Integrand size = 61, antiderivative size = 22 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=-2 x-\frac {\left (5+x^2\right )^2}{64 \log \left (\frac {1}{\log (x)}\right )} \] Output:
-2*x-1/8*(x^2+5)*(1/8*x^2+5/8)/ln(1/ln(x))
Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=-2 x-\frac {\left (5+x^2\right )^2}{64 \log \left (\frac {1}{\log (x)}\right )} \] Input:
Integrate[(-25 - 10*x^2 - x^4 + (-20*x^2 - 4*x^4)*Log[x]*Log[Log[x]^(-1)] - 128*x*Log[x]*Log[Log[x]^(-1)]^2)/(64*x*Log[x]*Log[Log[x]^(-1)]^2),x]
Output:
-2*x - (5 + x^2)^2/(64*Log[Log[x]^(-1)])
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.
\(\displaystyle \int \frac {-x^4-10 x^2+\left (-4 x^4-20 x^2\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )-25}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{64} \int -\frac {x^4+10 x^2+128 \log (x) \log ^2\left (\frac {1}{\log (x)}\right ) x+4 \left (x^4+5 x^2\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )+25}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{64} \int \frac {x^4+10 x^2+128 \log (x) \log ^2\left (\frac {1}{\log (x)}\right ) x+4 \left (x^4+5 x^2\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )+25}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{64} \int \left (\frac {\left (x^2+5\right )^2}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}+\frac {4 x \left (x^2+5\right )}{\log \left (\frac {1}{\log (x)}\right )}+128\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{64} \left (-\int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )}dx-4 \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )}dx-10 \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )}dx-20 \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )}dx-128 x-\frac {25}{\log \left (\frac {1}{\log (x)}\right )}\right )\) |
Input:
Int[(-25 - 10*x^2 - x^4 + (-20*x^2 - 4*x^4)*Log[x]*Log[Log[x]^(-1)] - 128* x*Log[x]*Log[Log[x]^(-1)]^2)/(64*x*Log[x]*Log[Log[x]^(-1)]^2),x]
Output:
$Aborted
Time = 0.61 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-2 x +\frac {x^{4}+10 x^{2}+25}{64 \ln \left (\ln \left (x \right )\right )}\) | \(22\) |
parallelrisch | \(-\frac {x^{4}+25+10 x^{2}+128 \ln \left (\frac {1}{\ln \left (x \right )}\right ) x}{64 \ln \left (\frac {1}{\ln \left (x \right )}\right )}\) | \(28\) |
Input:
int(1/64*(-128*x*ln(x)*ln(1/ln(x))^2+(-4*x^4-20*x^2)*ln(x)*ln(1/ln(x))-x^4 -10*x^2-25)/x/ln(x)/ln(1/ln(x))^2,x,method=_RETURNVERBOSE)
Output:
-2*x+1/64*(x^4+10*x^2+25)/ln(ln(x))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=-\frac {x^{4} + 10 \, x^{2} + 128 \, x \log \left (\frac {1}{\log \left (x\right )}\right ) + 25}{64 \, \log \left (\frac {1}{\log \left (x\right )}\right )} \] Input:
integrate(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1 /log(x))-x^4-10*x^2-25)/x/log(x)/log(1/log(x))^2,x, algorithm="fricas")
Output:
-1/64*(x^4 + 10*x^2 + 128*x*log(1/log(x)) + 25)/log(1/log(x))
Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=- 2 x + \frac {- x^{4} - 10 x^{2} - 25}{64 \log {\left (\frac {1}{\log {\left (x \right )}} \right )}} \] Input:
integrate(1/64*(-128*x*ln(x)*ln(1/ln(x))**2+(-4*x**4-20*x**2)*ln(x)*ln(1/l n(x))-x**4-10*x**2-25)/x/ln(x)/ln(1/ln(x))**2,x)
Output:
-2*x + (-x**4 - 10*x**2 - 25)/(64*log(1/log(x)))
Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=-2 \, x + \frac {x^{4} + 10 \, x^{2}}{64 \, \log \left (\log \left (x\right )\right )} + \frac {25}{64 \, \log \left (\log \left (x\right )\right )} \] Input:
integrate(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1 /log(x))-x^4-10*x^2-25)/x/log(x)/log(1/log(x))^2,x, algorithm="maxima")
Output:
-2*x + 1/64*(x^4 + 10*x^2)/log(log(x)) + 25/64/log(log(x))
Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=-2 \, x + \frac {x^{4} + 10 \, x^{2} + 25}{64 \, \log \left (\log \left (x\right )\right )} \] Input:
integrate(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1 /log(x))-x^4-10*x^2-25)/x/log(x)/log(1/log(x))^2,x, algorithm="giac")
Output:
-2*x + 1/64*(x^4 + 10*x^2 + 25)/log(log(x))
Time = 3.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=-\frac {128\,x\,\ln \left (\frac {1}{\ln \left (x\right )}\right )+10\,x^2+x^4+25}{64\,\ln \left (\frac {1}{\ln \left (x\right )}\right )} \] Input:
int(-((5*x^2)/32 + x^4/64 + (log(1/log(x))*log(x)*(20*x^2 + 4*x^4))/64 + 2 *x*log(1/log(x))^2*log(x) + 25/64)/(x*log(1/log(x))^2*log(x)),x)
Output:
-(128*x*log(1/log(x)) + 10*x^2 + x^4 + 25)/(64*log(1/log(x)))
Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx=\frac {-128 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right ) x +x^{4}+10 x^{2}+25}{64 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )} \] Input:
int(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1/log(x ))-x^4-10*x^2-25)/x/log(x)/log(1/log(x))^2,x)
Output:
( - 128*log(log(x))*x + x**4 + 10*x**2 + 25)/(64*log(log(x)))