Integrand size = 105, antiderivative size = 28 \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\frac {\log (x)}{4 x \left (4+5 \left (2 x+\frac {2 \log (x)}{x}\right )^2\right )} \]
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\frac {x \log (x)}{16 \left (x^2+5 x^4+10 x^2 \log (x)+5 \log ^2(x)\right )} \]
Integrate[(x^2 + 5*x^4 + (-x^2 - 15*x^4)*Log[x] + (-5 - 10*x^2)*Log[x]^2 + 5*Log[x]^3)/(16*x^4 + 160*x^6 + 400*x^8 + (320*x^4 + 1600*x^6)*Log[x] + ( 160*x^2 + 2400*x^4)*Log[x]^2 + 1600*x^2*Log[x]^3 + 400*Log[x]^4),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.
\(\displaystyle \int \frac {5 x^4+x^2+\left (-10 x^2-5\right ) \log ^2(x)+\left (-15 x^4-x^2\right ) \log (x)+5 \log ^3(x)}{400 x^8+160 x^6+16 x^4+1600 x^2 \log ^3(x)+\left (1600 x^6+320 x^4\right ) \log (x)+\left (2400 x^4+160 x^2\right ) \log ^2(x)+400 \log ^4(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 x^4+x^2+\left (-10 x^2-5\right ) \log ^2(x)+\left (-15 x^4-x^2\right ) \log (x)+5 \log ^3(x)}{16 \left (5 x^4+x^2+10 x^2 \log (x)+5 \log ^2(x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} \int \frac {5 x^4+x^2+5 \log ^3(x)-5 \left (2 x^2+1\right ) \log ^2(x)-\left (15 x^4+x^2\right ) \log (x)}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{16} \int \left (\frac {2 \left (10 x^4+10 \log (x) x^2+7 x^2+4 \log (x)+1\right ) x^2}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}+\frac {-4 x^2+\log (x)-1}{5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{16} \left (\int \frac {1}{-5 x^4-10 \log (x) x^2-x^2-5 \log ^2(x)}dx+2 \int \frac {x^2}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}dx+14 \int \frac {x^4}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}dx+8 \int \frac {x^2 \log (x)}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}dx+20 \int \frac {x^4 \log (x)}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}dx-4 \int \frac {x^2}{5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)}dx+\int \frac {\log (x)}{5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)}dx+20 \int \frac {x^6}{\left (5 x^4+10 \log (x) x^2+x^2+5 \log ^2(x)\right )^2}dx\right )\) |
Int[(x^2 + 5*x^4 + (-x^2 - 15*x^4)*Log[x] + (-5 - 10*x^2)*Log[x]^2 + 5*Log [x]^3)/(16*x^4 + 160*x^6 + 400*x^8 + (320*x^4 + 1600*x^6)*Log[x] + (160*x^ 2 + 2400*x^4)*Log[x]^2 + 1600*x^2*Log[x]^3 + 400*Log[x]^4),x]
3.22.33.3.1 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 = 0.50 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {x \ln \left (x \right )}{80 x^{4}+160 x^{2} \ln \left (x \right )+80 \ln \left (x \right )^{2}+16 x^{2}}\) | \(30\) |
risch | \(\frac {x \ln \left (x \right )}{80 x^{4}+160 x^{2} \ln \left (x \right )+80 \ln \left (x \right )^{2}+16 x^{2}}\) | \(30\) |
