Integrand size = 107, antiderivative size = 29 \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\frac {\log ^2\left (-\frac {16}{x}-x+4 x \left (x-x^2\right )\right )}{4 x} \]
Timed out. \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\text {\$Aborted} \]
Integrate[((-32 + 2*x^2 - 16*x^3 + 24*x^4)*Log[(-16 - x^2 + 4*x^3 - 4*x^4) /x] + (-16 - x^2 + 4*x^3 - 4*x^4)*Log[(-16 - x^2 + 4*x^3 - 4*x^4)/x]^2)/(6 4*x^2 + 4*x^4 - 16*x^5 + 16*x^6),x]
Time = 5.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {2026, 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.
\(\displaystyle \int \frac {\left (-4 x^4+4 x^3-x^2-16\right ) \log ^2\left (\frac {-4 x^4+4 x^3-x^2-16}{x}\right )+\left (24 x^4-16 x^3+2 x^2-32\right ) \log \left (\frac {-4 x^4+4 x^3-x^2-16}{x}\right )}{16 x^6-16 x^5+4 x^4+64 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-4 x^4+4 x^3-x^2-16\right ) \log ^2\left (\frac {-4 x^4+4 x^3-x^2-16}{x}\right )+\left (24 x^4-16 x^3+2 x^2-32\right ) \log \left (\frac {-4 x^4+4 x^3-x^2-16}{x}\right )}{x^2 \left (16 x^4-16 x^3+4 x^2+64\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (12 x^4-8 x^3+x^2-16\right ) \log \left (\frac {-4 x^4+4 x^3-x^2-16}{x}\right )}{2 x^2 \left (4 x^4-4 x^3+x^2+16\right )}-\frac {\log ^2\left (\frac {-4 x^4+4 x^3-x^2-16}{x}\right )}{4 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log ^2\left (-\frac {4 x^4-4 x^3+x^2+16}{x}\right )}{4 x}\) |
Int[((-32 + 2*x^2 - 16*x^3 + 24*x^4)*Log[(-16 - x^2 + 4*x^3 - 4*x^4)/x] + (-16 - x^2 + 4*x^3 - 4*x^4)*Log[(-16 - x^2 + 4*x^3 - 4*x^4)/x]^2)/(64*x^2 + 4*x^4 - 16*x^5 + 16*x^6),x]
3.2.41.3.1 Defintions of rubi rules used
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])
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \(\frac {\ln \left (-\frac {4 x^{4}-4 x^{3}+x^{2}+16}{x}\right )^{2}}{4 x}\) | \(29\) |
norman | \(\frac {\ln \left (\frac {-4 x^{4}+4 x^{3}-x^{2}-16}{x}\right )^{2}}{4 x}\) | \(30\) |
risch | \(\frac {\ln \left (\frac {-4 x^{4}+4 x^{3}-x^{2}-16}{x}\right )^{2}}{4 x}\) | \(30\) |
int(((-4*x^4+4*x^3-x^2-16)*ln((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16*x^3+2* x^2-32)*ln((-4*x^4+4*x^3-x^2-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2),x,method =_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\frac {\log \left (-\frac {4 \, x^{4} - 4 \, x^{3} + x^{2} + 16}{x}\right )^{2}}{4 \, x} \]
integrate(((-4*x^4+4*x^3-x^2-16)*log((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16 *x^3+2*x^2-32)*log((-4*x^4+4*x^3-x^2-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2), x, algorithm=\
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\frac {\log {\left (\frac {- 4 x^{4} + 4 x^{3} - x^{2} - 16}{x} \right )}^{2}}{4 x} \]
integrate(((-4*x**4+4*x**3-x**2-16)*ln((-4*x**4+4*x**3-x**2-16)/x)**2+(24* x**4-16*x**3+2*x**2-32)*ln((-4*x**4+4*x**3-x**2-16)/x))/(16*x**6-16*x**5+4 *x**4+64*x**2),x)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\frac {\log \left (-4 \, x^{4} + 4 \, x^{3} - x^{2} - 16\right )^{2} - 2 \, \log \left (-4 \, x^{4} + 4 \, x^{3} - x^{2} - 16\right ) \log \left (x\right ) + \log \left (x\right )^{2}}{4 \, x} \]
integrate(((-4*x^4+4*x^3-x^2-16)*log((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16 *x^3+2*x^2-32)*log((-4*x^4+4*x^3-x^2-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2), x, algorithm=\
1/4*(log(-4*x^4 + 4*x^3 - x^2 - 16)^2 - 2*log(-4*x^4 + 4*x^3 - x^2 - 16)*l og(x) + log(x)^2)/x
Exception generated. \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\text {Exception raised: TypeError} \]
integrate(((-4*x^4+4*x^3-x^2-16)*log((-4*x^4+4*x^3-x^2-16)/x)^2+(24*x^4-16 *x^3+2*x^2-32)*log((-4*x^4+4*x^3-x^2-16)/x))/(16*x^6-16*x^5+4*x^4+64*x^2), x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%{poly1[4645089346227255570024772312188510155765170031139 394747419
Time = 9.43 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-32+2 x^2-16 x^3+24 x^4\right ) \log \left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )+\left (-16-x^2+4 x^3-4 x^4\right ) \log ^2\left (\frac {-16-x^2+4 x^3-4 x^4}{x}\right )}{64 x^2+4 x^4-16 x^5+16 x^6} \, dx=\frac {{\ln \left (-\frac {4\,x^4-4\,x^3+x^2+16}{x}\right )}^2}{4\,x} \]