Integrand size = 97, antiderivative size = 29 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=\frac {x+5 \left (-3+\frac {4+x}{x \log \left ((-3+x)^2\right )}\right )}{1+2 x} \] Output:
(x+5*(4+x)/ln((-3+x)^2)/x-15)/(1+2*x)
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=\frac {-31+\frac {10 (4+x)}{x \log \left ((-3+x)^2\right )}}{2 (1+2 x)} \] Input:
Integrate[(-40*x - 90*x^2 - 20*x^3 + (60 + 220*x - 50*x^2 - 10*x^3)*Log[9 - 6*x + x^2] + (-93*x^2 + 31*x^3)*Log[9 - 6*x + x^2]^2)/((-3*x^2 - 11*x^3 - 8*x^4 + 4*x^5)*Log[9 - 6*x + x^2]^2),x]
Output:
(-31 + (10*(4 + x))/(x*Log[(-3 + x)^2]))/(2*(1 + 2*x))
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.
\(\displaystyle \int \frac {-20 x^3-90 x^2+\left (31 x^3-93 x^2\right ) \log ^2\left (x^2-6 x+9\right )+\left (-10 x^3-50 x^2+220 x+60\right ) \log \left (x^2-6 x+9\right )-40 x}{\left (4 x^5-8 x^4-11 x^3-3 x^2\right ) \log ^2\left (x^2-6 x+9\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-20 x^3-90 x^2+\left (31 x^3-93 x^2\right ) \log ^2\left (x^2-6 x+9\right )+\left (-10 x^3-50 x^2+220 x+60\right ) \log \left (x^2-6 x+9\right )-40 x}{x^2 \left (4 x^3-8 x^2-11 x-3\right ) \log ^2\left (x^2-6 x+9\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (-\frac {2 \left (-20 x^3-90 x^2+\left (31 x^3-93 x^2\right ) \log ^2\left (x^2-6 x+9\right )+\left (-10 x^3-50 x^2+220 x+60\right ) \log \left (x^2-6 x+9\right )-40 x\right )}{49 x^2 (2 x+1) \log ^2\left (x^2-6 x+9\right )}+\frac {-20 x^3-90 x^2+\left (31 x^3-93 x^2\right ) \log ^2\left (x^2-6 x+9\right )+\left (-10 x^3-50 x^2+220 x+60\right ) \log \left (x^2-6 x+9\right )-40 x}{49 (x-3) x^2 \log ^2\left (x^2-6 x+9\right )}-\frac {2 \left (-20 x^3-90 x^2+\left (31 x^3-93 x^2\right ) \log ^2\left (x^2-6 x+9\right )+\left (-10 x^3-50 x^2+220 x+60\right ) \log \left (x^2-6 x+9\right )-40 x\right )}{7 x^2 (2 x+1)^2 \log ^2\left (x^2-6 x+9\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \int \frac {1}{x^2 \log \left ((x-3)^2\right )}dx+\frac {40}{3} \int \frac {1}{x \log ^2\left ((x-3)^2\right )}dx-20 \int \frac {1}{(2 x+1) \log ^2\left ((x-3)^2\right )}dx+70 \int \frac {1}{(2 x+1)^2 \log \left ((x-3)^2\right )}dx-\frac {31}{2 (2 x+1)}+\frac {5}{3 \log \left ((x-3)^2\right )}\) |
Input:
Int[(-40*x - 90*x^2 - 20*x^3 + (60 + 220*x - 50*x^2 - 10*x^3)*Log[9 - 6*x + x^2] + (-93*x^2 + 31*x^3)*Log[9 - 6*x + x^2]^2)/((-3*x^2 - 11*x^3 - 8*x^ 4 + 4*x^5)*Log[9 - 6*x + x^2]^2),x]
Output:
$Aborted
Time = 4.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28
method | result | size |
risch | \(-\frac {31}{2 \left (1+2 x \right )}+\frac {5 x +20}{x \left (1+2 x \right ) \ln \left (x^{2}-6 x +9\right )}\) | \(37\) |
norman | \(\frac {20-\frac {31 x \ln \left (x^{2}-6 x +9\right )}{2}+5 x}{x \left (1+2 x \right ) \ln \left (x^{2}-6 x +9\right )}\) | \(40\) |
parallelrisch | \(\frac {40-31 x \ln \left (x^{2}-6 x +9\right )+10 x}{2 \ln \left (x^{2}-6 x +9\right ) x \left (1+2 x \right )}\) | \(41\) |
Input:
