Integrand size = 75, antiderivative size = 31 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=4+x^2+\frac {x \left (2+\frac {2 (3+x)}{5-x}\right )^2 \log (x)}{\log \left (x^2\right )} \] Output:
x^2+4+x*((3+x)/(5/2-1/2*x)+2)^2*ln(x)/ln(x^2)
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=\frac {x \left (256 \log (x)+(-5+x)^2 x \log \left (x^2\right )\right )}{(-5+x)^2 \log \left (x^2\right )} \] Input:
Integrate[((2560 - 512*x)*Log[x] + (-1280 + 256*x + (-1280 - 256*x)*Log[x] )*Log[x^2] + (-250*x + 150*x^2 - 30*x^3 + 2*x^4)*Log[x^2]^2)/((-125 + 75*x - 15*x^2 + x^3)*Log[x^2]^2),x]
Output:
(x*(256*Log[x] + (-5 + x)^2*x*Log[x^2]))/((-5 + x)^2*Log[x^2])
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.
\(\displaystyle \int \frac {(256 x+(-256 x-1280) \log (x)-1280) \log \left (x^2\right )+\left (2 x^4-30 x^3+150 x^2-250 x\right ) \log ^2\left (x^2\right )+(2560-512 x) \log (x)}{\left (x^3-15 x^2+75 x-125\right ) \log ^2\left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {(256 x+(-256 x-1280) \log (x)-1280) \log \left (x^2\right )+\left (2 x^4-30 x^3+150 x^2-250 x\right ) \log ^2\left (x^2\right )+(2560-512 x) \log (x)}{(x-5)^3 \log ^2\left (x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {512 \log (x)}{(x-5)^2 \log ^2\left (x^2\right )}-\frac {256 (-x+x \log (x)+5 \log (x)+5)}{(x-5)^3 \log \left (x^2\right )}+2 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -512 \int \frac {\log (x)}{(x-5)^2 \log ^2\left (x^2\right )}dx-1280 \int \frac {1}{(x-5)^3 \log \left (x^2\right )}dx+256 \int \frac {x}{(x-5)^3 \log \left (x^2\right )}dx-2560 \int \frac {\log (x)}{(x-5)^3 \log \left (x^2\right )}dx-256 \int \frac {\log (x)}{(x-5)^2 \log \left (x^2\right )}dx+x^2\) |
Input:
Int[((2560 - 512*x)*Log[x] + (-1280 + 256*x + (-1280 - 256*x)*Log[x])*Log[ x^2] + (-250*x + 150*x^2 - 30*x^3 + 2*x^4)*Log[x^2]^2)/((-125 + 75*x - 15* x^2 + x^3)*Log[x^2]^2),x]
Output:
$Aborted
Time = 11.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68
method | result | size |
parallelrisch | \(\frac {-2400 x^{3} \ln \left (x^{2}\right )+6000 x^{2} \ln \left (x^{2}\right )+240 x^{4} \ln \left (x^{2}\right )+61440 x \ln \left (x \right )}{240 \ln \left (x^{2}\right ) \left (x^{2}-10 x +25\right )}\) | \(52\) |
risch | \(\frac {x \left (x^{3}-10 x^{2}+25 x +128\right )}{x^{2}-10 x +25}-\frac {128 x \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right )}{\left (x^{2}-10 x +25\right ) \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )\right )}\) | \(131\) |
Input:
int(((2*x^4-30*x^3+150*x^2-250*x)*ln(x^2)^2+((-256*x-1280)*ln(x)+256*x-128 0)*ln(x^2)+(-512*x+2560)*ln(x))/(x^3-15*x^2+75*x-125)/ln(x^2)^2,x,method=_ RETURNVERBOSE)
Output:
