Integrand size = 100, antiderivative size = 22 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\frac {5}{2} (3+x-\log (3+x)) \left (5+\log ^2(1+\log (x))\right ) \]
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\frac {5}{2} \left (5 x-5 \log (3+x)+(3+x-\log (3+x)) \log ^2(1+\log (x))\right ) \]
Integrate[(50*x + 25*x^2 + (50*x + 25*x^2)*Log[x] + (90 + 60*x + 10*x^2 + (-30 - 10*x)*Log[3 + x])*Log[1 + Log[x]] + (10*x + 5*x^2 + (10*x + 5*x^2)* Log[x])*Log[1 + Log[x]]^2)/(6*x + 2*x^2 + (6*x + 2*x^2)*Log[x]),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.
\(\displaystyle \int \frac {25 x^2+\left (5 x^2+\left (5 x^2+10 x\right ) \log (x)+10 x\right ) \log ^2(\log (x)+1)+\left (25 x^2+50 x\right ) \log (x)+\left (10 x^2+60 x+(-10 x-30) \log (x+3)+90\right ) \log (\log (x)+1)+50 x}{2 x^2+\left (2 x^2+6 x\right ) \log (x)+6 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {25 x^2+\left (5 x^2+\left (5 x^2+10 x\right ) \log (x)+10 x\right ) \log ^2(\log (x)+1)+\left (25 x^2+50 x\right ) \log (x)+\left (10 x^2+60 x+(-10 x-30) \log (x+3)+90\right ) \log (\log (x)+1)+50 x}{2 x (x+3) (\log (x)+1)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {5 \left (5 x^2+10 x+\left (x^2+2 x+\left (x^2+2 x\right ) \log (x)\right ) \log ^2(\log (x)+1)+5 \left (x^2+2 x\right ) \log (x)+2 \left (x^2+6 x-(x+3) \log (x+3)+9\right ) \log (\log (x)+1)\right )}{x (x+3) (\log (x)+1)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{2} \int \frac {5 x^2+10 x+\left (x^2+2 x+\left (x^2+2 x\right ) \log (x)\right ) \log ^2(\log (x)+1)+5 \left (x^2+2 x\right ) \log (x)+2 \left (x^2+6 x-(x+3) \log (x+3)+9\right ) \log (\log (x)+1)}{x (x+3) (\log (x)+1)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5}{2} \int \left (\frac {(x+2) \log ^2(\log (x)+1)}{x+3}+\frac {2 (x-\log (x+3)+3) \log (\log (x)+1)}{x (\log (x)+1)}+\frac {5 (x+2)}{x+3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{2} \left (\int \log ^2(\log (x)+1)dx-\int \frac {\log ^2(\log (x)+1)}{x+3}dx+2 \int \frac {\log (\log (x)+1)}{\log (x)+1}dx-2 \int \frac {\log (x+3) \log (\log (x)+1)}{x (\log (x)+1)}dx+5 x+3 \log ^2(\log (x)+1)-5 \log (x+3)\right )\) |
Int[(50*x + 25*x^2 + (50*x + 25*x^2)*Log[x] + (90 + 60*x + 10*x^2 + (-30 - 10*x)*Log[3 + x])*Log[1 + Log[x]] + (10*x + 5*x^2 + (10*x + 5*x^2)*Log[x] )*Log[1 + Log[x]]^2)/(6*x + 2*x^2 + (6*x + 2*x^2)*Log[x]),x]
3.25.4.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 2.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36
method | result | size |
risch | \(\left (\frac {5 x}{2}-\frac {5 \ln \left (3+x \right )}{2}+\frac {15}{2}\right ) \ln \left (\ln \left (x \right )+1\right )^{2}+\frac {25 x}{2}-\frac {25 \ln \left (3+x \right )}{2}\) | \(30\) |
parallelrisch | \(\frac {5 x \ln \left (\ln \left (x \right )+1\right )^{2}}{2}-\frac {5 \ln \left (\ln \left (x \right )+1\right )^{2} \ln \left (3+x \right )}{2}+\frac {15 \ln \left (\ln \left (x \right )+1\right )^{2}}{2}-\frac {25 \ln \left (3+x \right )}{2}+\frac {25 x}{2}-\frac {75}{4}\) | \(44\) |