int(((31*x^3-93*x^2)*ln(x^2-6*x+9)^2+(-10*x^3-50*x^2+220*x+60)*ln(x^2-6*x+ 9)-20*x^3-90*x^2-40*x)/(4*x^5-8*x^4-11*x^3-3*x^2)/ln(x^2-6*x+9)^2,x,method =_RETURNVERBOSE)
Output:
-31/2/(1+2*x)+5*(4+x)/(1+2*x)/x/ln(x^2-6*x+9)
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=-\frac {31 \, x \log \left (x^{2} - 6 \, x + 9\right ) - 10 \, x - 40}{2 \, {\left (2 \, x^{2} + x\right )} \log \left (x^{2} - 6 \, x + 9\right )} \] Input:
integrate(((31*x^3-93*x^2)*log(x^2-6*x+9)^2+(-10*x^3-50*x^2+220*x+60)*log( x^2-6*x+9)-20*x^3-90*x^2-40*x)/(4*x^5-8*x^4-11*x^3-3*x^2)/log(x^2-6*x+9)^2 ,x, algorithm="fricas")
Output:
-1/2*(31*x*log(x^2 - 6*x + 9) - 10*x - 40)/((2*x^2 + x)*log(x^2 - 6*x + 9) )
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=\frac {5 x + 20}{\left (2 x^{2} + x\right ) \log {\left (x^{2} - 6 x + 9 \right )}} - \frac {31}{4 x + 2} \] Input:
integrate(((31*x**3-93*x**2)*ln(x**2-6*x+9)**2+(-10*x**3-50*x**2+220*x+60) *ln(x**2-6*x+9)-20*x**3-90*x**2-40*x)/(4*x**5-8*x**4-11*x**3-3*x**2)/ln(x* *2-6*x+9)**2,x)
Output:
(5*x + 20)/((2*x**2 + x)*log(x**2 - 6*x + 9)) - 31/(4*x + 2)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=-\frac {31 \, x \log \left (x - 3\right ) - 5 \, x - 20}{2 \, {\left (2 \, x^{2} + x\right )} \log \left (x - 3\right )} \] Input:
integrate(((31*x^3-93*x^2)*log(x^2-6*x+9)^2+(-10*x^3-50*x^2+220*x+60)*log( x^2-6*x+9)-20*x^3-90*x^2-40*x)/(4*x^5-8*x^4-11*x^3-3*x^2)/log(x^2-6*x+9)^2 ,x, algorithm="maxima")
Output:
-1/2*(31*x*log(x - 3) - 5*x - 20)/((2*x^2 + x)*log(x - 3))
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=\frac {5 \, {\left (x + 4\right )}}{2 \, x^{2} \log \left (x^{2} - 6 \, x + 9\right ) + x \log \left (x^{2} - 6 \, x + 9\right )} - \frac {31}{2 \, {\left (2 \, x + 1\right )}} \] Input:
integrate(((31*x^3-93*x^2)*log(x^2-6*x+9)^2+(-10*x^3-50*x^2+220*x+60)*log( x^2-6*x+9)-20*x^3-90*x^2-40*x)/(4*x^5-8*x^4-11*x^3-3*x^2)/log(x^2-6*x+9)^2 ,x, algorithm="giac")
Output:
5*(x + 4)/(2*x^2*log(x^2 - 6*x + 9) + x*log(x^2 - 6*x + 9)) - 31/2/(2*x + 1)
Time = 3.52 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=\frac {31\,x}{2\,x+1}+\frac {5\,x+20}{x\,\ln \left (x^2-6\,x+9\right )\,\left (2\,x+1\right )} \] Input:
int((40*x + log(x^2 - 6*x + 9)^2*(93*x^2 - 31*x^3) - log(x^2 - 6*x + 9)*(2 20*x - 50*x^2 - 10*x^3 + 60) + 90*x^2 + 20*x^3)/(log(x^2 - 6*x + 9)^2*(3*x ^2 + 11*x^3 + 8*x^4 - 4*x^5)),x)
Output:
(31*x)/(2*x + 1) + (5*x + 20)/(x*log(x^2 - 6*x + 9)*(2*x + 1))
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {-40 x-90 x^2-20 x^3+\left (60+220 x-50 x^2-10 x^3\right ) \log \left (9-6 x+x^2\right )+\left (-93 x^2+31 x^3\right ) \log ^2\left (9-6 x+x^2\right )}{\left (-3 x^2-11 x^3-8 x^4+4 x^5\right ) \log ^2\left (9-6 x+x^2\right )} \, dx=\frac {31 \,\mathrm {log}\left (x^{2}-6 x +9\right ) x^{2}+5 x +20}{\mathrm {log}\left (x^{2}-6 x +9\right ) x \left (2 x +1\right )} \] Input:
int(((31*x^3-93*x^2)*log(x^2-6*x+9)^2+(-10*x^3-50*x^2+220*x+60)*log(x^2-6* x+9)-20*x^3-90*x^2-40*x)/(4*x^5-8*x^4-11*x^3-3*x^2)/log(x^2-6*x+9)^2,x)
Output:
(31*log(x**2 - 6*x + 9)*x**2 + 5*x + 20)/(log(x**2 - 6*x + 9)*x*(2*x + 1))