Integrand size = 77, antiderivative size = 22 \[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\frac {5 \left (x+\frac {x}{\log (4)}\right ) \log ^2\left (x+\log ^2(x)\right )}{x^2} \]
\[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx \]
Integrate[((10*x + 10*x*Log[4] + (20 + 20*Log[4])*Log[x])*Log[x + Log[x]^2 ] + (-5*x - 5*x*Log[4] + (-5 - 5*Log[4])*Log[x]^2)*Log[x + Log[x]^2]^2)/(x ^3*Log[4] + x^2*Log[4]*Log[x]^2),x]
Integrate[((10*x + 10*x*Log[4] + (20 + 20*Log[4])*Log[x])*Log[x + Log[x]^2 ] + (-5*x - 5*x*Log[4] + (-5 - 5*Log[4])*Log[x]^2)*Log[x + Log[x]^2]^2)/(x ^3*Log[4] + x^2*Log[4]*Log[x]^2), 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.
\(\displaystyle \int \frac {\left (-5 x+(-5-5 \log (4)) \log ^2(x)-5 x \log (4)\right ) \log ^2\left (x+\log ^2(x)\right )+(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {\left (-5 x+(-5-5 \log (4)) \log ^2(x)-5 x \log (4)\right ) \log ^2\left (x+\log ^2(x)\right )+(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )}{x^2 \left (\log (4) \log ^2(x)+x \log (4)\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 (1+\log (4)) \log \left (x+\log ^2(x)\right ) \left (2 x-\log \left (x+\log ^2(x)\right ) \log ^2(x)-x \log \left (x+\log ^2(x)\right )+4 \log (x)\right )}{x^2 \log (4) \left (x+\log ^2(x)\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 (1+\log (4)) \int \frac {\log \left (\log ^2(x)+x\right ) \left (-\log \left (\log ^2(x)+x\right ) \log ^2(x)+4 \log (x)+2 x-x \log \left (\log ^2(x)+x\right )\right )}{x^2 \left (\log ^2(x)+x\right )}dx}{\log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5 (1+\log (4)) \int \left (\frac {2 (x+2 \log (x)) \log \left (\log ^2(x)+x\right )}{x^2 \left (\log ^2(x)+x\right )}-\frac {\log ^2\left (\log ^2(x)+x\right )}{x^2}\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 (1+\log (4)) \left (4 \int \frac {\log (x) \log \left (\log ^2(x)+x\right )}{x^2 \left (\log ^2(x)+x\right )}dx-\int \frac {\log ^2\left (\log ^2(x)+x\right )}{x^2}dx+2 \int \frac {\log \left (\log ^2(x)+x\right )}{x \left (\log ^2(x)+x\right )}dx\right )}{\log (4)}\) |
Int[((10*x + 10*x*Log[4] + (20 + 20*Log[4])*Log[x])*Log[x + Log[x]^2] + (- 5*x - 5*x*Log[4] + (-5 - 5*Log[4])*Log[x]^2)*Log[x + Log[x]^2]^2)/(x^3*Log [4] + x^2*Log[4]*Log[x]^2),x]
3.6.66.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 3.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {5 \left (1+2 \ln \left (2\right )\right ) \ln \left (\ln \left (x \right )^{2}+x \right )^{2}}{2 x \ln \left (2\right )}\) | \(25\) |
parallelrisch | \(\frac {10 \ln \left (2\right ) \ln \left (\ln \left (x \right )^{2}+x \right )^{2}+5 \ln \left (\ln \left (x \right )^{2}+x \right )^{2}}{2 x \ln \left (2\right )}\) | \(35\) |
int((((-10*ln(2)-5)*ln(x)^2-10*x*ln(2)-5*x)*ln(ln(x)^2+x)^2+((40*ln(2)+20) *ln(x)+20*x*ln(2)+10*x)*ln(ln(x)^2+x))/(2*x^2*ln(2)*ln(x)^2+2*x^3*ln(2)),x ,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\frac {5 \, {\left (2 \, \log \left (2\right ) + 1\right )} \log \left (\log \left (x\right )^{2} + x\right )^{2}}{2 \, x \log \left (2\right )} \]
integrate((((-10*log(2)-5)*log(x)^2-10*x*log(2)-5*x)*log(log(x)^2+x)^2+((4 0*log(2)+20)*log(x)+20*x*log(2)+10*x)*log(log(x)^2+x))/(2*x^2*log(2)*log(x )^2+2*x^3*log(2)),x, algorithm=\
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\frac {\left (5 + 10 \log {\left (2 \right )}\right ) \log {\left (x + \log {\left (x \right )}^{2} \right )}^{2}}{2 x \log {\left (2 \right )}} \]
integrate((((-10*ln(2)-5)*ln(x)**2-10*x*ln(2)-5*x)*ln(ln(x)**2+x)**2+((40* ln(2)+20)*ln(x)+20*x*ln(2)+10*x)*ln(ln(x)**2+x))/(2*x**2*ln(2)*ln(x)**2+2* x**3*ln(2)),x)
Time = 0.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\frac {5 \, {\left (2 \, \log \left (2\right ) + 1\right )} \log \left (\log \left (x\right )^{2} + x\right )^{2}}{2 \, x \log \left (2\right )} \]
integrate((((-10*log(2)-5)*log(x)^2-10*x*log(2)-5*x)*log(log(x)^2+x)^2+((4 0*log(2)+20)*log(x)+20*x*log(2)+10*x)*log(log(x)^2+x))/(2*x^2*log(2)*log(x )^2+2*x^3*log(2)),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\frac {5 \, {\left (2 \, \log \left (2\right ) + 1\right )} \log \left (\log \left (x\right )^{2} + x\right )^{2}}{2 \, x \log \left (2\right )} \]
integrate((((-10*log(2)-5)*log(x)^2-10*x*log(2)-5*x)*log(log(x)^2+x)^2+((4 0*log(2)+20)*log(x)+20*x*log(2)+10*x)*log(log(x)^2+x))/(2*x^2*log(2)*log(x )^2+2*x^3*log(2)),x, algorithm=\
Time = 9.78 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {(10 x+10 x \log (4)+(20+20 \log (4)) \log (x)) \log \left (x+\log ^2(x)\right )+\left (-5 x-5 x \log (4)+(-5-5 \log (4)) \log ^2(x)\right ) \log ^2\left (x+\log ^2(x)\right )}{x^3 \log (4)+x^2 \log (4) \log ^2(x)} \, dx=\frac {{\ln \left ({\ln \left (x\right )}^2+x\right )}^2\,\left (10\,\ln \left (2\right )+5\right )}{2\,x\,\ln \left (2\right )} \]