Integrand size = 76, antiderivative size = 21 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=28+\frac {\log (4)}{x \log (x) (5+2 x+\log (x))} \] Output:
28+2*ln(2)/x/ln(x)/(2*x+5+ln(x))
Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=\frac {\log (4)}{x \log (x) (5+2 x+\log (x))} \] Input:
Integrate[((-5 - 2*x)*Log[4] + (-7 - 4*x)*Log[4]*Log[x] - Log[4]*Log[x]^2) /((25*x^2 + 20*x^3 + 4*x^4)*Log[x]^2 + (10*x^2 + 4*x^3)*Log[x]^3 + x^2*Log [x]^4),x]
Output:
Log[4]/(x*Log[x]*(5 + 2*x + Log[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.
\(\displaystyle \int \frac {-\log (4) \log ^2(x)+(-4 x-7) \log (4) \log (x)+(-2 x-5) \log (4)}{x^2 \log ^4(x)+\left (4 x^3+10 x^2\right ) \log ^3(x)+\left (4 x^4+20 x^3+25 x^2\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log (4) \left (-2 x-\log ^2(x)-(4 x+7) \log (x)-5\right )}{x^2 \log ^2(x) (2 x+\log (x)+5)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (4) \int -\frac {\log ^2(x)+(4 x+7) \log (x)+2 x+5}{x^2 \log ^2(x) (2 x+\log (x)+5)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\log (4) \int \frac {\log ^2(x)+(4 x+7) \log (x)+2 x+5}{x^2 \log ^2(x) (2 x+\log (x)+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\log (4) \int \left (\frac {-4 x-5}{x^2 (2 x+5)^2 (2 x+\log (x)+5)}+\frac {4 x+5}{x^2 (2 x+5)^2 \log (x)}+\frac {1}{x^2 (2 x+5) \log ^2(x)}+\frac {-2 x-1}{x^2 (2 x+5) (2 x+\log (x)+5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\log (4) \left (\int \frac {1}{x^2 (2 x+5) \log ^2(x)}dx+\int \frac {4 x+5}{x^2 (2 x+5)^2 \log (x)}dx-\frac {1}{5} \int \frac {1}{x^2 (2 x+\log (x)+5)^2}dx-\frac {1}{5} \int \frac {1}{x^2 (2 x+\log (x)+5)}dx-\frac {8}{25} \int \frac {1}{x (2 x+\log (x)+5)^2}dx+\frac {16}{25} \int \frac {1}{(2 x+5) (2 x+\log (x)+5)^2}dx+\frac {4}{5} \int \frac {1}{(2 x+5)^2 (2 x+\log (x)+5)}dx\right )\) |
Input:
Int[((-5 - 2*x)*Log[4] + (-7 - 4*x)*Log[4]*Log[x] - Log[4]*Log[x]^2)/((25* x^2 + 20*x^3 + 4*x^4)*Log[x]^2 + (10*x^2 + 4*x^3)*Log[x]^3 + x^2*Log[x]^4) ,x]
Output:
$Aborted
Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\) | \(21\) |
norman | \(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\) | \(21\) |
risch | \(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\) | \(21\) |
parallelrisch | \(\frac {2 \ln \left (2\right )}{x \ln \left (x \right ) \left (2 x +5+\ln \left (x \right )\right )}\) | \(21\) |
Input:
int((-2*ln(2)*ln(x)^2+2*(-4*x-7)*ln(2)*ln(x)+2*(-2*x-5)*ln(2))/(x^2*ln(x)^ 4+(4*x^3+10*x^2)*ln(x)^3+(4*x^4+20*x^3+25*x^2)*ln(x)^2),x,method=_RETURNVE RBOSE)
Output:
2*ln(2)/x/ln(x)/(2*x+5+ln(x))
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=\frac {2 \, \log \left (2\right )}{x \log \left (x\right )^{2} + {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x\right )} \] Input:
integrate((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/ (x^2*log(x)^4+(4*x^3+10*x^2)*log(x)^3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x, a lgorithm="fricas")
Output:
2*log(2)/(x*log(x)^2 + (2*x^2 + 5*x)*log(x))
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=\frac {2 \log {\left (2 \right )}}{x \log {\left (x \right )}^{2} + \left (2 x^{2} + 5 x\right ) \log {\left (x \right )}} \] Input:
integrate((-2*ln(2)*ln(x)**2+2*(-4*x-7)*ln(2)*ln(x)+2*(-2*x-5)*ln(2))/(x** 2*ln(x)**4+(4*x**3+10*x**2)*ln(x)**3+(4*x**4+20*x**3+25*x**2)*ln(x)**2),x)
Output:
2*log(2)/(x*log(x)**2 + (2*x**2 + 5*x)*log(x))
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=\frac {2 \, \log \left (2\right )}{x \log \left (x\right )^{2} + {\left (2 \, x^{2} + 5 \, x\right )} \log \left (x\right )} \] Input:
integrate((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/ (x^2*log(x)^4+(4*x^3+10*x^2)*log(x)^3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x, a lgorithm="maxima")
Output:
2*log(2)/(x*log(x)^2 + (2*x^2 + 5*x)*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.48 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=-\frac {2 \, \log \left (2\right )}{4 \, x^{3} + 2 \, x^{2} \log \left (x\right ) + 20 \, x^{2} + 5 \, x \log \left (x\right ) + 25 \, x} + \frac {2 \, \log \left (2\right )}{2 \, x^{2} \log \left (x\right ) + 5 \, x \log \left (x\right )} \] Input:
integrate((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/ (x^2*log(x)^4+(4*x^3+10*x^2)*log(x)^3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x, a lgorithm="giac")
Output:
-2*log(2)/(4*x^3 + 2*x^2*log(x) + 20*x^2 + 5*x*log(x) + 25*x) + 2*log(2)/( 2*x^2*log(x) + 5*x*log(x))
Time = 0.38 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.76 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=\frac {2\,\left (5\,\ln \left (2\right )+2\,x\,\ln \left (2\right )\right )}{x\,\left ({\ln \left (x\right )}^2+\left (2\,x+5\right )\,\ln \left (x\right )\right )\,\left (2\,x+5\right )} \] Input:
int(-(2*log(2)*(2*x + 5) + 2*log(2)*log(x)^2 + 2*log(2)*log(x)*(4*x + 7))/ (log(x)^3*(10*x^2 + 4*x^3) + x^2*log(x)^4 + log(x)^2*(25*x^2 + 20*x^3 + 4* x^4)),x)
Output:
(2*(5*log(2) + 2*x*log(2)))/(x*(log(x)^2 + log(x)*(2*x + 5))*(2*x + 5))
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {(-5-2 x) \log (4)+(-7-4 x) \log (4) \log (x)-\log (4) \log ^2(x)}{\left (25 x^2+20 x^3+4 x^4\right ) \log ^2(x)+\left (10 x^2+4 x^3\right ) \log ^3(x)+x^2 \log ^4(x)} \, dx=\frac {2 \,\mathrm {log}\left (2\right )}{\mathrm {log}\left (x \right ) x \left (\mathrm {log}\left (x \right )+2 x +5\right )} \] Input:
int((-2*log(2)*log(x)^2+2*(-4*x-7)*log(2)*log(x)+2*(-2*x-5)*log(2))/(x^2*l og(x)^4+(4*x^3+10*x^2)*log(x)^3+(4*x^4+20*x^3+25*x^2)*log(x)^2),x)
Output:
(2*log(2))/(log(x)*x*(log(x) + 2*x + 5))