Integrand size = 67, antiderivative size = 26 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {x (5+25 x)}{5 \left (\frac {3}{5}+x \log (3)-4 \log ^2(x)\right )} \] Output:
1/5*(25*x+5)*x/(3/5+x*ln(3)-4*ln(x)^2)
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 x (1+5 x)}{3+5 x \log (3)-20 \log ^2(x)} \] Input:
Integrate[(15 + 150*x + 125*x^2*Log[3] + (200 + 1000*x)*Log[x] + (-100 - 1 000*x)*Log[x]^2)/(9 + 30*x*Log[3] + 25*x^2*Log[3]^2 + (-120 - 200*x*Log[3] )*Log[x]^2 + 400*Log[x]^4),x]
Output:
(5*x*(1 + 5*x))/(3 + 5*x*Log[3] - 20*Log[x]^2)
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.
\(\displaystyle \int \frac {125 x^2 \log (3)+150 x+(-1000 x-100) \log ^2(x)+(1000 x+200) \log (x)+15}{25 x^2 \log ^2(3)+400 \log ^4(x)+(-200 x \log (3)-120) \log ^2(x)+30 x \log (3)+9} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {125 x^2 \log (3)+150 x+(-1000 x-100) \log ^2(x)+(1000 x+200) \log (x)+15}{\left (-20 \log ^2(x)+5 x \log (3)+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 (10 x+1)}{-20 \log ^2(x)+5 x \log (3)+3}-\frac {25 (5 x+1) (x \log (3)-8 \log (x))}{\left (-20 \log ^2(x)+5 x \log (3)+3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -125 \log (3) \int \frac {x^2}{\left (-20 \log ^2(x)+5 x \log (3)+3\right )^2}dx-25 \log (3) \int \frac {x}{\left (-20 \log ^2(x)+5 x \log (3)+3\right )^2}dx+5 \int \frac {1}{-20 \log ^2(x)+5 x \log (3)+3}dx+50 \int \frac {x}{-20 \log ^2(x)+5 x \log (3)+3}dx+200 \int \frac {\log (x)}{\left (20 \log ^2(x)-5 x \log (3)-3\right )^2}dx+1000 \int \frac {x \log (x)}{\left (20 \log ^2(x)-5 x \log (3)-3\right )^2}dx\) |
Input:
Int[(15 + 150*x + 125*x^2*Log[3] + (200 + 1000*x)*Log[x] + (-100 - 1000*x) *Log[x]^2)/(9 + 30*x*Log[3] + 25*x^2*Log[3]^2 + (-120 - 200*x*Log[3])*Log[ x]^2 + 400*Log[x]^4),x]
Output:
$Aborted
Time = 1.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {5 \left (1+5 x \right ) x}{5 x \ln \left (3\right )-20 \ln \left (x \right )^{2}+3}\) | \(24\) |
default | \(\frac {25 x^{2}+5 x}{5 x \ln \left (3\right )-20 \ln \left (x \right )^{2}+3}\) | \(25\) |
norman | \(\frac {25 x^{2}+5 x}{5 x \ln \left (3\right )-20 \ln \left (x \right )^{2}+3}\) | \(26\) |
parallelrisch | \(\frac {75 x^{2}+15 x}{15 x \ln \left (3\right )-60 \ln \left (x \right )^{2}+9}\) | \(27\) |
Input:
int(((-1000*x-100)*ln(x)^2+(1000*x+200)*ln(x)+125*x^2*ln(3)+150*x+15)/(400 *ln(x)^4+(-200*x*ln(3)-120)*ln(x)^2+25*x^2*ln(3)^2+30*x*ln(3)+9),x,method= _RETURNVERBOSE)
Output:
5*(1+5*x)*x/(5*x*ln(3)-20*ln(x)^2+3)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 \, {\left (5 \, x^{2} + x\right )}}{5 \, x \log \left (3\right ) - 20 \, \log \left (x\right )^{2} + 3} \] Input:
