Integrand size = 64, antiderivative size = 19 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=\log \left (\frac {3 \log \left (x (-x-\log (2))^2\right )}{x}\right ) \] Output:
ln(3*ln(x*(-x-ln(2))^2)/x)
Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=-\log (x)+\log \left (\log \left (x (x+\log (2))^2\right )\right ) \] Input:
Integrate[(3*x + Log[2] + (-x - Log[2])*Log[x^3 + 2*x^2*Log[2] + x*Log[2]^ 2])/((x^2 + x*Log[2])*Log[x^3 + 2*x^2*Log[2] + x*Log[2]^2]),x]
Output:
-Log[x] + Log[Log[x*(x + Log[2])^2]]
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.
\(\displaystyle \int \frac {(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )+3 x+\log (2)}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )+3 x+\log (2)}{x (x+\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )+3 x+\log (2)}{x (x+\log (2)) \log \left (x \left (x^2+2 x \log (2)+\log ^2(2)\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 x+\log (2)}{x (x+\log (2)) \log \left (x (x+\log (2))^2\right )}-\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {3 x+\log (2)}{x (x+\log (2)) \log \left (x (x+\log (2))^2\right )}dx-\log (x)\) |
Input:
Int[(3*x + Log[2] + (-x - Log[2])*Log[x^3 + 2*x^2*Log[2] + x*Log[2]^2])/(( x^2 + x*Log[2])*Log[x^3 + 2*x^2*Log[2] + x*Log[2]^2]),x]
Output:
$Aborted
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32
method | result | size |
default | \(\ln \left (\ln \left (x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+x^{3}\right )\right )-\ln \left (x \right )\) | \(25\) |
norman | \(\ln \left (\ln \left (x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+x^{3}\right )\right )-\ln \left (x \right )\) | \(25\) |
risch | \(\ln \left (\ln \left (x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+x^{3}\right )\right )-\ln \left (x \right )\) | \(25\) |
parallelrisch | \(\ln \left (\ln \left (x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+x^{3}\right )\right )-\ln \left (x \right )\) | \(25\) |
parts | \(\ln \left (\ln \left (x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+x^{3}\right )\right )-\ln \left (x \right )\) | \(25\) |
Input:
int(((-x-ln(2))*ln(x*ln(2)^2+2*x^2*ln(2)+x^3)+ln(2)+3*x)/(x*ln(2)+x^2)/ln( x*ln(2)^2+2*x^2*ln(2)+x^3),x,method=_RETURNVERBOSE)
Output:
ln(ln(x*ln(2)^2+2*x^2*ln(2)+x^3))-ln(x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=-\log \left (x\right ) + \log \left (\log \left (x^{3} + 2 \, x^{2} \log \left (2\right ) + x \log \left (2\right )^{2}\right )\right ) \] Input:
integrate(((-x-log(2))*log(x*log(2)^2+2*x^2*log(2)+x^3)+log(2)+3*x)/(x*log (2)+x^2)/log(x*log(2)^2+2*x^2*log(2)+x^3),x, algorithm="fricas")
Output:
-log(x) + log(log(x^3 + 2*x^2*log(2) + x*log(2)^2))
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=- \log {\left (x \right )} + \log {\left (\log {\left (x^{3} + 2 x^{2} \log {\left (2 \right )} + x \log {\left (2 \right )}^{2} \right )} \right )} \] Input:
integrate(((-x-ln(2))*ln(x*ln(2)**2+2*x**2*ln(2)+x**3)+ln(2)+3*x)/(x*ln(2) +x**2)/ln(x*ln(2)**2+2*x**2*ln(2)+x**3),x)
Output:
-log(x) + log(log(x**3 + 2*x**2*log(2) + x*log(2)**2))
Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=-\log \left (x\right ) + \log \left (\log \left (x + \log \left (2\right )\right ) + \frac {1}{2} \, \log \left (x\right )\right ) \] Input:
integrate(((-x-log(2))*log(x*log(2)^2+2*x^2*log(2)+x^3)+log(2)+3*x)/(x*log (2)+x^2)/log(x*log(2)^2+2*x^2*log(2)+x^3),x, algorithm="maxima")
Output:
-log(x) + log(log(x + log(2)) + 1/2*log(x))
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=-\log \left (x\right ) + \log \left (\log \left (x^{3} + 2 \, x^{2} \log \left (2\right ) + x \log \left (2\right )^{2}\right )\right ) \] Input:
integrate(((-x-log(2))*log(x*log(2)^2+2*x^2*log(2)+x^3)+log(2)+3*x)/(x*log (2)+x^2)/log(x*log(2)^2+2*x^2*log(2)+x^3),x, algorithm="giac")
Output:
-log(x) + log(log(x^3 + 2*x^2*log(2) + x*log(2)^2))
Time = 3.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=\ln \left (\ln \left (x^3+2\,\ln \left (2\right )\,x^2+{\ln \left (2\right )}^2\,x\right )\right )-\ln \left (x\right ) \] Input:
int((3*x + log(2) - log(x*log(2)^2 + 2*x^2*log(2) + x^3)*(x + log(2)))/(lo g(x*log(2)^2 + 2*x^2*log(2) + x^3)*(x*log(2) + x^2)),x)
Output:
log(log(x*log(2)^2 + 2*x^2*log(2) + x^3)) - log(x)
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {3 x+\log (2)+(-x-\log (2)) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )}{\left (x^2+x \log (2)\right ) \log \left (x^3+2 x^2 \log (2)+x \log ^2(2)\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )^{2} x +2 \,\mathrm {log}\left (2\right ) x^{2}+x^{3}\right )\right )-\mathrm {log}\left (x \right ) \] Input:
int(((-x-log(2))*log(x*log(2)^2+2*x^2*log(2)+x^3)+log(2)+3*x)/(x*log(2)+x^ 2)/log(x*log(2)^2+2*x^2*log(2)+x^3),x)
Output:
log(log(log(2)**2*x + 2*log(2)*x**2 + x**3)) - log(x)