int((((5*x^2+10*x)*ln(x)+5*x^2+10*x)*ln(ln(x)+1)^2+((-10*x-30)*ln(3+x)+10* x^2+60*x+90)*ln(ln(x)+1)+(25*x^2+50*x)*ln(x)+25*x^2+50*x)/((2*x^2+6*x)*ln( x)+2*x^2+6*x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\frac {5}{2} \, {\left (x - \log \left (x + 3\right ) + 3\right )} \log \left (\log \left (x\right ) + 1\right )^{2} + \frac {25}{2} \, x - \frac {25}{2} \, \log \left (x + 3\right ) \]
integrate((((5*x^2+10*x)*log(x)+5*x^2+10*x)*log(log(x)+1)^2+((-10*x-30)*lo g(3+x)+10*x^2+60*x+90)*log(log(x)+1)+(25*x^2+50*x)*log(x)+25*x^2+50*x)/((2 *x^2+6*x)*log(x)+2*x^2+6*x),x, algorithm=\
Time = 0.41 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\frac {25 x}{2} + \left (\frac {5 x}{2} - \frac {5 \log {\left (x + 3 \right )}}{2} + \frac {15}{2}\right ) \log {\left (\log {\left (x \right )} + 1 \right )}^{2} - \frac {25 \log {\left (x + 3 \right )}}{2} \]
integrate((((5*x**2+10*x)*ln(x)+5*x**2+10*x)*ln(ln(x)+1)**2+((-10*x-30)*ln (3+x)+10*x**2+60*x+90)*ln(ln(x)+1)+(25*x**2+50*x)*ln(x)+25*x**2+50*x)/((2* x**2+6*x)*ln(x)+2*x**2+6*x),x)
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\frac {5}{2} \, {\left (x - \log \left (x + 3\right ) + 3\right )} \log \left (\log \left (x\right ) + 1\right )^{2} + \frac {25}{2} \, x - \frac {25}{2} \, \log \left (x + 3\right ) \]
integrate((((5*x^2+10*x)*log(x)+5*x^2+10*x)*log(log(x)+1)^2+((-10*x-30)*lo g(3+x)+10*x^2+60*x+90)*log(log(x)+1)+(25*x^2+50*x)*log(x)+25*x^2+50*x)/((2 *x^2+6*x)*log(x)+2*x^2+6*x),x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\frac {5}{2} \, {\left (x - \log \left (x + 3\right )\right )} \log \left (\log \left (x\right ) + 1\right )^{2} + \frac {15}{2} \, \log \left (\log \left (x\right ) + 1\right )^{2} + \frac {25}{2} \, x - \frac {25}{2} \, \log \left (x + 3\right ) \]
integrate((((5*x^2+10*x)*log(x)+5*x^2+10*x)*log(log(x)+1)^2+((-10*x-30)*lo g(3+x)+10*x^2+60*x+90)*log(log(x)+1)+(25*x^2+50*x)*log(x)+25*x^2+50*x)/((2 *x^2+6*x)*log(x)+2*x^2+6*x),x, algorithm=\
Time = 13.39 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.14 \[ \int \frac {50 x+25 x^2+\left (50 x+25 x^2\right ) \log (x)+\left (90+60 x+10 x^2+(-30-10 x) \log (3+x)\right ) \log (1+\log (x))+\left (10 x+5 x^2+\left (10 x+5 x^2\right ) \log (x)\right ) \log ^2(1+\log (x))}{6 x+2 x^2+\left (6 x+2 x^2\right ) \log (x)} \, dx=\left (\frac {5\,x^3+15\,x^2}{2\,x\,\left (x+3\right )}-\frac {5\,\ln \left (x+3\right )}{2}+\frac {15}{2}\right )\,{\ln \left (\ln \left (x\right )+1\right )}^2+\frac {25\,x}{2}-\frac {25\,\ln \left (x+3\right )}{2} \]
int((50*x + log(log(x) + 1)*(60*x + 10*x^2 - log(x + 3)*(10*x + 30) + 90) + log(log(x) + 1)^2*(10*x + log(x)*(10*x + 5*x^2) + 5*x^2) + log(x)*(50*x + 25*x^2) + 25*x^2)/(6*x + log(x)*(6*x + 2*x^2) + 2*x^2),x)