integrate(((-1000*x-100)*log(x)^2+(1000*x+200)*log(x)+125*x^2*log(3)+150*x +15)/(400*log(x)^4+(-200*x*log(3)-120)*log(x)^2+25*x^2*log(3)^2+30*x*log(3 )+9),x, algorithm="fricas")
Output:
5*(5*x^2 + x)/(5*x*log(3) - 20*log(x)^2 + 3)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {- 25 x^{2} - 5 x}{- 5 x \log {\left (3 \right )} + 20 \log {\left (x \right )}^{2} - 3} \] Input:
integrate(((-1000*x-100)*ln(x)**2+(1000*x+200)*ln(x)+125*x**2*ln(3)+150*x+ 15)/(400*ln(x)**4+(-200*x*ln(3)-120)*ln(x)**2+25*x**2*ln(3)**2+30*x*ln(3)+ 9),x)
Output:
(-25*x**2 - 5*x)/(-5*x*log(3) + 20*log(x)**2 - 3)
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 \, {\left (5 \, x^{2} + x\right )}}{5 \, x \log \left (3\right ) - 20 \, \log \left (x\right )^{2} + 3} \] Input:
integrate(((-1000*x-100)*log(x)^2+(1000*x+200)*log(x)+125*x^2*log(3)+150*x +15)/(400*log(x)^4+(-200*x*log(3)-120)*log(x)^2+25*x^2*log(3)^2+30*x*log(3 )+9),x, algorithm="maxima")
Output:
5*(5*x^2 + x)/(5*x*log(3) - 20*log(x)^2 + 3)
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {5 \, {\left (5 \, x^{2} + x\right )}}{5 \, x \log \left (3\right ) - 20 \, \log \left (x\right )^{2} + 3} \] Input:
integrate(((-1000*x-100)*log(x)^2+(1000*x+200)*log(x)+125*x^2*log(3)+150*x +15)/(400*log(x)^4+(-200*x*log(3)-120)*log(x)^2+25*x^2*log(3)^2+30*x*log(3 )+9),x, algorithm="giac")
Output:
5*(5*x^2 + x)/(5*x*log(3) - 20*log(x)^2 + 3)
Timed out. \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\int \frac {150\,x+\ln \left (x\right )\,\left (1000\,x+200\right )+125\,x^2\,\ln \left (3\right )-{\ln \left (x\right )}^2\,\left (1000\,x+100\right )+15}{25\,x^2\,{\ln \left (3\right )}^2+30\,x\,\ln \left (3\right )+400\,{\ln \left (x\right )}^4-{\ln \left (x\right )}^2\,\left (200\,x\,\ln \left (3\right )+120\right )+9} \,d x \] Input:
int((150*x + log(x)*(1000*x + 200) + 125*x^2*log(3) - log(x)^2*(1000*x + 1 00) + 15)/(25*x^2*log(3)^2 + 30*x*log(3) + 400*log(x)^4 - log(x)^2*(200*x* log(3) + 120) + 9),x)
Output:
int((150*x + log(x)*(1000*x + 200) + 125*x^2*log(3) - log(x)^2*(1000*x + 1 00) + 15)/(25*x^2*log(3)^2 + 30*x*log(3) + 400*log(x)^4 - log(x)^2*(200*x* log(3) + 120) + 9), x)
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {15+150 x+125 x^2 \log (3)+(200+1000 x) \log (x)+(-100-1000 x) \log ^2(x)}{9+30 x \log (3)+25 x^2 \log ^2(3)+(-120-200 x \log (3)) \log ^2(x)+400 \log ^4(x)} \, dx=\frac {-20 \mathrm {log}\left (x \right )^{2}-25 \,\mathrm {log}\left (3\right ) x^{2}+3}{\mathrm {log}\left (3\right ) \left (20 \mathrm {log}\left (x \right )^{2}-5 \,\mathrm {log}\left (3\right ) x -3\right )} \] Input:
int(((-1000*x-100)*log(x)^2+(1000*x+200)*log(x)+125*x^2*log(3)+150*x+15)/( 400*log(x)^4+(-200*x*log(3)-120)*log(x)^2+25*x^2*log(3)^2+30*x*log(3)+9),x )
Output:
( - 20*log(x)**2 - 25*log(3)*x**2 + 3)/(log(3)*(20*log(x)**2 - 5*log(3)*x - 3))