Integrand size = 69, antiderivative size = 18 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=-\frac {5 \log (3)}{32 (-3+\log (x)) \log (1+x)} \] Output:
5/32*ln(3)/ln(1+x)/(3-ln(x))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=\frac {5 \log (3)}{32 (3-\log (x)) \log (1+x)} \] Input:
Integrate[(-15*x*Log[3] + 5*x*Log[3]*Log[x] + (5 + 5*x)*Log[3]*Log[1 + x]) /((288*x + 288*x^2 + (-192*x - 192*x^2)*Log[x] + (32*x + 32*x^2)*Log[x]^2) *Log[1 + x]^2),x]
Output:
(5*Log[3])/(32*(3 - Log[x])*Log[1 + 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.
\(\displaystyle \int \frac {5 x \log (3) \log (x)-15 x \log (3)+(5 x+5) \log (3) \log (x+1)}{\left (288 x^2+\left (32 x^2+32 x\right ) \log ^2(x)+\left (-192 x^2-192 x\right ) \log (x)+288 x\right ) \log ^2(x+1)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 \log (3) (-3 x+x \log (x)+x \log (x+1)+\log (x+1))}{32 x (x+1) (3-\log (x))^2 \log ^2(x+1)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{32} \log (3) \int -\frac {-\log (x) x-\log (x+1) x+3 x-\log (x+1)}{x (x+1) (3-\log (x))^2 \log ^2(x+1)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {5}{32} \log (3) \int \frac {-\log (x) x-\log (x+1) x+3 x-\log (x+1)}{x (x+1) (3-\log (x))^2 \log ^2(x+1)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5}{32} \log (3) \int \left (-\frac {1}{(x+1) \log ^2(x+1) (\log (x)-3)}-\frac {1}{x \log (x+1) (\log (x)-3)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{32} \log (3) \left (-\int \frac {1}{(x+1) (\log (x)-3) \log ^2(x+1)}dx-\int \frac {1}{x (\log (x)-3)^2 \log (x+1)}dx\right )\) |
Input:
Int[(-15*x*Log[3] + 5*x*Log[3]*Log[x] + (5 + 5*x)*Log[3]*Log[1 + x])/((288 *x + 288*x^2 + (-192*x - 192*x^2)*Log[x] + (32*x + 32*x^2)*Log[x]^2)*Log[1 + x]^2),x]
Output:
$Aborted
Time = 2.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {5 \ln \left (3\right )}{32 \left (\ln \left (x \right )-3\right ) \ln \left (1+x \right )}\) | \(17\) |
parallelrisch | \(-\frac {5 \ln \left (3\right )}{32 \left (\ln \left (x \right )-3\right ) \ln \left (1+x \right )}\) | \(17\) |
Input:
int(((5*x+5)*ln(3)*ln(1+x)+5*x*ln(3)*ln(x)-15*x*ln(3))/((32*x^2+32*x)*ln(x )^2+(-192*x^2-192*x)*ln(x)+288*x^2+288*x)/ln(1+x)^2,x,method=_RETURNVERBOS E)
Output:
-5/32*ln(3)/(ln(x)-3)/ln(1+x)
Time = 0.12 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=-\frac {5 \, \log \left (3\right )}{32 \, {\left (\log \left (x\right ) - 3\right )} \log \left (x + 1\right )} \] Input:
integrate(((5*x+5)*log(3)*log(1+x)+5*x*log(3)*log(x)-15*x*log(3))/((32*x^2 +32*x)*log(x)^2+(-192*x^2-192*x)*log(x)+288*x^2+288*x)/log(1+x)^2,x, algor ithm="fricas")
Output:
-5/32*log(3)/((log(x) - 3)*log(x + 1))
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=- \frac {5 \log {\left (3 \right )}}{\left (32 \log {\left (x \right )} - 96\right ) \log {\left (x + 1 \right )}} \] Input:
