Integrand size = 92, antiderivative size = 22 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {5 \left (x+\log \left (x^2\right )\right )^2}{10 x-\frac {16}{\log ^2(4)}} \] Output:
5/(10*x-4/ln(2)^2)*(ln(x^2)+x)^2
Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(22)=44\).
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.73 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {64-40 x \log ^2(4)+25 x^2 \log ^4(4)+50 x \log ^4(4) \log \left (x^2\right )+25 \log ^4(4) \log ^2\left (x^2\right )}{10 \log ^2(4) \left (-8+5 x \log ^2(4)\right )} \] Input:
Integrate[((-160*x - 80*x^2)*Log[4]^2 + (100*x^2 + 25*x^3)*Log[4]^4 + ((-1 60 - 80*x)*Log[4]^2 + 100*x*Log[4]^4)*Log[x^2] - 25*x*Log[4]^4*Log[x^2]^2) /(128*x - 160*x^2*Log[4]^2 + 50*x^3*Log[4]^4),x]
Output:
(64 - 40*x*Log[4]^2 + 25*x^2*Log[4]^4 + 50*x*Log[4]^4*Log[x^2] + 25*Log[4] ^4*Log[x^2]^2)/(10*Log[4]^2*(-8 + 5*x*Log[4]^2))
Leaf count is larger than twice the leaf count of optimal. \(91\) vs. \(2(22)=44\).
Time = 1.50 (sec) , antiderivative size = 91, normalized size of antiderivative = 4.14, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {2026, 7277, 27, 7292, 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.
\(\displaystyle \int \frac {\left (-80 x^2-160 x\right ) \log ^2(4)-25 x \log ^4(4) \log ^2\left (x^2\right )+\left (100 x \log ^4(4)+(-80 x-160) \log ^2(4)\right ) \log \left (x^2\right )+\left (25 x^3+100 x^2\right ) \log ^4(4)}{50 x^3 \log ^4(4)-160 x^2 \log ^2(4)+128 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-80 x^2-160 x\right ) \log ^2(4)-25 x \log ^4(4) \log ^2\left (x^2\right )+\left (100 x \log ^4(4)+(-80 x-160) \log ^2(4)\right ) \log \left (x^2\right )+\left (25 x^3+100 x^2\right ) \log ^4(4)}{x \left (50 x^2 \log ^4(4)-160 x \log ^2(4)+128\right )}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 200 \log ^4(4) \int -\frac {5 x \log ^4(4) \log ^2\left (x^2\right )+4 \left (4 (x+2) \log ^2(4)-5 x \log ^4(4)\right ) \log \left (x^2\right )-5 \left (x^3+4 x^2\right ) \log ^4(4)+16 \left (x^2+2 x\right ) \log ^2(4)}{80 x \log ^4(4) \left (8-5 x \log ^2(4)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{2} \int \frac {5 x \log ^4(4) \log ^2\left (x^2\right )+4 \left (4 (x+2) \log ^2(4)-5 x \log ^4(4)\right ) \log \left (x^2\right )-5 \left (x^3+4 x^2\right ) \log ^4(4)+16 \left (x^2+2 x\right ) \log ^2(4)}{x \left (8-5 x \log ^2(4)\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {5}{2} \int \frac {\log ^2(4) \left (x+\log \left (x^2\right )\right ) \left (-5 \log ^2(4) x^2+5 \log ^2(4) \log \left (x^2\right ) x+16 \left (1-\frac {5 \log ^2(4)}{4}\right ) x+32\right )}{x \left (8-5 x \log ^2(4)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{2} \log ^2(4) \int \frac {\left (x+\log \left (x^2\right )\right ) \left (-5 \log ^2(4) x^2+5 \log ^2(4) \log \left (x^2\right ) x+4 \left (4-5 \log ^2(4)\right ) x+32\right )}{x \left (8-5 x \log ^2(4)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5}{2} \log ^2(4) \int \left (\frac {5 \log ^2(4) \log ^2\left (x^2\right )}{\left (5 x \log ^2(4)-8\right )^2}+\frac {4 \left (\left (4-5 \log ^2(4)\right ) x+8\right ) \log \left (x^2\right )}{x \left (8-5 x \log ^2(4)\right )^2}+\frac {-5 \log ^2(4) x^2+4 \left (4-5 \log ^2(4)\right ) x+32}{\left (8-5 x \log ^2(4)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{2} \log ^2(4) \left (\frac {5 x \log ^2(4) \log ^2\left (x^2\right )}{8 \left (8-5 x \log ^2(4)\right )}+\frac {1}{8} \log ^2\left (x^2\right )+\frac {2 x \log \left (x^2\right )}{8-5 x \log ^2(4)}-\frac {x}{5 \log ^2(4)}+\frac {64}{25 \log ^4(4) \left (8-5 x \log ^2(4)\right )}\right )\) |
Input:
Int[((-160*x - 80*x^2)*Log[4]^2 + (100*x^2 + 25*x^3)*Log[4]^4 + ((-160 - 8 0*x)*Log[4]^2 + 100*x*Log[4]^4)*Log[x^2] - 25*x*Log[4]^4*Log[x^2]^2)/(128* x - 160*x^2*Log[4]^2 + 50*x^3*Log[4]^4),x]
Output:
(-5*Log[4]^2*(-1/5*x/Log[4]^2 + 64/(25*Log[4]^4*(8 - 5*x*Log[4]^2)) + (2*x *Log[x^2])/(8 - 5*x*Log[4]^2) + Log[x^2]^2/8 + (5*x*Log[4]^2*Log[x^2]^2)/( 8*(8 - 5*x*Log[4]^2))))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, 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]
Leaf count of result is larger than twice the leaf count of optimal. \(45\) vs. \(2(22)=44\).
Time = 1.30 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.09
method | result | size |
norman | \(\frac {5 \ln \left (2\right )^{2} \ln \left (x^{2}\right ) x +\frac {5 x^{2} \ln \left (2\right )^{2}}{2}+\frac {5 \ln \left (2\right )^{2} \ln \left (x^{2}\right )^{2}}{2}}{5 x \ln \left (2\right )^{2}-2}\) | \(46\) |
parallelrisch | \(\frac {10 x^{2} \ln \left (2\right )^{2}+20 \ln \left (2\right )^{2} \ln \left (x^{2}\right ) x +10 \ln \left (2\right )^{2} \ln \left (x^{2}\right )^{2}}{20 x \ln \left (2\right )^{2}-8}\) | \(47\) |
risch | \(\frac {5 \ln \left (2\right )^{2} \ln \left (x^{2}\right )^{2}}{2 \left (5 x \ln \left (2\right )^{2}-2\right )}+\frac {2 \ln \left (x^{2}\right )}{5 x \ln \left (2\right )^{2}-2}+\frac {100 \ln \left (2\right )^{4} \ln \left (-x \right ) x +25 \ln \left (2\right )^{4} x^{2}-40 \ln \left (-x \right ) \ln \left (2\right )^{2}-10 x \ln \left (2\right )^{2}+4}{10 \ln \left (2\right )^{2} \left (5 x \ln \left (2\right )^{2}-2\right )}\) | \(98\) |
Input:
int((-400*x*ln(2)^4*ln(x^2)^2+(1600*x*ln(2)^4+4*(-80*x-160)*ln(2)^2)*ln(x^ 2)+16*(25*x^3+100*x^2)*ln(2)^4+4*(-80*x^2-160*x)*ln(2)^2)/(800*x^3*ln(2)^4 -640*x^2*ln(2)^2+128*x),x,method=_RETURNVERBOSE)
Output:
(5*ln(2)^2*ln(x^2)*x+5/2*x^2*ln(2)^2+5/2*ln(2)^2*ln(x^2)^2)/(5*x*ln(2)^2-2 )
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (22) = 44\).
