Integrand size = 44, antiderivative size = 23 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\log (2 x)-\frac {1296 \left (x+\log \left (x^2\right )\right )}{5 x (1+x)} \] Output:
ln(2*x)-1296*(ln(x^2)+x)/(5*x+5)/x
Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\frac {1}{5} \left (5 \log (x)-\frac {1296 \left (x+\log \left (x^2\right )\right )}{x (1+x)}\right ) \] Input:
Integrate[(-2592 - 2587*x + 1306*x^2 + 5*x^3 + (1296 + 2592*x)*Log[x^2])/( 5*x^2 + 10*x^3 + 5*x^4),x]
Output:
(5*Log[x] - (1296*(x + Log[x^2]))/(x*(1 + x)))/5
Time = 0.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2026, 2007, 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.
\(\displaystyle \int \frac {5 x^3+1306 x^2+(2592 x+1296) \log \left (x^2\right )-2587 x-2592}{5 x^4+10 x^3+5 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {5 x^3+1306 x^2+(2592 x+1296) \log \left (x^2\right )-2587 x-2592}{x^2 \left (5 x^2+10 x+5\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {5 x^3+1306 x^2+(2592 x+1296) \log \left (x^2\right )-2587 x-2592}{x^2 \left (\sqrt {5} x+\sqrt {5}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2592}{5 (x+1)^2 x^2}+\frac {1296 (2 x+1) \log \left (x^2\right )}{5 (x+1)^2 x^2}+\frac {x}{(x+1)^2}+\frac {1306}{5 (x+1)^2}-\frac {2587}{5 (x+1)^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1296 \log \left (x^2\right )}{5 x}-\frac {1296 x \log \left (x^2\right )}{5 (x+1)}-\frac {1296}{5 (x+1)}+\frac {2597 \log (x)}{5}\) |
Input:
Int[(-2592 - 2587*x + 1306*x^2 + 5*x^3 + (1296 + 2592*x)*Log[x^2])/(5*x^2 + 10*x^3 + 5*x^4),x]
Output:
-1296/(5*(1 + x)) + (2597*Log[x])/5 - (1296*Log[x^2])/(5*x) - (1296*x*Log[ x^2])/(5*(1 + x))
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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 = 0.54 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48
method | result | size |
risch | \(-\frac {1296 \ln \left (x^{2}\right )}{5 x \left (1+x \right )}+\frac {5 x \ln \left (x \right )+5 \ln \left (x \right )-1296}{5 x +5}\) | \(34\) |
norman | \(\frac {-\frac {1296 x}{5}+\frac {x \ln \left (x^{2}\right )}{2}+\frac {x^{2} \ln \left (x^{2}\right )}{2}-\frac {1296 \ln \left (x^{2}\right )}{5}}{\left (1+x \right ) x}\) | \(36\) |
parallelrisch | \(\frac {5 x^{2} \ln \left (x^{2}\right )+5 x \ln \left (x^{2}\right )-2592 x -2592 \ln \left (x^{2}\right )}{10 x \left (1+x \right )}\) | \(37\) |
parts | \(-\frac {1296}{5 \left (1+x \right )}+\frac {2592}{5 x}+\frac {2597 \ln \left (x \right )}{5}+\frac {-\frac {2592}{5}-\frac {2592 x}{5}-\frac {1296 x \ln \left (x^{2}\right )}{5}-\frac {1296 x^{2} \ln \left (x^{2}\right )}{5}-\frac {1296 \ln \left (x^{2}\right )}{5}}{\left (1+x \right ) x}\) | \(54\) |
default | \(-\frac {1296}{5 \left (1+x \right )}+\frac {2592}{5 x}+\frac {2597 \ln \left (x \right )}{5}+\frac {-2592-2592 x -1296 x \ln \left (x^{2}\right )-1296 x^{2} \ln \left (x^{2}\right )-1296 \ln \left (x^{2}\right )}{5 \left (1+x \right ) x}\) | \(55\) |