parallelrisch | \(\frac {x \ln \left (x \right )}{80 x^{4}+160 x^{2} \ln \left (x \right )+80 \ln \left (x \right )^{2}+16 x^{2}}\) | \(30\) |
int((5*ln(x)^3+(-10*x^2-5)*ln(x)^2+(-15*x^4-x^2)*ln(x)+5*x^4+x^2)/(400*ln( x)^4+1600*x^2*ln(x)^3+(2400*x^4+160*x^2)*ln(x)^2+(1600*x^6+320*x^4)*ln(x)+ 400*x^8+160*x^6+16*x^4),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\frac {x \log \left (x\right )}{16 \, {\left (5 \, x^{4} + 10 \, x^{2} \log \left (x\right ) + x^{2} + 5 \, \log \left (x\right )^{2}\right )}} \]
integrate((5*log(x)^3+(-10*x^2-5)*log(x)^2+(-15*x^4-x^2)*log(x)+5*x^4+x^2) /(400*log(x)^4+1600*x^2*log(x)^3+(2400*x^4+160*x^2)*log(x)^2+(1600*x^6+320 *x^4)*log(x)+400*x^8+160*x^6+16*x^4),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\frac {x \log {\left (x \right )}}{80 x^{4} + 160 x^{2} \log {\left (x \right )} + 16 x^{2} + 80 \log {\left (x \right )}^{2}} \]
integrate((5*ln(x)**3+(-10*x**2-5)*ln(x)**2+(-15*x**4-x**2)*ln(x)+5*x**4+x **2)/(400*ln(x)**4+1600*x**2*ln(x)**3+(2400*x**4+160*x**2)*ln(x)**2+(1600* x**6+320*x**4)*ln(x)+400*x**8+160*x**6+16*x**4),x)
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\frac {x \log \left (x\right )}{16 \, {\left (5 \, x^{4} + 10 \, x^{2} \log \left (x\right ) + x^{2} + 5 \, \log \left (x\right )^{2}\right )}} \]
integrate((5*log(x)^3+(-10*x^2-5)*log(x)^2+(-15*x^4-x^2)*log(x)+5*x^4+x^2) /(400*log(x)^4+1600*x^2*log(x)^3+(2400*x^4+160*x^2)*log(x)^2+(1600*x^6+320 *x^4)*log(x)+400*x^8+160*x^6+16*x^4),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\frac {x \log \left (x\right )}{16 \, {\left (5 \, x^{4} + 10 \, x^{2} \log \left (x\right ) + x^{2} + 5 \, \log \left (x\right )^{2}\right )}} \]
integrate((5*log(x)^3+(-10*x^2-5)*log(x)^2+(-15*x^4-x^2)*log(x)+5*x^4+x^2) /(400*log(x)^4+1600*x^2*log(x)^3+(2400*x^4+160*x^2)*log(x)^2+(1600*x^6+320 *x^4)*log(x)+400*x^8+160*x^6+16*x^4),x, algorithm=\
Timed out. \[ \int \frac {x^2+5 x^4+\left (-x^2-15 x^4\right ) \log (x)+\left (-5-10 x^2\right ) \log ^2(x)+5 \log ^3(x)}{16 x^4+160 x^6+400 x^8+\left (320 x^4+1600 x^6\right ) \log (x)+\left (160 x^2+2400 x^4\right ) \log ^2(x)+1600 x^2 \log ^3(x)+400 \log ^4(x)} \, dx=\int \frac {5\,{\ln \left (x\right )}^3-{\ln \left (x\right )}^2\,\left (10\,x^2+5\right )-\ln \left (x\right )\,\left (15\,x^4+x^2\right )+x^2+5\,x^4}{\ln \left (x\right )\,\left (1600\,x^6+320\,x^4\right )+400\,{\ln \left (x\right )}^4+{\ln \left (x\right )}^2\,\left (2400\,x^4+160\,x^2\right )+1600\,x^2\,{\ln \left (x\right )}^3+16\,x^4+160\,x^6+400\,x^8} \,d x \]
int((5*log(x)^3 - log(x)^2*(10*x^2 + 5) - log(x)*(x^2 + 15*x^4) + x^2 + 5* x^4)/(log(x)*(320*x^4 + 1600*x^6) + 400*log(x)^4 + log(x)^2*(160*x^2 + 240 0*x^4) + 1600*x^2*log(x)^3 + 16*x^4 + 160*x^6 + 400*x^8),x)