Integrand size = 83, antiderivative size = 25 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=-4+\frac {4}{\log (x)}-\log \left (2+\frac {\log \left (x+5 x^2\right )}{x}\right ) \] Output:
4/ln(x)-4-ln(2+ln(5*x^2+x)/x)
Time = 0.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=\frac {4}{\log (x)}+\log (x)-\log (2 x+\log (x (1+5 x))) \] Input:
Integrate[(-8*x - 40*x^2 + (-1 - 10*x)*Log[x]^2 + (-4 - 20*x + (1 + 5*x)*L og[x]^2)*Log[x + 5*x^2])/((2*x^2 + 10*x^3)*Log[x]^2 + (x + 5*x^2)*Log[x]^2 *Log[x + 5*x^2]),x]
Output:
4/Log[x] + Log[x] - Log[2*x + Log[x*(1 + 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.
\(\displaystyle \int \frac {-40 x^2+\left (-20 x+(5 x+1) \log ^2(x)-4\right ) \log \left (5 x^2+x\right )-8 x+(-10 x-1) \log ^2(x)}{\left (5 x^2+x\right ) \log \left (5 x^2+x\right ) \log ^2(x)+\left (10 x^3+2 x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-40 x^2+\left (-20 x+(5 x+1) \log ^2(x)-4\right ) \log \left (5 x^2+x\right )-8 x+(-10 x-1) \log ^2(x)}{x (5 x+1) \log ^2(x) (2 x+\log (x (5 x+1)))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-40 x^2-8 x-10 x \log ^2(x)-\log ^2(x)}{x (5 x+1) \log ^2(x) (2 x+\log (x (5 x+1)))}+\frac {(\log (x)-2) (\log (x)+2) \log (x (5 x+1))}{x \log ^2(x) (2 x+\log (x (5 x+1)))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{2 x+\log (x (5 x+1))}dx-\int \frac {1}{x (2 x+\log (x (5 x+1)))}dx-5 \int \frac {1}{(5 x+1) (2 x+\log (x (5 x+1)))}dx+\log (x)+\frac {4}{\log (x)}\) |
Input:
Int[(-8*x - 40*x^2 + (-1 - 10*x)*Log[x]^2 + (-4 - 20*x + (1 + 5*x)*Log[x]^ 2)*Log[x + 5*x^2])/((2*x^2 + 10*x^3)*Log[x]^2 + (x + 5*x^2)*Log[x]^2*Log[x + 5*x^2]),x]
Output:
$Aborted
Time = 3.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
default | \(\ln \left (x \right )-\ln \left (2 x +\ln \left (5 x^{2}+x \right )\right )+\frac {4}{\ln \left (x \right )}\) | \(25\) |
parts | \(\ln \left (x \right )-\ln \left (2 x +\ln \left (5 x^{2}+x \right )\right )+\frac {4}{\ln \left (x \right )}\) | \(25\) |
parallelrisch | \(\frac {100-25 \ln \left (x +\frac {\ln \left (5 x^{2}+x \right )}{2}\right ) \ln \left (x \right )+25 \ln \left (x \right )^{2}}{25 \ln \left (x \right )}\) | \(32\) |
risch | \(\frac {\ln \left (x \right )^{2}+4}{\ln \left (x \right )}-\ln \left (\ln \left (\frac {1}{5}+x \right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i \left (\frac {1}{5}+x \right )\right ) \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (\frac {1}{5}+x \right )\right ) \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{3}+\pi \operatorname {csgn}\left (i x \left (\frac {1}{5}+x \right )\right )^{2} \operatorname {csgn}\left (i x \right )-4 i x -2 i \ln \left (x \right )\right )}{2}\right )\) | \(106\) |
Input:
int((((1+5*x)*ln(x)^2-20*x-4)*ln(5*x^2+x)+(-10*x-1)*ln(x)^2-40*x^2-8*x)/(( 5*x^2+x)*ln(x)^2*ln(5*x^2+x)+(10*x^3+2*x^2)*ln(x)^2),x,method=_RETURNVERBO SE)