integrate(((5*x+5)*ln(3)*ln(1+x)+5*x*ln(3)*ln(x)-15*x*ln(3))/((32*x**2+32* x)*ln(x)**2+(-192*x**2-192*x)*ln(x)+288*x**2+288*x)/ln(1+x)**2,x)
Output:
-5*log(3)/((32*log(x) - 96)*log(x + 1))
Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=-\frac {5 \, \log \left (3\right )}{32 \, {\left (\log \left (x\right ) - 3\right )} \log \left (x + 1\right )} \] Input:
integrate(((5*x+5)*log(3)*log(1+x)+5*x*log(3)*log(x)-15*x*log(3))/((32*x^2 +32*x)*log(x)^2+(-192*x^2-192*x)*log(x)+288*x^2+288*x)/log(1+x)^2,x, algor ithm="maxima")
Output:
-5/32*log(3)/((log(x) - 3)*log(x + 1))
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=-\frac {5 \, \log \left (3\right )}{32 \, {\left (\log \left (x + 1\right ) \log \left (x\right ) - 3 \, \log \left (x + 1\right )\right )}} \] Input:
integrate(((5*x+5)*log(3)*log(1+x)+5*x*log(3)*log(x)-15*x*log(3))/((32*x^2 +32*x)*log(x)^2+(-192*x^2-192*x)*log(x)+288*x^2+288*x)/log(1+x)^2,x, algor ithm="giac")
Output:
-5/32*log(3)/(log(x + 1)*log(x) - 3*log(x + 1))
Time = 4.66 (sec) , antiderivative size = 110, normalized size of antiderivative = 6.11 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=\frac {\frac {5\,\ln \left (3\right )}{32\,x}-\frac {5\,\ln \left (3\right )\,\ln \left (x\right )}{64\,x}}{\ln \left (x\right )-3}-\frac {\frac {5\,\ln \left (3\right )}{32\,\left (\ln \left (x\right )-3\right )}+\frac {5\,\ln \left (x+1\right )\,\ln \left (3\right )\,\left (x+1\right )}{32\,x\,{\left (\ln \left (x\right )-3\right )}^2}}{\ln \left (x+1\right )}+\frac {5\,\ln \left (3\right )}{64\,x}-\frac {\frac {5\,\left (\ln \left (3\right )-2\,x\,\ln \left (3\right )\right )}{64\,x}-\frac {5\,\ln \left (3\right )\,\ln \left (x\right )}{64\,x}}{{\ln \left (x\right )}^2-6\,\ln \left (x\right )+9} \] Input:
int((5*x*log(3)*log(x) - 15*x*log(3) + log(x + 1)*log(3)*(5*x + 5))/(log(x + 1)^2*(288*x + log(x)^2*(32*x + 32*x^2) - log(x)*(192*x + 192*x^2) + 288 *x^2)),x)
Output:
((5*log(3))/(32*x) - (5*log(3)*log(x))/(64*x))/(log(x) - 3) - ((5*log(3))/ (32*(log(x) - 3)) + (5*log(x + 1)*log(3)*(x + 1))/(32*x*(log(x) - 3)^2))/l og(x + 1) + (5*log(3))/(64*x) - ((5*(log(3) - 2*x*log(3)))/(64*x) - (5*log (3)*log(x))/(64*x))/(log(x)^2 - 6*log(x) + 9)
Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-15 x \log (3)+5 x \log (3) \log (x)+(5+5 x) \log (3) \log (1+x)}{\left (288 x+288 x^2+\left (-192 x-192 x^2\right ) \log (x)+\left (32 x+32 x^2\right ) \log ^2(x)\right ) \log ^2(1+x)} \, dx=-\frac {5 \,\mathrm {log}\left (3\right )}{32 \,\mathrm {log}\left (x +1\right ) \left (\mathrm {log}\left (x \right )-3\right )} \] Input:
int(((5*x+5)*log(3)*log(1+x)+5*x*log(3)*log(x)-15*x*log(3))/((32*x^2+32*x) *log(x)^2+(-192*x^2-192*x)*log(x)+288*x^2+288*x)/log(1+x)^2,x)
Output:
( - 5*log(3))/(32*log(x + 1)*(log(x) - 3))