Integrand size = 81, antiderivative size = 22 \[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\frac {\log ^2\left (\frac {4 \left (-5+\frac {\log (x)}{x}\right )^2}{x^2}\right )}{x^2} \]
\[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx \]
Integrate[((4 + 20*x - 8*Log[x])*Log[(100*x^2 - 40*x*Log[x] + 4*Log[x]^2)/ x^4] + (10*x - 2*Log[x])*Log[(100*x^2 - 40*x*Log[x] + 4*Log[x]^2)/x^4]^2)/ (-5*x^4 + x^3*Log[x]),x]
Integrate[((4 + 20*x - 8*Log[x])*Log[(100*x^2 - 40*x*Log[x] + 4*Log[x]^2)/ x^4] + (10*x - 2*Log[x])*Log[(100*x^2 - 40*x*Log[x] + 4*Log[x]^2)/x^4]^2)/ (-5*x^4 + x^3*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.
\(\displaystyle \int \frac {(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2+4 \log ^2(x)-40 x \log (x)}{x^4}\right )+(20 x-8 \log (x)+4) \log \left (\frac {100 x^2+4 \log ^2(x)-40 x \log (x)}{x^4}\right )}{x^3 \log (x)-5 x^4} \, dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2+4 \log ^2(x)-40 x \log (x)}{x^4}\right )+(20 x-8 \log (x)+4) \log \left (\frac {100 x^2+4 \log ^2(x)-40 x \log (x)}{x^4}\right )}{x^3 (\log (x)-5 x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right ) \left (-5 x \log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )+\log (x) \log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )-10 x+4 \log (x)-2\right )}{x^3 (5 x-\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {\log \left (\frac {4 (5 x-\log (x))^2}{x^4}\right ) \left (5 \log \left (\frac {4 (5 x-\log (x))^2}{x^4}\right ) x+10 x-4 \log (x)-\log (x) \log \left (\frac {4 (5 x-\log (x))^2}{x^4}\right )+2\right )}{x^3 (5 x-\log (x))}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {\log \left (\frac {4 (5 x-\log (x))^2}{x^4}\right ) \left (5 \log \left (\frac {4 (5 x-\log (x))^2}{x^4}\right ) x+10 x-4 \log (x)-\log (x) \log \left (\frac {4 (5 x-\log (x))^2}{x^4}\right )+2\right )}{x^3 (5 x-\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {\log ^2\left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )}{x^3}+\frac {2 (5 x-2 \log (x)+1) \log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )}{x^3 (5 x-\log (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\int \frac {\log ^2\left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )}{x^3}dx+2 \int \frac {\log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )}{x^3 (5 x-\log (x))}dx-4 \int \frac {\log (x) \log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )}{x^3 (5 x-\log (x))}dx+10 \int \frac {\log \left (\frac {4 (\log (x)-5 x)^2}{x^4}\right )}{x^2 (5 x-\log (x))}dx\right )\) |
Int[((4 + 20*x - 8*Log[x])*Log[(100*x^2 - 40*x*Log[x] + 4*Log[x]^2)/x^4] + (10*x - 2*Log[x])*Log[(100*x^2 - 40*x*Log[x] + 4*Log[x]^2)/x^4]^2)/(-5*x^ 4 + x^3*Log[x]),x]
3.23.45.3.1 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 = 1.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {4 \ln \left (x \right )^{2}-40 x \ln \left (x \right )+100 x^{2}}{x^{4}}\right )^{2}}{x^{2}}\) | \(28\) |
risch | \(\text {Expression too large to display}\) | \(7795\) |
int(((-2*ln(x)+10*x)*ln((4*ln(x)^2-40*x*ln(x)+100*x^2)/x^4)^2+(-8*ln(x)+20 *x+4)*ln((4*ln(x)^2-40*x*ln(x)+100*x^2)/x^4))/(x^3*ln(x)-5*x^4),x,method=_ RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\frac {\log \left (\frac {4 \, {\left (25 \, x^{2} - 10 \, x \log \left (x\right ) + \log \left (x\right )^{2}\right )}}{x^{4}}\right )^{2}}{x^{2}} \]
integrate(((-2*log(x)+10*x)*log((4*log(x)^2-40*x*log(x)+100*x^2)/x^4)^2+(- 8*log(x)+20*x+4)*log((4*log(x)^2-40*x*log(x)+100*x^2)/x^4))/(x^3*log(x)-5* x^4),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\frac {\log {\left (\frac {100 x^{2} - 40 x \log {\left (x \right )} + 4 \log {\left (x \right )}^{2}}{x^{4}} \right )}^{2}}{x^{2}} \]
integrate(((-2*ln(x)+10*x)*ln((4*ln(x)**2-40*x*ln(x)+100*x**2)/x**4)**2+(- 8*ln(x)+20*x+4)*ln((4*ln(x)**2-40*x*ln(x)+100*x**2)/x**4))/(x**3*ln(x)-5*x **4),x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (22) = 44\).
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.14 \[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\frac {4 \, {\left (\log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (x\right ) + 4 \, \log \left (x\right )^{2} + 2 \, {\left (\log \left (2\right ) - 2 \, \log \left (x\right )\right )} \log \left (-5 \, x + \log \left (x\right )\right ) + \log \left (-5 \, x + \log \left (x\right )\right )^{2}\right )}}{x^{2}} \]
integrate(((-2*log(x)+10*x)*log((4*log(x)^2-40*x*log(x)+100*x^2)/x^4)^2+(- 8*log(x)+20*x+4)*log((4*log(x)^2-40*x*log(x)+100*x^2)/x^4))/(x^3*log(x)-5* x^4),x, algorithm=\
4*(log(2)^2 - 4*log(2)*log(x) + 4*log(x)^2 + 2*(log(2) - 2*log(x))*log(-5* x + log(x)) + log(-5*x + log(x))^2)/x^2
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (22) = 44\).
Time = 0.82 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\frac {\log \left (100 \, x^{2} - 40 \, x \log \left (x\right ) + 4 \, \log \left (x\right )^{2}\right )^{2}}{x^{2}} - \frac {8 \, \log \left (100 \, x^{2} - 40 \, x \log \left (x\right ) + 4 \, \log \left (x\right )^{2}\right ) \log \left (x\right )}{x^{2}} + \frac {16 \, \log \left (x\right )^{2}}{x^{2}} \]
integrate(((-2*log(x)+10*x)*log((4*log(x)^2-40*x*log(x)+100*x^2)/x^4)^2+(- 8*log(x)+20*x+4)*log((4*log(x)^2-40*x*log(x)+100*x^2)/x^4))/(x^3*log(x)-5* x^4),x, algorithm=\
log(100*x^2 - 40*x*log(x) + 4*log(x)^2)^2/x^2 - 8*log(100*x^2 - 40*x*log(x ) + 4*log(x)^2)*log(x)/x^2 + 16*log(x)^2/x^2
Time = 14.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {(4+20 x-8 \log (x)) \log \left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )+(10 x-2 \log (x)) \log ^2\left (\frac {100 x^2-40 x \log (x)+4 \log ^2(x)}{x^4}\right )}{-5 x^4+x^3 \log (x)} \, dx=\frac {{\ln \left (\frac {4\,\left (25\,x^2-10\,x\,\ln \left (x\right )+{\ln \left (x\right )}^2\right )}{x^4}\right )}^2}{x^2} \]