Output:
ln(x)-ln(2*x+ln(5*x^2+x))+4/ln(x)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=-\frac {\log \left (2 \, x + \log \left (5 \, x^{2} + x\right )\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 4}{\log \left (x\right )} \] Input:
integrate((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^ 2-8*x)/((5*x^2+x)*log(x)^2*log(5*x^2+x)+(10*x^3+2*x^2)*log(x)^2),x, algori thm="fricas")
Output:
-(log(2*x + log(5*x^2 + x))*log(x) - log(x)^2 - 4)/log(x)
Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=\log {\left (x \right )} - \log {\left (2 x + \log {\left (5 x^{2} + x \right )} \right )} + \frac {4}{\log {\left (x \right )}} \] Input:
integrate((((1+5*x)*ln(x)**2-20*x-4)*ln(5*x**2+x)+(-10*x-1)*ln(x)**2-40*x* *2-8*x)/((5*x**2+x)*ln(x)**2*ln(5*x**2+x)+(10*x**3+2*x**2)*ln(x)**2),x)
Output:
log(x) - log(2*x + log(5*x**2 + x)) + 4/log(x)
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=\frac {4}{\log \left (x\right )} - \log \left (2 \, x + \log \left (5 \, x + 1\right ) + \log \left (x\right )\right ) + \log \left (x\right ) \] Input:
integrate((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^ 2-8*x)/((5*x^2+x)*log(x)^2*log(5*x^2+x)+(10*x^3+2*x^2)*log(x)^2),x, algori thm="maxima")
Output:
4/log(x) - log(2*x + log(5*x + 1) + log(x)) + log(x)
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=\frac {4}{\log \left (x\right )} - \log \left (2 \, x + \log \left (5 \, x + 1\right ) + \log \left (x\right )\right ) + \log \left (x\right ) \] Input:
integrate((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^ 2-8*x)/((5*x^2+x)*log(x)^2*log(5*x^2+x)+(10*x^3+2*x^2)*log(x)^2),x, algori thm="giac")
Output:
4/log(x) - log(2*x + log(5*x + 1) + log(x)) + log(x)
Time = 7.88 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=\ln \left (x\right )-\ln \left (2\,x+\ln \left (x\,\left (5\,x+1\right )\right )\right )+\frac {4}{\ln \left (x\right )} \] Input:
int(-(8*x + 40*x^2 + log(x)^2*(10*x + 1) + log(x + 5*x^2)*(20*x - log(x)^2 *(5*x + 1) + 4))/(log(x)^2*(2*x^2 + 10*x^3) + log(x + 5*x^2)*log(x)^2*(x + 5*x^2)),x)
Output:
log(x) - log(2*x + log(x*(5*x + 1))) + 4/log(x)
Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {-8 x-40 x^2+(-1-10 x) \log ^2(x)+\left (-4-20 x+(1+5 x) \log ^2(x)\right ) \log \left (x+5 x^2\right )}{\left (2 x^2+10 x^3\right ) \log ^2(x)+\left (x+5 x^2\right ) \log ^2(x) \log \left (x+5 x^2\right )} \, dx=\frac {-\mathrm {log}\left (\mathrm {log}\left (5 x^{2}+x \right )+2 x \right ) \mathrm {log}\left (x \right )+\mathrm {log}\left (x \right )^{2}+4}{\mathrm {log}\left (x \right )} \] Input:
int((((1+5*x)*log(x)^2-20*x-4)*log(5*x^2+x)+(-10*x-1)*log(x)^2-40*x^2-8*x) /((5*x^2+x)*log(x)^2*log(5*x^2+x)+(10*x^3+2*x^2)*log(x)^2),x)
Output:
( - log(log(5*x**2 + x) + 2*x)*log(x) + log(x)**2 + 4)/log(x)