1/240*(-2400*x^3*ln(x^2)+6000*x^2*ln(x^2)+240*x^4*ln(x^2)+61440*x*ln(x))/l n(x^2)/(x^2-10*x+25)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=\frac {x^{4} - 10 \, x^{3} + 25 \, x^{2} + 128 \, x}{x^{2} - 10 \, x + 25} \] Input:
integrate(((2*x^4-30*x^3+150*x^2-250*x)*log(x^2)^2+((-256*x-1280)*log(x)+2 56*x-1280)*log(x^2)+(-512*x+2560)*log(x))/(x^3-15*x^2+75*x-125)/log(x^2)^2 ,x, algorithm="fricas")
Output:
(x^4 - 10*x^3 + 25*x^2 + 128*x)/(x^2 - 10*x + 25)
Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.45 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=x^{2} + \frac {128 x}{x^{2} - 10 x + 25} \] Input:
integrate(((2*x**4-30*x**3+150*x**2-250*x)*ln(x**2)**2+((-256*x-1280)*ln(x )+256*x-1280)*ln(x**2)+(-512*x+2560)*ln(x))/(x**3-15*x**2+75*x-125)/ln(x** 2)**2,x)
Output:
x**2 + 128*x/(x**2 - 10*x + 25)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=\frac {x^{4} - 10 \, x^{3} + 25 \, x^{2} + 128 \, x}{x^{2} - 10 \, x + 25} \] Input:
integrate(((2*x^4-30*x^3+150*x^2-250*x)*log(x^2)^2+((-256*x-1280)*log(x)+2 56*x-1280)*log(x^2)+(-512*x+2560)*log(x))/(x^3-15*x^2+75*x-125)/log(x^2)^2 ,x, algorithm="maxima")
Output:
(x^4 - 10*x^3 + 25*x^2 + 128*x)/(x^2 - 10*x + 25)
Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=x^{2} + \frac {128 \, x}{x^{2} - 10 \, x + 25} \] Input:
integrate(((2*x^4-30*x^3+150*x^2-250*x)*log(x^2)^2+((-256*x-1280)*log(x)+2 56*x-1280)*log(x^2)+(-512*x+2560)*log(x))/(x^3-15*x^2+75*x-125)/log(x^2)^2 ,x, algorithm="giac")
Output:
x^2 + 128*x/(x^2 - 10*x + 25)
Time = 3.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=\frac {x\,\left (256\,\ln \left (x\right )+25\,x\,\ln \left (x^2\right )-10\,x^2\,\ln \left (x^2\right )+x^3\,\ln \left (x^2\right )\right )}{\ln \left (x^2\right )\,{\left (x-5\right )}^2} \] Input:
int(-(log(x^2)*(log(x)*(256*x + 1280) - 256*x + 1280) + log(x)*(512*x - 25 60) + log(x^2)^2*(250*x - 150*x^2 + 30*x^3 - 2*x^4))/(log(x^2)^2*(75*x - 1 5*x^2 + x^3 - 125)),x)
Output:
(x*(256*log(x) + 25*x*log(x^2) - 10*x^2*log(x^2) + x^3*log(x^2)))/(log(x^2 )*(x - 5)^2)
Time = 0.17 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {(2560-512 x) \log (x)+(-1280+256 x+(-1280-256 x) \log (x)) \log \left (x^2\right )+\left (-250 x+150 x^2-30 x^3+2 x^4\right ) \log ^2\left (x^2\right )}{\left (-125+75 x-15 x^2+x^3\right ) \log ^2\left (x^2\right )} \, dx=\frac {x \left (\mathrm {log}\left (x^{2}\right ) x^{3}-10 \,\mathrm {log}\left (x^{2}\right ) x^{2}+25 \,\mathrm {log}\left (x^{2}\right ) x +256 \,\mathrm {log}\left (x \right )\right )}{\mathrm {log}\left (x^{2}\right ) \left (x^{2}-10 x +25\right )} \] Input:
int(((2*x^4-30*x^3+150*x^2-250*x)*log(x^2)^2+((-256*x-1280)*log(x)+256*x-1 280)*log(x^2)+(-512*x+2560)*log(x))/(x^3-15*x^2+75*x-125)/log(x^2)^2,x)
Output:
(x*(log(x**2)*x**3 - 10*log(x**2)*x**2 + 25*log(x**2)*x + 256*log(x)))/(lo g(x**2)*(x**2 - 10*x + 25))