Integrand size = 63, antiderivative size = 24 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=\frac {10 \left (4+\frac {3}{-\frac {x}{4}+\log (x)}\right )}{\log ^2(5) \log (x)} \] Output:
10*(4+3/(ln(x)-1/4*x))/ln(5)^2/ln(x)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=-\frac {40 (3-x+4 \log (x))}{\log ^2(5) \left (x \log (x)-4 \log ^2(x)\right )} \] Input:
Integrate[(120*x - 40*x^2 + (-960 + 440*x)*Log[x] - 640*Log[x]^2)/(x^3*Log [5]^2*Log[x]^2 - 8*x^2*Log[5]^2*Log[x]^3 + 16*x*Log[5]^2*Log[x]^4),x]
Output:
(-40*(3 - x + 4*Log[x]))/(Log[5]^2*(x*Log[x] - 4*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.
\(\displaystyle \int \frac {-40 x^2+120 x-640 \log ^2(x)+(440 x-960) \log (x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {40 \left (-x^2+3 x-16 \log ^2(x)+11 x \log (x)-24 \log (x)\right )}{x \log ^2(5) (x-4 \log (x))^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {40 \int \frac {-x^2+11 \log (x) x+3 x-16 \log ^2(x)-24 \log (x)}{x (x-4 \log (x))^2 \log ^2(x)}dx}{\log ^2(5)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {40 \int \left (\frac {3-x}{x^2 \log ^2(x)}+\frac {12}{x^2 (x-4 \log (x))}+\frac {3}{x^2 \log (x)}+\frac {12 (x-4)}{x^2 (x-4 \log (x))^2}\right )dx}{\log ^2(5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {40 \left (-48 \int \frac {1}{x^2 (x-4 \log (x))^2}dx+12 \int \frac {1}{x^2 (x-4 \log (x))}dx+12 \int \frac {1}{x (x-4 \log (x))^2}dx-\frac {3}{x \log (x)}+\frac {1}{\log (x)}\right )}{\log ^2(5)}\) |
Input:
Int[(120*x - 40*x^2 + (-960 + 440*x)*Log[x] - 640*Log[x]^2)/(x^3*Log[5]^2* Log[x]^2 - 8*x^2*Log[5]^2*Log[x]^3 + 16*x*Log[5]^2*Log[x]^4),x]
Output:
$Aborted
Time = 0.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\frac {-120-160 \ln \left (x \right )+40 x}{\ln \left (5\right )^{2} \left (x -4 \ln \left (x \right )\right ) \ln \left (x \right )}\) | \(26\) |
default | \(-\frac {40 \left (-3-4 \ln \left (x \right )+x \right )}{\ln \left (5\right )^{2} \left (4 \ln \left (x \right )-x \right ) \ln \left (x \right )}\) | \(28\) |
parallelrisch | \(\frac {-480+160 x -640 \ln \left (x \right )}{4 \ln \left (5\right )^{2} \ln \left (x \right ) \left (x -4 \ln \left (x \right )\right )}\) | \(28\) |
norman | \(\frac {-\frac {120}{\ln \left (5\right )}+\frac {40 x}{\ln \left (5\right )}-\frac {160 \ln \left (x \right )}{\ln \left (5\right )}}{\ln \left (5\right ) \ln \left (x \right ) \left (x -4 \ln \left (x \right )\right )}\) | \(40\) |
Input:
int((-640*ln(x)^2+(440*x-960)*ln(x)-40*x^2+120*x)/(16*x*ln(5)^2*ln(x)^4-8* x^2*ln(5)^2*ln(x)^3+x^3*ln(5)^2*ln(x)^2),x,method=_RETURNVERBOSE)
Output:
40*(-3-4*ln(x)+x)/ln(5)^2/(x-4*ln(x))/ln(x)
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=\frac {40 \, {\left (x - 4 \, \log \left (x\right ) - 3\right )}}{x \log \left (5\right )^{2} \log \left (x\right ) - 4 \, \log \left (5\right )^{2} \log \left (x\right )^{2}} \] Input:
integrate((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*l og(x)^4-8*x^2*log(5)^2*log(x)^3+x^3*log(5)^2*log(x)^2),x, algorithm="frica s")
Output:
40*(x - 4*log(x) - 3)/(x*log(5)^2*log(x) - 4*log(5)^2*log(x)^2)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=\frac {- 40 x + 160 \log {\left (x \right )} + 120}{- x \log {\left (5 \right )}^{2} \log {\left (x \right )} + 4 \log {\left (5 \right )}^{2} \log {\left (x \right )}^{2}} \] Input:
integrate((-640*ln(x)**2+(440*x-960)*ln(x)-40*x**2+120*x)/(16*x*ln(5)**2*l n(x)**4-8*x**2*ln(5)**2*ln(x)**3+x**3*ln(5)**2*ln(x)**2),x)
Output:
(-40*x + 160*log(x) + 120)/(-x*log(5)**2*log(x) + 4*log(5)**2*log(x)**2)
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=\frac {40 \, {\left (x - 4 \, \log \left (x\right ) - 3\right )}}{x \log \left (5\right )^{2} \log \left (x\right ) - 4 \, \log \left (5\right )^{2} \log \left (x\right )^{2}} \] Input:
integrate((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*l og(x)^4-8*x^2*log(5)^2*log(x)^3+x^3*log(5)^2*log(x)^2),x, algorithm="maxim a")
Output:
40*(x - 4*log(x) - 3)/(x*log(5)^2*log(x) - 4*log(5)^2*log(x)^2)
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=-\frac {480}{x^{2} \log \left (5\right )^{2} - 4 \, x \log \left (5\right )^{2} \log \left (x\right )} + \frac {40 \, {\left (x - 3\right )}}{x \log \left (5\right )^{2} \log \left (x\right )} \] Input:
integrate((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*l og(x)^4-8*x^2*log(5)^2*log(x)^3+x^3*log(5)^2*log(x)^2),x, algorithm="giac" )
Output:
-480/(x^2*log(5)^2 - 4*x*log(5)^2*log(x)) + 40*(x - 3)/(x*log(5)^2*log(x))
Time = 3.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=-\frac {40\,\left (4\,\ln \left (x\right )-x+3\right )}{{\ln \left (5\right )}^2\,\ln \left (x\right )\,\left (x-4\,\ln \left (x\right )\right )} \] Input:
int((120*x - 640*log(x)^2 + log(x)*(440*x - 960) - 40*x^2)/(x^3*log(5)^2*l og(x)^2 - 8*x^2*log(5)^2*log(x)^3 + 16*x*log(5)^2*log(x)^4),x)
Output:
-(40*(4*log(x) - x + 3))/(log(5)^2*log(x)*(x - 4*log(x)))
Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {120 x-40 x^2+(-960+440 x) \log (x)-640 \log ^2(x)}{x^3 \log ^2(5) \log ^2(x)-8 x^2 \log ^2(5) \log ^3(x)+16 x \log ^2(5) \log ^4(x)} \, dx=\frac {160 \,\mathrm {log}\left (x \right )-40 x +120}{\mathrm {log}\left (x \right ) \mathrm {log}\left (5\right )^{2} \left (4 \,\mathrm {log}\left (x \right )-x \right )} \] Input:
int((-640*log(x)^2+(440*x-960)*log(x)-40*x^2+120*x)/(16*x*log(5)^2*log(x)^ 4-8*x^2*log(5)^2*log(x)^3+x^3*log(5)^2*log(x)^2),x)
Output:
(40*(4*log(x) - x + 3))/(log(x)*log(5)**2*(4*log(x) - x))