Integrand size = 58, antiderivative size = 25 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\frac {x \log (x)}{5 \left (4+\frac {4 (2+3 x-\log (6))}{x}\right )} \] Output:
1/5*x*ln(x)/(4/x*(3*x-ln(6)+2)+4)
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\frac {x^2 \log (x)}{20 (2+4 x-\log (6))} \] Input:
Integrate[(2*x + 4*x^2 - x*Log[6] + (4*x + 4*x^2 - 2*x*Log[6])*Log[x])/(80 + 320*x + 320*x^2 + (-80 - 160*x)*Log[6] + 20*Log[6]^2),x]
Output:
(x^2*Log[x])/(20*(2 + 4*x - Log[6]))
Time = 0.66 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6, 7292, 7277, 27, 7293, 2009}
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.
\(\displaystyle \int \frac {4 x^2+\left (4 x^2+4 x-2 x \log (6)\right ) \log (x)+2 x-x \log (6)}{320 x^2+320 x+(-160 x-80) \log (6)+80+20 \log ^2(6)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {4 x^2+\left (4 x^2+4 x-2 x \log (6)\right ) \log (x)+x (2-\log (6))}{320 x^2+320 x+(-160 x-80) \log (6)+80+20 \log ^2(6)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 x^2+\left (4 x^2+4 x-2 x \log (6)\right ) \log (x)+x (2-\log (6))}{320 x^2+160 x (2-\log (6))+20 (2-\log (6))^2}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 1280 \int \frac {4 x^2+(2-\log (6)) x+2 \left (2 x^2-\log (6) x+2 x\right ) \log (x)}{25600 (4 x-\log (6)+2)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \int \frac {4 x^2+(2-\log (6)) x+2 \left (2 x^2-\log (6) x+2 x\right ) \log (x)}{(4 x-\log (6)+2)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{20} \int \left (\frac {2 (2 x-\log (6)+2) \log (x) x}{(4 x-\log (6)+2)^2}+\frac {x}{4 x-\log (6)+2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{20} \left (\frac {1}{4} x \log (x)-\frac {x (2-\log (6)) \log (x)}{4 (4 x+2-\log (6))}\right )\) |
Input:
Int[(2*x + 4*x^2 - x*Log[6] + (4*x + 4*x^2 - 2*x*Log[6])*Log[x])/(80 + 320 *x + 320*x^2 + (-80 - 160*x)*Log[6] + 20*Log[6]^2),x]
Output:
((x*Log[x])/4 - (x*(2 - Log[6])*Log[x])/(4*(2 + 4*x - Log[6])))/20
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.89 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68
method | result | size |
norman | \(-\frac {x^{2} \ln \left (x \right )}{20 \left (\ln \left (6\right )-4 x -2\right )}\) | \(17\) |
parallelrisch | \(-\frac {x^{2} \ln \left (x \right )}{20 \left (\ln \left (6\right )-4 x -2\right )}\) | \(17\) |
default | \(-\frac {\left (-\frac {\ln \left (6\right )}{16}+\frac {1}{8}\right ) \ln \left (-\ln \left (6\right )+4 x +2\right )}{20}+\frac {x \ln \left (x \right )}{80}-\frac {\left (\frac {\ln \left (\ln \left (6\right )-4 x -2\right )}{4 \ln \left (6\right )-8}+\frac {\ln \left (x \right ) x}{\left (\ln \left (6\right )-2\right ) \left (\ln \left (6\right )-4 x -2\right )}\right ) \left (\ln \left (6\right )^{2}-4 \ln \left (6\right )+4\right )}{80}\) | \(73\) |
parts | \(-\frac {\left (-\frac {\ln \left (6\right )}{16}+\frac {1}{8}\right ) \ln \left (-\ln \left (6\right )+4 x +2\right )}{20}+\frac {x \ln \left (x \right )}{80}-\frac {\left (\frac {\ln \left (\ln \left (6\right )-4 x -2\right )}{4 \ln \left (6\right )-8}+\frac {\ln \left (x \right ) x}{\left (\ln \left (6\right )-2\right ) \left (\ln \left (6\right )-4 x -2\right )}\right ) \left (\ln \left (6\right )^{2}-4 \ln \left (6\right )+4\right )}{80}\) | \(73\) |
