Integrand size = 109, antiderivative size = 34 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=e^5-x+x \log \left (\frac {5 \left (\frac {1}{5 (4-x) x}+\log (x)\right )}{3 x}\right ) \] Output:
x*ln(5/3*(ln(x)+1/5/x/(4-x))/x)+exp(5)-x
Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=-x+x \log \left (\frac {-1+5 (-4+x) x \log (x)}{3 (-4+x) x^2}\right ) \] Input:
Integrate[(-12 + 84*x - 40*x^2 + 5*x^3 + (-160*x + 80*x^2 - 10*x^3)*Log[x] + (4 - x + (80*x - 40*x^2 + 5*x^3)*Log[x])*Log[(-1 + (-20*x + 5*x^2)*Log[ x])/(-12*x^2 + 3*x^3)])/(4 - x + (80*x - 40*x^2 + 5*x^3)*Log[x]),x]
Output:
-x + x*Log[(-1 + 5*(-4 + x)*x*Log[x])/(3*(-4 + x)*x^2)]
Time = 2.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7292, 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.
\(\displaystyle \int \frac {5 x^3-40 x^2+\left (-10 x^3+80 x^2-160 x\right ) \log (x)+\left (\left (5 x^3-40 x^2+80 x\right ) \log (x)-x+4\right ) \log \left (\frac {\left (5 x^2-20 x\right ) \log (x)-1}{3 x^3-12 x^2}\right )+84 x-12}{\left (5 x^3-40 x^2+80 x\right ) \log (x)-x+4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 x^3-40 x^2+\left (-10 x^3+80 x^2-160 x\right ) \log (x)+\left (\left (5 x^3-40 x^2+80 x\right ) \log (x)-x+4\right ) \log \left (\frac {\left (5 x^2-20 x\right ) \log (x)-1}{3 x^3-12 x^2}\right )+84 x-12}{(4-x) \left (-5 x^2 \log (x)+20 x \log (x)+1\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {40 x^2}{(x-4) \left (5 x^2 \log (x)-20 x \log (x)-1\right )}-\frac {10 (x-4) x \log (x)}{5 x^2 \log (x)-20 x \log (x)-1}+\frac {84 x}{(x-4) \left (5 x^2 \log (x)-20 x \log (x)-1\right )}+\log \left (\frac {5 (x-4) x \log (x)-1}{3 (x-4) x^2}\right )-\frac {12}{(x-4) \left (5 x^2 \log (x)-20 x \log (x)-1\right )}+\frac {5 x^3}{(x-4) \left (5 x^2 \log (x)-20 x \log (x)-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x \log \left (\frac {5 (4-x) x \log (x)+1}{3 (4-x) x^2}\right )-x\) |
Input:
Int[(-12 + 84*x - 40*x^2 + 5*x^3 + (-160*x + 80*x^2 - 10*x^3)*Log[x] + (4 - x + (80*x - 40*x^2 + 5*x^3)*Log[x])*Log[(-1 + (-20*x + 5*x^2)*Log[x])/(- 12*x^2 + 3*x^3)])/(4 - x + (80*x - 40*x^2 + 5*x^3)*Log[x]),x]
Output:
-x + x*Log[(1 + 5*(4 - x)*x*Log[x])/(3*(4 - x)*x^2)]
Time = 2.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-12+\ln \left (\frac {\left (5 x^{2}-20 x \right ) \ln \left (x \right )-1}{3 x^{2} \left (x -4\right )}\right ) x -x\) | \(33\) |
risch | \(x \ln \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )-x \ln \left (x -4\right )-2 x \ln \left (x \right )-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x -4}\right ) \operatorname {csgn}\left (i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x -4}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x^{2} \left (x -4\right )}\right )}^{2}}{2}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x^{2} \left (x -4\right )}\right )}^{3}}{2}+\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x -4}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x -4}\right ) \operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x^{2} \left (x -4\right )}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x -4}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x -4}\right )}^{2}}{2}-i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x -4}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x^{2} \left (x -4\right )}\right )}^{2}}{2}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left (x^{2} \ln \left (x \right )-4 x \ln \left (x \right )-\frac {1}{5}\right )}{x -4}\right )}^{3}}{2}+\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-x \ln \left (3\right )+x \ln \left (5\right )-x\) | \(444\) |
Input:
int((((5*x^3-40*x^2+80*x)*ln(x)-x+4)*ln(((5*x^2-20*x)*ln(x)-1)/(3*x^3-12*x ^2))+(-10*x^3+80*x^2-160*x)*ln(x)+5*x^3-40*x^2+84*x-12)/((5*x^3-40*x^2+80* x)*ln(x)-x+4),x,method=_RETURNVERBOSE)
Output:
-12+ln(1/3*((5*x^2-20*x)*ln(x)-1)/x^2/(x-4))*x-x
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=x \log \left (\frac {5 \, {\left (x^{2} - 4 \, x\right )} \log \left (x\right ) - 1}{3 \, {\left (x^{3} - 4 \, x^{2}\right )}}\right ) - x \] Input:
integrate((((5*x^3-40*x^2+80*x)*log(x)-x+4)*log(((5*x^2-20*x)*log(x)-1)/(3 *x^3-12*x^2))+(-10*x^3+80*x^2-160*x)*log(x)+5*x^3-40*x^2+84*x-12)/((5*x^3- 40*x^2+80*x)*log(x)-x+4),x, algorithm="fricas")
Output:
x*log(1/3*(5*(x^2 - 4*x)*log(x) - 1)/(x^3 - 4*x^2)) - x
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (24) = 48\).
