Integrand size = 113, antiderivative size = 24 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=\frac {12}{x (4+\log (x) (1+x-2 x (5+\log (5 x))))} \] Output:
12/((1+x-2*(ln(5*x)+5)*x)*ln(x)+4)/x
Time = 5.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=-\frac {12}{x (-4+\log (x) (-1+9 x+2 x \log (5 x)))} \] Input:
Integrate[(-60 + 108*x + (-12 + 240*x)*Log[x] + (24*x + 48*x*Log[x])*Log[5 *x])/(16*x^2 + (8*x^2 - 72*x^3)*Log[x] + (x^2 - 18*x^3 + 81*x^4)*Log[x]^2 + (-16*x^3*Log[x] + (-4*x^3 + 36*x^4)*Log[x]^2)*Log[5*x] + 4*x^4*Log[x]^2* Log[5*x]^2),x]
Output:
-12/(x*(-4 + Log[x]*(-1 + 9*x + 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.
\(\displaystyle \int \frac {108 x+(240 x-12) \log (x)+(24 x+48 x \log (x)) \log (5 x)-60}{4 x^4 \log ^2(x) \log ^2(5 x)+16 x^2+\left (\left (36 x^4-4 x^3\right ) \log ^2(x)-16 x^3 \log (x)\right ) \log (5 x)+\left (8 x^2-72 x^3\right ) \log (x)+\left (81 x^4-18 x^3+x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {12 (9 x+2 x \log (5 x)+\log (x) (20 x+4 x \log (5 x)-1)-5)}{x^2 (4-\log (x) (9 x+2 x \log (5 x)-1))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 12 \int -\frac {-2 \log (5 x) x-9 x+\log (x) (-4 \log (5 x) x-20 x+1)+5}{x^2 (\log (x) (-2 \log (5 x) x-9 x+1)+4)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -12 \int \frac {-2 \log (5 x) x-9 x+\log (x) (-4 \log (5 x) x-20 x+1)+5}{x^2 (\log (x) (-2 \log (5 x) x-9 x+1)+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -12 \int \left (\frac {-2 \log (x)-1}{x^2 \log (x) (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)}+\frac {-2 x \log ^2(x)-\log ^2(x)-4 \log (x)-4}{x^2 \log (x) (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -12 \left (-4 \int \frac {1}{x^2 (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)^2}dx-4 \int \frac {1}{x^2 \log (x) (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)^2}dx-\int \frac {\log (x)}{x^2 (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)^2}dx-2 \int \frac {1}{x^2 (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)}dx-\int \frac {1}{x^2 \log (x) (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)}dx-2 \int \frac {\log (x)}{x (9 x \log (x)+2 x \log (5 x) \log (x)-\log (x)-4)^2}dx\right )\) |
Input:
Int[(-60 + 108*x + (-12 + 240*x)*Log[x] + (24*x + 48*x*Log[x])*Log[5*x])/( 16*x^2 + (8*x^2 - 72*x^3)*Log[x] + (x^2 - 18*x^3 + 81*x^4)*Log[x]^2 + (-16 *x^3*Log[x] + (-4*x^3 + 36*x^4)*Log[x]^2)*Log[5*x] + 4*x^4*Log[x]^2*Log[5* x]^2),x]
Output:
$Aborted
Time = 1.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17
method | result | size |
parallelrisch | \(-\frac {12}{x \left (2 \ln \left (5 x \right ) x \ln \left (x \right )+9 x \ln \left (x \right )-\ln \left (x \right )-4\right )}\) | \(28\) |
default | \(-\frac {12}{x \left (2 x \ln \left (5\right ) \ln \left (x \right )+2 x \ln \left (x \right )^{2}+9 x \ln \left (x \right )-\ln \left (x \right )-4\right )}\) | \(33\) |
risch | \(-\frac {12 i}{x \left (2 i \ln \left (5\right ) x \ln \left (x \right )+2 i x \ln \left (x \right )^{2}+9 i x \ln \left (x \right )-i \ln \left (x \right )-4 i\right )}\) | \(39\) |
Input:
int(((48*x*ln(x)+24*x)*ln(5*x)+(240*x-12)*ln(x)+108*x-60)/(4*x^4*ln(x)^2*l n(5*x)^2+((36*x^4-4*x^3)*ln(x)^2-16*x^3*ln(x))*ln(5*x)+(81*x^4-18*x^3+x^2) *ln(x)^2+(-72*x^3+8*x^2)*ln(x)+16*x^2),x,method=_RETURNVERBOSE)
Output:
-12/x/(2*ln(5*x)*x*ln(x)+9*x*ln(x)-ln(x)-4)
Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=-\frac {12}{2 \, x^{2} \log \left (x\right )^{2} + {\left (2 \, x^{2} \log \left (5\right ) + 9 \, x^{2} - x\right )} \log \left (x\right ) - 4 \, x} \] Input:
integrate(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4* log(x)^2*log(5*x)^2+((36*x^4-4*x^3)*log(x)^2-16*x^3*log(x))*log(5*x)+(81*x ^4-18*x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x, algorithm="frica s")
Output:
-12/(2*x^2*log(x)^2 + (2*x^2*log(5) + 9*x^2 - x)*log(x) - 4*x)
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=- \frac {12}{2 x^{2} \log {\left (x \right )}^{2} - 4 x + \left (2 x^{2} \log {\left (5 \right )} + 9 x^{2} - x\right ) \log {\left (x \right )}} \] Input:
integrate(((48*x*ln(x)+24*x)*ln(5*x)+(240*x-12)*ln(x)+108*x-60)/(4*x**4*ln (x)**2*ln(5*x)**2+((36*x**4-4*x**3)*ln(x)**2-16*x**3*ln(x))*ln(5*x)+(81*x* *4-18*x**3+x**2)*ln(x)**2+(-72*x**3+8*x**2)*ln(x)+16*x**2),x)
Output:
-12/(2*x**2*log(x)**2 - 4*x + (2*x**2*log(5) + 9*x**2 - x)*log(x))
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=-\frac {12}{2 \, x^{2} \log \left (x\right )^{2} + {\left (x^{2} {\left (2 \, \log \left (5\right ) + 9\right )} - x\right )} \log \left (x\right ) - 4 \, x} \] Input:
integrate(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4* log(x)^2*log(5*x)^2+((36*x^4-4*x^3)*log(x)^2-16*x^3*log(x))*log(5*x)+(81*x ^4-18*x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x, algorithm="maxim a")
Output:
-12/(2*x^2*log(x)^2 + (x^2*(2*log(5) + 9) - x)*log(x) - 4*x)
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=-\frac {12}{2 \, x^{2} \log \left (5\right ) \log \left (x\right ) + 2 \, x^{2} \log \left (x\right )^{2} + 9 \, x^{2} \log \left (x\right ) - x \log \left (x\right ) - 4 \, x} \] Input:
integrate(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4* log(x)^2*log(5*x)^2+((36*x^4-4*x^3)*log(x)^2-16*x^3*log(x))*log(5*x)+(81*x ^4-18*x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x, algorithm="giac" )
Output:
-12/(2*x^2*log(5)*log(x) + 2*x^2*log(x)^2 + 9*x^2*log(x) - x*log(x) - 4*x)
Timed out. \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=\int \frac {108\,x+\ln \left (x\right )\,\left (240\,x-12\right )+\ln \left (5\,x\right )\,\left (24\,x+48\,x\,\ln \left (x\right )\right )-60}{\ln \left (x\right )\,\left (8\,x^2-72\,x^3\right )+{\ln \left (x\right )}^2\,\left (81\,x^4-18\,x^3+x^2\right )-\ln \left (5\,x\right )\,\left (16\,x^3\,\ln \left (x\right )+{\ln \left (x\right )}^2\,\left (4\,x^3-36\,x^4\right )\right )+16\,x^2+4\,x^4\,{\ln \left (5\,x\right )}^2\,{\ln \left (x\right )}^2} \,d x \] Input:
int((108*x + log(x)*(240*x - 12) + log(5*x)*(24*x + 48*x*log(x)) - 60)/(lo g(x)*(8*x^2 - 72*x^3) + log(x)^2*(x^2 - 18*x^3 + 81*x^4) - log(5*x)*(16*x^ 3*log(x) + log(x)^2*(4*x^3 - 36*x^4)) + 16*x^2 + 4*x^4*log(5*x)^2*log(x)^2 ),x)
Output:
int((108*x + log(x)*(240*x - 12) + log(5*x)*(24*x + 48*x*log(x)) - 60)/(lo g(x)*(8*x^2 - 72*x^3) + log(x)^2*(x^2 - 18*x^3 + 81*x^4) - log(5*x)*(16*x^ 3*log(x) + log(x)^2*(4*x^3 - 36*x^4)) + 16*x^2 + 4*x^4*log(5*x)^2*log(x)^2 ), x)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx=-\frac {12}{x \left (2 \,\mathrm {log}\left (5 x \right ) \mathrm {log}\left (x \right ) x +9 \,\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )-4\right )} \] Input:
int(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4*log(x) ^2*log(5*x)^2+((36*x^4-4*x^3)*log(x)^2-16*x^3*log(x))*log(5*x)+(81*x^4-18* x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x)
Output:
( - 12)/(x*(2*log(5*x)*log(x)*x + 9*log(x)*x - log(x) - 4))