Integrand size = 110, antiderivative size = 27 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=2 x^2+\frac {1}{4+5 \left (\frac {3}{2 x}-x\right )+\log (5 x)} \] Output:
1/(15/2/x-5*x+ln(5*x)+4)+2*x^2
Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=2 \left (x^2+\frac {x}{15+8 x-10 x^2+2 x \log (5 x)}\right ) \] Input:
Integrate[(30 + 896*x + 980*x^2 - 944*x^3 - 640*x^4 + 400*x^5 + (240*x^2 + 128*x^3 - 160*x^4)*Log[5*x] + 16*x^3*Log[5*x]^2)/(225 + 240*x - 236*x^2 - 160*x^3 + 100*x^4 + (60*x + 32*x^2 - 40*x^3)*Log[5*x] + 4*x^2*Log[5*x]^2) ,x]
Output:
2*(x^2 + x/(15 + 8*x - 10*x^2 + 2*x*Log[5*x]))
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.
\(\displaystyle \int \frac {400 x^5-640 x^4-944 x^3+16 x^3 \log ^2(5 x)+980 x^2+\left (-160 x^4+128 x^3+240 x^2\right ) \log (5 x)+896 x+30}{100 x^4-160 x^3-236 x^2+4 x^2 \log ^2(5 x)+\left (-40 x^3+32 x^2+60 x\right ) \log (5 x)+240 x+225} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {400 x^5-640 x^4-944 x^3+16 x^3 \log ^2(5 x)+980 x^2+\left (-160 x^4+128 x^3+240 x^2\right ) \log (5 x)+896 x+30}{\left (-10 x^2+8 x+2 x \log (5 x)+15\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (10 x^2-2 x+15\right )}{\left (10 x^2-8 x-2 x \log (5 x)-15\right )^2}+4 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 30 \int \frac {1}{\left (10 x^2-2 \log (5 x) x-8 x-15\right )^2}dx-4 \int \frac {x}{\left (10 x^2-2 \log (5 x) x-8 x-15\right )^2}dx+20 \int \frac {x^2}{\left (10 x^2-2 \log (5 x) x-8 x-15\right )^2}dx+2 x^2\) |
Input:
Int[(30 + 896*x + 980*x^2 - 944*x^3 - 640*x^4 + 400*x^5 + (240*x^2 + 128*x ^3 - 160*x^4)*Log[5*x] + 16*x^3*Log[5*x]^2)/(225 + 240*x - 236*x^2 - 160*x ^3 + 100*x^4 + (60*x + 32*x^2 - 40*x^3)*Log[5*x] + 4*x^2*Log[5*x]^2),x]
Output:
$Aborted
Time = 0.73 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07
method | result | size |
risch | \(2 x^{2}-\frac {2 x}{10 x^{2}-2 x \ln \left (5 x \right )-8 x -15}\) | \(29\) |
derivativedivides | \(\frac {10 x +150 x^{2}+80 x^{3}-100 x^{4}+20 x^{3} \ln \left (5 x \right )}{10 x \ln \left (5 x \right )-50 x^{2}+40 x +75}\) | \(50\) |
default | \(\frac {10 x +150 x^{2}+80 x^{3}-100 x^{4}+20 x^{3} \ln \left (5 x \right )}{10 x \ln \left (5 x \right )-50 x^{2}+40 x +75}\) | \(50\) |
parallelrisch | \(\frac {40 x^{4}-8 x^{3} \ln \left (5 x \right )-32 x^{3}-60 x^{2}-4 x}{20 x^{2}-4 x \ln \left (5 x \right )-16 x -30}\) | \(50\) |
norman | \(\frac {-26 x -6 x \ln \left (5 x \right )-16 x^{3}+20 x^{4}-4 x^{3} \ln \left (5 x \right )-45}{10 x^{2}-2 x \ln \left (5 x \right )-8 x -15}\) | \(52\) |
Input:
int((16*x^3*ln(5*x)^2+(-160*x^4+128*x^3+240*x^2)*ln(5*x)+400*x^5-640*x^4-9 44*x^3+980*x^2+896*x+30)/(4*x^2*ln(5*x)^2+(-40*x^3+32*x^2+60*x)*ln(5*x)+10 0*x^4-160*x^3-236*x^2+240*x+225),x,method=_RETURNVERBOSE)
Output:
2*x^2-2*x/(10*x^2-2*x*ln(5*x)-8*x-15)
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=\frac {2 \, {\left (10 \, x^{4} - 2 \, x^{3} \log \left (5 \, x\right ) - 8 \, x^{3} - 15 \, x^{2} - x\right )}}{10 \, x^{2} - 2 \, x \log \left (5 \, x\right ) - 8 \, x - 15} \] Input:
integrate((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-6 40*x^4-944*x^3+980*x^2+896*x+30)/(4*x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*l og(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x, algorithm="fricas")
Output:
2*(10*x^4 - 2*x^3*log(5*x) - 8*x^3 - 15*x^2 - x)/(10*x^2 - 2*x*log(5*x) - 8*x - 15)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=2 x^{2} + \frac {2 x}{- 10 x^{2} + 2 x \log {\left (5 x \right )} + 8 x + 15} \] Input:
integrate((16*x**3*ln(5*x)**2+(-160*x**4+128*x**3+240*x**2)*ln(5*x)+400*x* *5-640*x**4-944*x**3+980*x**2+896*x+30)/(4*x**2*ln(5*x)**2+(-40*x**3+32*x* *2+60*x)*ln(5*x)+100*x**4-160*x**3-236*x**2+240*x+225),x)
Output:
2*x**2 + 2*x/(-10*x**2 + 2*x*log(5*x) + 8*x + 15)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=\frac {2 \, {\left (10 \, x^{4} - 2 \, x^{3} {\left (\log \left (5\right ) + 4\right )} - 2 \, x^{3} \log \left (x\right ) - 15 \, x^{2} - x\right )}}{10 \, x^{2} - 2 \, x {\left (\log \left (5\right ) + 4\right )} - 2 \, x \log \left (x\right ) - 15} \] Input:
integrate((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-6 40*x^4-944*x^3+980*x^2+896*x+30)/(4*x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*l og(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x, algorithm="maxima")
Output:
2*(10*x^4 - 2*x^3*(log(5) + 4) - 2*x^3*log(x) - 15*x^2 - x)/(10*x^2 - 2*x* (log(5) + 4) - 2*x*log(x) - 15)
Time = 0.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=2 \, x^{2} - \frac {2 \, x}{10 \, x^{2} - 2 \, x \log \left (5 \, x\right ) - 8 \, x - 15} \] Input:
integrate((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-6 40*x^4-944*x^3+980*x^2+896*x+30)/(4*x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*l og(5*x)+100*x^4-160*x^3-236*x^2+240*x+225),x, algorithm="giac")
Output:
2*x^2 - 2*x/(10*x^2 - 2*x*log(5*x) - 8*x - 15)
Time = 3.56 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=\frac {2\,x}{8\,x+2\,x\,\ln \left (5\,x\right )-10\,x^2+15}+2\,x^2 \] Input:
int((896*x + log(5*x)*(240*x^2 + 128*x^3 - 160*x^4) + 980*x^2 - 944*x^3 - 640*x^4 + 400*x^5 + 16*x^3*log(5*x)^2 + 30)/(240*x + log(5*x)*(60*x + 32*x ^2 - 40*x^3) - 236*x^2 - 160*x^3 + 100*x^4 + 4*x^2*log(5*x)^2 + 225),x)
Output:
(2*x)/(8*x + 2*x*log(5*x) - 10*x^2 + 15) + 2*x^2
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {30+896 x+980 x^2-944 x^3-640 x^4+400 x^5+\left (240 x^2+128 x^3-160 x^4\right ) \log (5 x)+16 x^3 \log ^2(5 x)}{225+240 x-236 x^2-160 x^3+100 x^4+\left (60 x+32 x^2-40 x^3\right ) \log (5 x)+4 x^2 \log ^2(5 x)} \, dx=\frac {4 \,\mathrm {log}\left (5 x \right ) x^{3}+6 \,\mathrm {log}\left (5 x \right ) x -20 x^{4}+16 x^{3}+26 x +45}{2 \,\mathrm {log}\left (5 x \right ) x -10 x^{2}+8 x +15} \] Input:
int((16*x^3*log(5*x)^2+(-160*x^4+128*x^3+240*x^2)*log(5*x)+400*x^5-640*x^4 -944*x^3+980*x^2+896*x+30)/(4*x^2*log(5*x)^2+(-40*x^3+32*x^2+60*x)*log(5*x )+100*x^4-160*x^3-236*x^2+240*x+225),x)
Output:
(4*log(5*x)*x**3 + 6*log(5*x)*x - 20*x**4 + 16*x**3 + 26*x + 45)/(2*log(5* x)*x - 10*x**2 + 8*x + 15)