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {25 \, x^{2} \log \left (2\right )^{4} + 50 \, x \log \left (2\right )^{4} \log \left (x^{2}\right ) + 25 \, \log \left (2\right )^{4} \log \left (x^{2}\right )^{2} - 10 \, x \log \left (2\right )^{2} + 4}{10 \, {\left (5 \, x \log \left (2\right )^{4} - 2 \, \log \left (2\right )^{2}\right )}} \] Input:
integrate((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2 )^2)*log(x^2)+16*(25*x^3+100*x^2)*log(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(80 0*x^3*log(2)^4-640*x^2*log(2)^2+128*x),x, algorithm="fricas")
Output:
1/10*(25*x^2*log(2)^4 + 50*x*log(2)^4*log(x^2) + 25*log(2)^4*log(x^2)^2 - 10*x*log(2)^2 + 4)/(5*x*log(2)^4 - 2*log(2)^2)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (19) = 38\).
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.95 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {x}{2} + 2 \log {\left (x \right )} + \frac {2}{25 x \log {\left (2 \right )}^{4} - 10 \log {\left (2 \right )}^{2}} + \frac {5 \log {\left (2 \right )}^{2} \log {\left (x^{2} \right )}^{2}}{10 x \log {\left (2 \right )}^{2} - 4} + \frac {2 \log {\left (x^{2} \right )}}{5 x \log {\left (2 \right )}^{2} - 2} \] Input:
integrate((-400*x*ln(2)**4*ln(x**2)**2+(1600*x*ln(2)**4+4*(-80*x-160)*ln(2 )**2)*ln(x**2)+16*(25*x**3+100*x**2)*ln(2)**4+4*(-80*x**2-160*x)*ln(2)**2) /(800*x**3*ln(2)**4-640*x**2*ln(2)**2+128*x),x)
Output:
x/2 + 2*log(x) + 2/(25*x*log(2)**4 - 10*log(2)**2) + 5*log(2)**2*log(x**2) **2/(10*x*log(2)**2 - 4) + 2*log(x**2)/(5*x*log(2)**2 - 2)
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (22) = 44\).
Time = 0.16 (sec) , antiderivative size = 197, normalized size of antiderivative = 8.95 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=-\frac {1}{10} \, {\left (\frac {4}{5 \, x \log \left (2\right )^{8} - 2 \, \log \left (2\right )^{6}} - \frac {5 \, x}{\log \left (2\right )^{4}} - \frac {4 \, \log \left (5 \, x \log \left (2\right )^{2} - 2\right )}{\log \left (2\right )^{6}}\right )} \log \left (2\right )^{4} - 2 \, {\left (\frac {2}{5 \, x \log \left (2\right )^{6} - 2 \, \log \left (2\right )^{4}} - \frac {\log \left (5 \, x \log \left (2\right )^{2} - 2\right )}{\log \left (2\right )^{4}}\right )} \log \left (2\right )^{4} + \frac {2}{5} \, {\left (\frac {2}{5 \, x \log \left (2\right )^{6} - 2 \, \log \left (2\right )^{4}} - \frac {\log \left (5 \, x \log \left (2\right )^{2} - 2\right )}{\log \left (2\right )^{4}}\right )} \log \left (2\right )^{2} + \frac {4 \, \log \left (2\right )^{2}}{5 \, x \log \left (2\right )^{4} - 2 \, \log \left (2\right )^{2}} + \frac {2 \, {\left (5 \, \log \left (2\right )^{2} \log \left (x\right )^{2} + 2 \, \log \left (x\right )\right )}}{5 \, x \log \left (2\right )^{2} - 2} - 2 \, \log \left (5 \, x \log \left (2\right )^{2} - 2\right ) + 2 \, \log \left (x\right ) \] Input:
integrate((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2 )^2)*log(x^2)+16*(25*x^3+100*x^2)*log(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(80 0*x^3*log(2)^4-640*x^2*log(2)^2+128*x),x, algorithm="maxima")
Output:
-1/10*(4/(5*x*log(2)^8 - 2*log(2)^6) - 5*x/log(2)^4 - 4*log(5*x*log(2)^2 - 2)/log(2)^6)*log(2)^4 - 2*(2/(5*x*log(2)^6 - 2*log(2)^4) - log(5*x*log(2) ^2 - 2)/log(2)^4)*log(2)^4 + 2/5*(2/(5*x*log(2)^6 - 2*log(2)^4) - log(5*x* log(2)^2 - 2)/log(2)^4)*log(2)^2 + 4*log(2)^2/(5*x*log(2)^4 - 2*log(2)^2) + 2*(5*log(2)^2*log(x)^2 + 2*log(x))/(5*x*log(2)^2 - 2) - 2*log(5*x*log(2) ^2 - 2) + 2*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (22) = 44\).