risch | \(-\frac {\left (\ln \left (3\right )^{2}+2 \ln \left (2\right ) \ln \left (3\right )-4 x \ln \left (3\right )+\ln \left (2\right )^{2}-4 x \ln \left (2\right )+16 x^{2}-4 \ln \left (3\right )-4 \ln \left (2\right )+8 x +4\right ) \ln \left (x \right )}{320 \left (\ln \left (3\right )+\ln \left (2\right )-4 x -2\right )}+\frac {\ln \left (2\right ) \ln \left (x \right )}{320}+\frac {\ln \left (3\right ) \ln \left (x \right )}{320}-\frac {\ln \left (x \right )}{160}\) | \(75\) |
Input:
int(((-2*x*ln(6)+4*x^2+4*x)*ln(x)-x*ln(6)+4*x^2+2*x)/(20*ln(6)^2+(-160*x-8 0)*ln(6)+320*x^2+320*x+80),x,method=_RETURNVERBOSE)
Output:
-1/20*x^2*ln(x)/(ln(6)-4*x-2)
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\frac {x^{2} \log \left (x\right )}{20 \, {\left (4 \, x - \log \left (6\right ) + 2\right )}} \] Input:
integrate(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2 +(-160*x-80)*log(6)+320*x^2+320*x+80),x, algorithm="fricas")
Output:
1/20*x^2*log(x)/(4*x - log(6) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).
Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\left (- \frac {1}{160} + \frac {\log {\left (6 \right )}}{320}\right ) \log {\left (x \right )} + \frac {\left (16 x^{2} - 4 x \log {\left (6 \right )} + 8 x - 4 \log {\left (6 \right )} + \log {\left (6 \right )}^{2} + 4\right ) \log {\left (x \right )}}{1280 x - 320 \log {\left (6 \right )} + 640} \] Input:
integrate(((-2*x*ln(6)+4*x**2+4*x)*ln(x)-x*ln(6)+4*x**2+2*x)/(20*ln(6)**2+ (-160*x-80)*ln(6)+320*x**2+320*x+80),x)
Output:
(-1/160 + log(6)/320)*log(x) + (16*x**2 - 4*x*log(6) + 8*x - 4*log(6) + lo g(6)**2 + 4)*log(x)/(1280*x - 320*log(6) + 640)
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (22) = 44\).
Time = 0.19 (sec) , antiderivative size = 209, normalized size of antiderivative = 8.36 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\frac {1}{320} \, {\left (\frac {\log \left (6\right ) - 2}{4 \, x - \log \left (6\right ) + 2} - \log \left (4 \, x - \log \left (6\right ) + 2\right )\right )} \log \left (6\right ) + \frac {1}{160} \, {\left (\log \left (6\right ) - 2\right )} \log \left (4 \, x - \log \left (6\right ) + 2\right ) - \frac {1}{320} \, {\left (\log \left (3\right ) + \log \left (2\right ) - 2\right )} \log \left (4 \, x - \log \left (3\right ) - \log \left (2\right ) + 2\right ) + \frac {1}{320} \, {\left (\log \left (3\right ) + \log \left (2\right ) - 2\right )} \log \left (x\right ) + \frac {1}{80} \, x - \frac {\log \left (6\right )^{2} - 4 \, \log \left (6\right ) + 4}{320 \, {\left (4 \, x - \log \left (6\right ) + 2\right )}} - \frac {16 \, x^{2} - 4 \, x {\left (\log \left (3\right ) + \log \left (2\right ) - 2\right )} - {\left (16 \, x^{2} - 4 \, x {\left (\log \left (3\right ) + \log \left (2\right ) - 2\right )} + \log \left (3\right )^{2} + 2 \, {\left (\log \left (3\right ) - 2\right )} \log \left (2\right ) + \log \left (2\right )^{2} - 4 \, \log \left (3\right ) + 4\right )} \log \left (x\right )}{320 \, {\left (4 \, x - \log \left (3\right ) - \log \left (2\right ) + 2\right )}} - \frac {\log \left (6\right ) - 2}{160 \, {\left (4 \, x - \log \left (6\right ) + 2\right )}} + \frac {1}{160} \, \log \left (4 \, x - \log \left (6\right ) + 2\right ) \] Input:
integrate(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2 +(-160*x-80)*log(6)+320*x^2+320*x+80),x, algorithm="maxima")
Output:
1/320*((log(6) - 2)/(4*x - log(6) + 2) - log(4*x - log(6) + 2))*log(6) + 1 /160*(log(6) - 2)*log(4*x - log(6) + 2) - 1/320*(log(3) + log(2) - 2)*log( 4*x - log(3) - log(2) + 2) + 1/320*(log(3) + log(2) - 2)*log(x) + 1/80*x - 1/320*(log(6)^2 - 4*log(6) + 4)/(4*x - log(6) + 2) - 1/320*(16*x^2 - 4*x* (log(3) + log(2) - 2) - (16*x^2 - 4*x*(log(3) + log(2) - 2) + log(3)^2 + 2 *(log(3) - 2)*log(2) + log(2)^2 - 4*log(3) + 4)*log(x))/(4*x - log(3) - lo g(2) + 2) - 1/160*(log(6) - 2)/(4*x - log(6) + 2) + 1/160*log(4*x - log(6) + 2)
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\frac {1}{320} \, {\left (4 \, x + \frac {\log \left (6\right )^{2} - 4 \, \log \left (6\right ) + 4}{4 \, x - \log \left (6\right ) + 2}\right )} \log \left (x\right ) + \frac {1}{320} \, {\left (\log \left (6\right ) - 2\right )} \log \left (x\right ) \] Input:
integrate(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2 +(-160*x-80)*log(6)+320*x^2+320*x+80),x, algorithm="giac")
Output:
1/320*(4*x + (log(6)^2 - 4*log(6) + 4)/(4*x - log(6) + 2))*log(x) + 1/320* (log(6) - 2)*log(x)
Time = 1.04 (sec) , antiderivative size = 135, normalized size of antiderivative = 5.40 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=\frac {\ln \left (\ln \left (6\right )-4\,x-2\right )}{160}-\frac {\frac {{\ln \left (6\right )}^2}{4}-\ln \left (6\right )+1}{320\,x-80\,\ln \left (6\right )+160}+\ln \left (\ln \left (6\right )-4\,x-2\right )\,\left (\frac {\ln \left (6\right )}{160}-\frac {1}{80}\right )-\frac {\ln \left (\ln \left (6\right )-4\,x-2\right )\,\ln \left (6\right )}{320}-\frac {\frac {\ln \left (6\right )}{16}-\frac {1}{8}}{40\,x-10\,\ln \left (6\right )+20}-\ln \left (4\,x-\ln \left (6\right )+2\right )\,\left (\frac {\ln \left (6\right )}{320}-\frac {1}{160}\right )+\frac {x^2\,\ln \left (x\right )}{20\,\left (4\,x-\ln \left (6\right )+2\right )}+\frac {\ln \left (6\right )\,\left (\ln \left (6\right )-2\right )}{16\,\left (80\,x-20\,\ln \left (6\right )+40\right )} \] Input:
int((2*x - x*log(6) + 4*x^2 + log(x)*(4*x - 2*x*log(6) + 4*x^2))/(320*x - log(6)*(160*x + 80) + 20*log(6)^2 + 320*x^2 + 80),x)
Output:
log(log(6) - 4*x - 2)/160 - (log(6)^2/4 - log(6) + 1)/(320*x - 80*log(6) + 160) + log(log(6) - 4*x - 2)*(log(6)/160 - 1/80) - (log(log(6) - 4*x - 2) *log(6))/320 - (log(6)/16 - 1/8)/(40*x - 10*log(6) + 20) - log(4*x - log(6 ) + 2)*(log(6)/320 - 1/160) + (x^2*log(x))/(20*(4*x - log(6) + 2)) + (log( 6)*(log(6) - 2))/(16*(80*x - 20*log(6) + 40))
Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2 x+4 x^2-x \log (6)+\left (4 x+4 x^2-2 x \log (6)\right ) \log (x)}{80+320 x+320 x^2+(-80-160 x) \log (6)+20 \log ^2(6)} \, dx=-\frac {\mathrm {log}\left (x \right ) x^{2}}{20 \,\mathrm {log}\left (6\right )-80 x -40} \] Input:
int(((-2*x*log(6)+4*x^2+4*x)*log(x)-x*log(6)+4*x^2+2*x)/(20*log(6)^2+(-160 *x-80)*log(6)+320*x^2+320*x+80),x)
Output:
( - log(x)*x**2)/(20*(log(6) - 4*x - 2))