Integrand size = 96, antiderivative size = 23 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=\frac {\log \left (\frac {1}{4 x \log \left (x^2 (x+\log (x))\right )}\right )}{x} \]
Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=\frac {\log \left (\frac {1}{4 x \log \left (x^2 (x+\log (x))\right )}\right )}{x} \]
Integrate[(-1 - 3*x - 2*Log[x] + (-x - Log[x])*Log[x^3 + x^2*Log[x]] + (-x - Log[x])*Log[x^3 + x^2*Log[x]]*Log[1/(4*x*Log[x^3 + x^2*Log[x]])])/((x^3 + x^2*Log[x])*Log[x^3 + x^2*Log[x]]),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.
\(\displaystyle \int \frac {(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )-3 x-2 \log (x)-1}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )-3 x-2 \log (x)-1}{x^2 (x+\log (x)) \log \left (x^3+x^2 \log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \left (-\log \left (x^2 (x+\log (x))\right )\right )-\log (x) \log \left (x^2 (x+\log (x))\right )-3 x-2 \log (x)-1}{x^2 (x+\log (x)) \log \left (x^3+x^2 \log (x)\right )}-\frac {\log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {1}{x^2 (-x-\log (x)) \log \left (x^3+\log (x) x^2\right )}dx+3 \int \frac {1}{x (-x-\log (x)) \log \left (x^3+\log (x) x^2\right )}dx+2 \int \frac {\log (x)}{x^2 (-x-\log (x)) \log \left (x^3+\log (x) x^2\right )}dx+\int \frac {1}{x^2 (x+\log (x)) \log \left (x^3+\log (x) x^2\right )}dx+3 \int \frac {1}{x (x+\log (x)) \log \left (x^3+\log (x) x^2\right )}dx+2 \int \frac {\log (x)}{x^2 (x+\log (x)) \log \left (x^3+\log (x) x^2\right )}dx+\frac {\log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{x}\) |
Int[(-1 - 3*x - 2*Log[x] + (-x - Log[x])*Log[x^3 + x^2*Log[x]] + (-x - Log [x])*Log[x^3 + x^2*Log[x]]*Log[1/(4*x*Log[x^3 + x^2*Log[x]])])/((x^3 + x^2 *Log[x])*Log[x^3 + x^2*Log[x]]),x]
3.20.58.3.1 Defintions of rubi rules used
Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.)) ^(p_.), x_Symbol] :> Int[u*x^(p*r)*(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]
Time = 1.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {1}{4 x \ln \left (x^{2} \left (x +\ln \left (x \right )\right )\right )}\right )}{x}\) | \(22\) |
risch | \(\text {Expression too large to display}\) | \(2102\) |
int(((-x-ln(x))*ln(x^2*ln(x)+x^3)*ln(1/4/x/ln(x^2*ln(x)+x^3))+(-x-ln(x))*l n(x^2*ln(x)+x^3)-2*ln(x)-3*x-1)/(x^2*ln(x)+x^3)/ln(x^2*ln(x)+x^3),x,method =_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=\frac {\log \left (\frac {1}{4 \, x \log \left (x^{3} + x^{2} \log \left (x\right )\right )}\right )}{x} \]
integrate(((-x-log(x))*log(x^2*log(x)+x^3)*log(1/4/x/log(x^2*log(x)+x^3))+ (-x-log(x))*log(x^2*log(x)+x^3)-2*log(x)-3*x-1)/(x^2*log(x)+x^3)/log(x^2*l og(x)+x^3),x, algorithm=\
Time = 0.84 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=\frac {\log {\left (\frac {1}{4 x \log {\left (x^{3} + x^{2} \log {\left (x \right )} \right )}} \right )}}{x} \]
integrate(((-x-ln(x))*ln(x**2*ln(x)+x**3)*ln(1/4/x/ln(x**2*ln(x)+x**3))+(- x-ln(x))*ln(x**2*ln(x)+x**3)-2*ln(x)-3*x-1)/(x**2*ln(x)+x**3)/ln(x**2*ln(x )+x**3),x)
Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=-\frac {2 \, \log \left (2\right ) + \log \left (x\right ) + \log \left (\log \left (x + \log \left (x\right )\right ) + 2 \, \log \left (x\right )\right )}{x} \]
integrate(((-x-log(x))*log(x^2*log(x)+x^3)*log(1/4/x/log(x^2*log(x)+x^3))+ (-x-log(x))*log(x^2*log(x)+x^3)-2*log(x)-3*x-1)/(x^2*log(x)+x^3)/log(x^2*l og(x)+x^3),x, algorithm=\
Time = 0.46 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=-\frac {\log \left (x\right )}{x} - \frac {\log \left (4 \, \log \left (x + \log \left (x\right )\right ) + 8 \, \log \left (x\right )\right )}{x} \]
integrate(((-x-log(x))*log(x^2*log(x)+x^3)*log(1/4/x/log(x^2*log(x)+x^3))+ (-x-log(x))*log(x^2*log(x)+x^3)-2*log(x)-3*x-1)/(x^2*log(x)+x^3)/log(x^2*l og(x)+x^3),x, algorithm=\
Time = 14.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-1-3 x-2 \log (x)+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right )+(-x-\log (x)) \log \left (x^3+x^2 \log (x)\right ) \log \left (\frac {1}{4 x \log \left (x^3+x^2 \log (x)\right )}\right )}{\left (x^3+x^2 \log (x)\right ) \log \left (x^3+x^2 \log (x)\right )} \, dx=\frac {\ln \left (\frac {1}{4\,x\,\ln \left (x^2\,\ln \left (x\right )+x^3\right )}\right )}{x} \]