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 3.00 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {5 \, \log \left (2\right )^{2} \log \left (x^{2}\right )^{2}}{2 \, {\left (5 \, x \log \left (2\right )^{2} - 2\right )}} + \frac {1}{2} \, x + \frac {2 \, \log \left (x^{2}\right )}{5 \, x \log \left (2\right )^{2} - 2} + \frac {2}{5 \, {\left (5 \, x \log \left (2\right )^{4} - 2 \, \log \left (2\right )^{2}\right )}} + 2 \, \log \left (x\right ) \] Input:
integrate((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2 )^2)*log(x^2)+16*(25*x^3+100*x^2)*log(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(80 0*x^3*log(2)^4-640*x^2*log(2)^2+128*x),x, algorithm="giac")
Output:
5/2*log(2)^2*log(x^2)^2/(5*x*log(2)^2 - 2) + 1/2*x + 2*log(x^2)/(5*x*log(2 )^2 - 2) + 2/5/(5*x*log(2)^4 - 2*log(2)^2) + 2*log(x)
Time = 4.17 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.95 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {x}{2}+2\,\ln \left (x\right )+\frac {2\,\ln \left (x^2\right )}{5\,x\,{\ln \left (2\right )}^2-2}+\frac {2}{5\,{\ln \left (2\right )}^2\,\left (5\,x\,{\ln \left (2\right )}^2-2\right )}+\frac {5\,{\ln \left (x^2\right )}^2\,{\ln \left (2\right )}^2}{2\,\left (5\,x\,{\ln \left (2\right )}^2-2\right )} \] Input:
int(-(log(x^2)*(4*log(2)^2*(80*x + 160) - 1600*x*log(2)^4) + 4*log(2)^2*(1 60*x + 80*x^2) - 16*log(2)^4*(100*x^2 + 25*x^3) + 400*x*log(x^2)^2*log(2)^ 4)/(128*x - 640*x^2*log(2)^2 + 800*x^3*log(2)^4),x)
Output:
x/2 + 2*log(x) + (2*log(x^2))/(5*x*log(2)^2 - 2) + 2/(5*log(2)^2*(5*x*log( 2)^2 - 2)) + (5*log(x^2)^2*log(2)^2)/(2*(5*x*log(2)^2 - 2))
Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {\left (-160 x-80 x^2\right ) \log ^2(4)+\left (100 x^2+25 x^3\right ) \log ^4(4)+\left ((-160-80 x) \log ^2(4)+100 x \log ^4(4)\right ) \log \left (x^2\right )-25 x \log ^4(4) \log ^2\left (x^2\right )}{128 x-160 x^2 \log ^2(4)+50 x^3 \log ^4(4)} \, dx=\frac {5 \mathrm {log}\left (2\right )^{2} \left (\mathrm {log}\left (x^{2}\right )^{2}+2 \,\mathrm {log}\left (x^{2}\right ) x +x^{2}\right )}{10 \mathrm {log}\left (2\right )^{2} x -4} \] Input:
int((-400*x*log(2)^4*log(x^2)^2+(1600*x*log(2)^4+4*(-80*x-160)*log(2)^2)*l og(x^2)+16*(25*x^3+100*x^2)*log(2)^4+4*(-80*x^2-160*x)*log(2)^2)/(800*x^3* log(2)^4-640*x^2*log(2)^2+128*x),x)
Output:
(5*log(2)**2*(log(x**2)**2 + 2*log(x**2)*x + x**2))/(2*(5*log(2)**2*x - 2) )