orering | \(-\frac {x \left (25 x^{5}+13035 x^{4}+13385038 x^{2}+40284839 x +20155392\right ) \left (\left (2592 x +1296\right ) \ln \left (x^{2}\right )+5 x^{3}+1306 x^{2}-2587 x -2592\right )}{2592 \left (20 x^{3}+2607 x^{2}+10373 x +2592\right ) \left (5 x^{4}+10 x^{3}+5 x^{2}\right )}-\frac {\left (50 x^{4}+13060 x^{3}+6692519 x +13436928\right ) \left (1+x \right ) x^{2} \left (\frac {2592 \ln \left (x^{2}\right )+\frac {5184 x +2592}{x}+15 x^{2}+2612 x -2587}{5 x^{4}+10 x^{3}+5 x^{2}}-\frac {\left (\left (2592 x +1296\right ) \ln \left (x^{2}\right )+5 x^{3}+1306 x^{2}-2587 x -2592\right ) \left (20 x^{3}+30 x^{2}+10 x \right )}{\left (5 x^{4}+10 x^{3}+5 x^{2}\right )^{2}}\right )}{5184 \left (20 x^{3}+2607 x^{2}+10373 x +2592\right )}\) | \(230\) |
Input:
int(((2592*x+1296)*ln(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^3+5*x^2 ),x,method=_RETURNVERBOSE)
Output:
-1296/5/x/(1+x)*ln(x^2)+1/5*(5*x*ln(x)+5*ln(x)-1296)/(1+x)
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\frac {{\left (5 \, x^{2} + 5 \, x - 2592\right )} \log \left (x^{2}\right ) - 2592 \, x}{10 \, {\left (x^{2} + x\right )}} \] Input:
integrate(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^ 3+5*x^2),x, algorithm="fricas")
Output:
1/10*((5*x^2 + 5*x - 2592)*log(x^2) - 2592*x)/(x^2 + x)
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\log {\left (x \right )} - \frac {1296 \log {\left (x^{2} \right )}}{5 x^{2} + 5 x} - \frac {1296}{5 x + 5} \] Input:
integrate(((2592*x+1296)*ln(x**2)+5*x**3+1306*x**2-2587*x-2592)/(5*x**4+10 *x**3+5*x**2),x)
Output:
log(x) - 1296*log(x**2)/(5*x**2 + 5*x) - 1296/(5*x + 5)
Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\frac {2592 \, {\left (2 \, x + 1\right )}}{5 \, {\left (x^{2} + x\right )}} - \frac {2592 \, {\left (x + \log \left (x\right ) + 1\right )}}{5 \, {\left (x^{2} + x\right )}} - \frac {3888}{5 \, {\left (x + 1\right )}} + \log \left (x\right ) \] Input:
integrate(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^ 3+5*x^2),x, algorithm="maxima")
Output:
2592/5*(2*x + 1)/(x^2 + x) - 2592/5*(x + log(x) + 1)/(x^2 + x) - 3888/5/(x + 1) + log(x)
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\frac {1296}{5} \, {\left (\frac {1}{x + 1} - \frac {1}{x}\right )} \log \left (x^{2}\right ) - \frac {1296}{5 \, {\left (x + 1\right )}} + \log \left (x\right ) \] Input:
integrate(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^ 3+5*x^2),x, algorithm="giac")
Output:
1296/5*(1/(x + 1) - 1/x)*log(x^2) - 1296/5/(x + 1) + log(x)
Time = 2.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\frac {\ln \left (x^2\right )}{2}-\frac {\frac {1296\,x}{5}+\frac {1296\,\ln \left (x^2\right )}{5}}{x\,\left (x+1\right )} \] Input:
int((1306*x^2 - 2587*x + 5*x^3 + log(x^2)*(2592*x + 1296) - 2592)/(5*x^2 + 10*x^3 + 5*x^4),x)
Output:
log(x^2)/2 - ((1296*x)/5 + (1296*log(x^2))/5)/(x*(x + 1))
Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.17 \[ \int \frac {-2592-2587 x+1306 x^2+5 x^3+(1296+2592 x) \log \left (x^2\right )}{5 x^2+10 x^3+5 x^4} \, dx=\frac {-1296 \,\mathrm {log}\left (x^{2}\right ) x^{2}-1296 \,\mathrm {log}\left (x^{2}\right ) x -1296 \,\mathrm {log}\left (x^{2}\right )+2597 \,\mathrm {log}\left (x \right ) x^{2}+2597 \,\mathrm {log}\left (x \right ) x +1296 x^{2}}{5 x \left (x +1\right )} \] Input:
int(((2592*x+1296)*log(x^2)+5*x^3+1306*x^2-2587*x-2592)/(5*x^4+10*x^3+5*x^ 2),x)
Output:
( - 1296*log(x**2)*x**2 - 1296*log(x**2)*x - 1296*log(x**2) + 2597*log(x)* x**2 + 2597*log(x)*x + 1296*x**2)/(5*x*(x + 1))