \log \left (\frac {4 x}{x-\log (2)}\right )+\log (x)-(1+\log (2)) \log (x-\log (2))+\log (2) \log (x-\log (2))-\frac {4+5 \log (5)}{x}\) |
Int[(-4*x - 15*x^3 + (4 + x + 16*x^2)*Log[2] + (-5*x + 5*Log[2])*Log[5] + (-x^3 + x^2*Log[2])*Log[(-4*x)/(-x + Log[2])])/(-x^3 + x^2*Log[2]),x]
15*x - (4 + 5*Log[5])/x + Log[x] + x*Log[(4*x)/(x - Log[2])] + Log[2]*Log[ x - Log[2]] - (1 + Log[2])*Log[x - Log[2]]
3.22.54.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.42 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39
method | result | size |
norman | \(\frac {x^{2} \ln \left (-\frac {4 x}{\ln \left (2\right )-x}\right )+x \ln \left (-\frac {4 x}{\ln \left (2\right )-x}\right )+15 x^{2}-5 \ln \left (5\right )-4}{x}\) | \(46\) |
risch | \(x \ln \left (-\frac {4 x}{\ln \left (2\right )-x}\right )-\frac {-x \ln \left (x \right )+\ln \left (x -\ln \left (2\right )\right ) x -15 x^{2}+5 \ln \left (5\right )+4}{x}\) | \(46\) |
parallelrisch | \(-\frac {2 x \ln \left (2\right ) \ln \left (x \right )-2 \ln \left (2\right ) \ln \left (x -\ln \left (2\right )\right ) x -2 \ln \left (2\right ) \ln \left (-\frac {4 x}{\ln \left (2\right )-x}\right ) x +4-x^{2} \ln \left (-\frac {4 x}{\ln \left (2\right )-x}\right )-30 x \ln \left (2\right )-x \ln \left (x \right )+\ln \left (x -\ln \left (2\right )\right ) x -15 x^{2}+5 \ln \left (5\right )}{x}\) | \(89\) |
parts | \(15 x +\ln \left (x \right )-\frac {5 \ln \left (5\right )+4}{x}+\left (-\ln \left (2\right )-1\right ) \ln \left (x -\ln \left (2\right )\right )-4 \ln \left (2\right ) \left (\frac {\ln \left (\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right )}{4}-\frac {\ln \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right ) \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right ) \left (x -\ln \left (2\right )\right )}{16 \ln \left (2\right )}\right )\) | \(93\) |
derivativedivides | \(-4 \ln \left (2\right ) \left (-\frac {\ln \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right ) \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right ) \left (x -\ln \left (2\right )\right )}{16 \ln \left (2\right )}+\frac {-\frac {4}{4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}}-\frac {5 \ln \left (5\right )}{4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}}}{\ln \left (2\right )^{2}}-\frac {\ln \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right )}{4 \ln \left (2\right )}-\frac {15 \left (x -\ln \left (2\right )\right )}{4 \ln \left (2\right )}\right )\) | \(124\) |
default | \(-4 \ln \left (2\right ) \left (-\frac {\ln \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right ) \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right ) \left (x -\ln \left (2\right )\right )}{16 \ln \left (2\right )}+\frac {-\frac {4}{4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}}-\frac {5 \ln \left (5\right )}{4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}}}{\ln \left (2\right )^{2}}-\frac {\ln \left (4+\frac {4 \ln \left (2\right )}{x -\ln \left (2\right )}\right )}{4 \ln \left (2\right )}-\frac {15 \left (x -\ln \left (2\right )\right )}{4 \ln \left (2\right )}\right )\) | \(124\) |
int(((x^2*ln(2)-x^3)*ln(-4*x/(ln(2)-x))+(5*ln(2)-5*x)*ln(5)+(16*x^2+x+4)*l n(2)-15*x^3-4*x)/(x^2*ln(2)-x^3),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=\frac {15 \, x^{2} + {\left (x^{2} + x\right )} \log \left (\frac {4 \, x}{x - \log \left (2\right )}\right ) - 5 \, \log \left (5\right ) - 4}{x} \]
integrate(((x^2*log(2)-x^3)*log(-4*x/(log(2)-x))+(5*log(2)-5*x)*log(5)+(16 *x^2+x+4)*log(2)-15*x^3-4*x)/(x^2*log(2)-x^3),x, algorithm=\
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=x \log {\left (- \frac {4 x}{- x + \log {\left (2 \right )}} \right )} + 15 x + \log {\left (x \right )} - \log {\left (x - \log {\left (2 \right )} \right )} + \frac {- 5 \log {\left (5 \right )} - 4}{x} \]
integrate(((x**2*ln(2)-x**3)*ln(-4*x/(ln(2)-x))+(5*ln(2)-5*x)*ln(5)+(16*x* *2+x+4)*ln(2)-15*x**3-4*x)/(x**2*ln(2)-x**3),x)
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (31) = 62\).
Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 5.39 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=-5 \, {\left (\frac {\log \left (x - \log \left (2\right )\right )}{\log \left (2\right )^{2}} - \frac {\log \left (x\right )}{\log \left (2\right )^{2}} + \frac {1}{x \log \left (2\right )}\right )} \log \left (5\right ) \log \left (2\right ) + 5 \, {\left (\frac {\log \left (x - \log \left (2\right )\right )}{\log \left (2\right )} - \frac {\log \left (x\right )}{\log \left (2\right )}\right )} \log \left (5\right ) + 2 \, x \log \left (2\right ) - {\left (\frac {\log \left (x - \log \left (2\right )\right )}{\log \left (2\right )} - \frac {\log \left (x\right )}{\log \left (2\right )}\right )} \log \left (2\right ) - 4 \, {\left (\frac {\log \left (x - \log \left (2\right )\right )}{\log \left (2\right )^{2}} - \frac {\log \left (x\right )}{\log \left (2\right )^{2}} + \frac {1}{x \log \left (2\right )}\right )} \log \left (2\right ) - {\left (x - \log \left (2\right )\right )} \log \left (x - \log \left (2\right )\right ) - \log \left (2\right ) \log \left (x - \log \left (2\right )\right ) + x \log \left (x\right ) + 15 \, x + \frac {4 \, \log \left (x - \log \left (2\right )\right )}{\log \left (2\right )} - \frac {4 \, \log \left (x\right )}{\log \left (2\right )} \]
integrate(((x^2*log(2)-x^3)*log(-4*x/(log(2)-x))+(5*log(2)-5*x)*log(5)+(16 *x^2+x+4)*log(2)-15*x^3-4*x)/(x^2*log(2)-x^3),x, algorithm=\
-5*(log(x - log(2))/log(2)^2 - log(x)/log(2)^2 + 1/(x*log(2)))*log(5)*log( 2) + 5*(log(x - log(2))/log(2) - log(x)/log(2))*log(5) + 2*x*log(2) - (log (x - log(2))/log(2) - log(x)/log(2))*log(2) - 4*(log(x - log(2))/log(2)^2 - log(x)/log(2)^2 + 1/(x*log(2)))*log(2) - (x - log(2))*log(x - log(2)) - log(2)*log(x - log(2)) + x*log(x) + 15*x + 4*log(x - log(2))/log(2) - 4*lo g(x)/log(2)
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=x {\left (2 \, \log \left (2\right ) + 15\right )} - x \log \left (x - \log \left (2\right )\right ) + x \log \left (x\right ) - \frac {5 \, \log \left (5\right ) + 4}{x} - \log \left (x - \log \left (2\right )\right ) + \log \left (x\right ) \]
integrate(((x^2*log(2)-x^3)*log(-4*x/(log(2)-x))+(5*log(2)-5*x)*log(5)+(16 *x^2+x+4)*log(2)-15*x^3-4*x)/(x^2*log(2)-x^3),x, algorithm=\
x*(2*log(2) + 15) - x*log(x - log(2)) + x*log(x) - (5*log(5) + 4)/x - log( x - log(2)) + log(x)
Time = 9.78 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {-4 x-15 x^3+\left (4+x+16 x^2\right ) \log (2)+(-5 x+5 \log (2)) \log (5)+\left (-x^3+x^2 \log (2)\right ) \log \left (-\frac {4 x}{-x+\log (2)}\right )}{-x^3+x^2 \log (2)} \, dx=15\,x+x\,\ln \left (\frac {x}{x-\ln \left (2\right )}\right )+2\,x\,\ln \left (2\right )-\frac {5\,\ln \left (5\right )}{x}-\frac {4}{x}-\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{\ln \left (2\right )}-\mathrm {i}\right )\,2{}\mathrm {i} \]