Time = 0.55 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=- x + \left (x - \frac {2}{3}\right ) \log {\left (\frac {\left (5 x^{2} - 20 x\right ) \log {\left (x \right )} - 1}{3 x^{3} - 12 x^{2}} \right )} - \frac {2 \log {\left (x \right )}}{3} + \frac {2 \log {\left (\log {\left (x \right )} - \frac {1}{5 x^{2} - 20 x} \right )}}{3} \] Input:
integrate((((5*x**3-40*x**2+80*x)*ln(x)-x+4)*ln(((5*x**2-20*x)*ln(x)-1)/(3 *x**3-12*x**2))+(-10*x**3+80*x**2-160*x)*ln(x)+5*x**3-40*x**2+84*x-12)/((5 *x**3-40*x**2+80*x)*ln(x)-x+4),x)
Output:
-x + (x - 2/3)*log(((5*x**2 - 20*x)*log(x) - 1)/(3*x**3 - 12*x**2)) - 2*lo g(x)/3 + 2*log(log(x) - 1/(5*x**2 - 20*x))/3
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=-x {\left (\log \left (3\right ) + 1\right )} + x \log \left (5 \, {\left (x^{2} - 4 \, x\right )} \log \left (x\right ) - 1\right ) - x \log \left (x - 4\right ) - 2 \, x \log \left (x\right ) \] Input:
integrate((((5*x^3-40*x^2+80*x)*log(x)-x+4)*log(((5*x^2-20*x)*log(x)-1)/(3 *x^3-12*x^2))+(-10*x^3+80*x^2-160*x)*log(x)+5*x^3-40*x^2+84*x-12)/((5*x^3- 40*x^2+80*x)*log(x)-x+4),x, algorithm="maxima")
Output:
-x*(log(3) + 1) + x*log(5*(x^2 - 4*x)*log(x) - 1) - x*log(x - 4) - 2*x*log (x)
Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=x \log \left (5 \, x^{2} \log \left (x\right ) - 20 \, x \log \left (x\right ) - 1\right ) - x \log \left (3 \, x - 12\right ) - 2 \, x \log \left (x\right ) - x \] Input:
integrate((((5*x^3-40*x^2+80*x)*log(x)-x+4)*log(((5*x^2-20*x)*log(x)-1)/(3 *x^3-12*x^2))+(-10*x^3+80*x^2-160*x)*log(x)+5*x^3-40*x^2+84*x-12)/((5*x^3- 40*x^2+80*x)*log(x)-x+4),x, algorithm="giac")
Output:
x*log(5*x^2*log(x) - 20*x*log(x) - 1) - x*log(3*x - 12) - 2*x*log(x) - x
Time = 1.96 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=x\,\left (\ln \left (\frac {\ln \left (x\right )\,\left (20\,x-5\,x^2\right )+1}{12\,x^2-3\,x^3}\right )-1\right ) \] Input:
int((84*x + log((log(x)*(20*x - 5*x^2) + 1)/(12*x^2 - 3*x^3))*(log(x)*(80* x - 40*x^2 + 5*x^3) - x + 4) - 40*x^2 + 5*x^3 - log(x)*(160*x - 80*x^2 + 1 0*x^3) - 12)/(log(x)*(80*x - 40*x^2 + 5*x^3) - x + 4),x)
Output:
x*(log((log(x)*(20*x - 5*x^2) + 1)/(12*x^2 - 3*x^3)) - 1)
Time = 0.23 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-12+84 x-40 x^2+5 x^3+\left (-160 x+80 x^2-10 x^3\right ) \log (x)+\left (4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)\right ) \log \left (\frac {-1+\left (-20 x+5 x^2\right ) \log (x)}{-12 x^2+3 x^3}\right )}{4-x+\left (80 x-40 x^2+5 x^3\right ) \log (x)} \, dx=x \left (\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (x \right ) x^{2}-20 \,\mathrm {log}\left (x \right ) x -1}{3 x^{3}-12 x^{2}}\right )-1\right ) \] Input:
int((((5*x^3-40*x^2+80*x)*log(x)-x+4)*log(((5*x^2-20*x)*log(x)-1)/(3*x^3-1 2*x^2))+(-10*x^3+80*x^2-160*x)*log(x)+5*x^3-40*x^2+84*x-12)/((5*x^3-40*x^2 +80*x)*log(x)-x+4),x)
Output:
x*(log((5*log(x)*x**2 - 20*log(x)*x - 1)/(3*x**3 - 12*x**2)) - 1)