Integrand size = 75, antiderivative size = 21 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {x}{4}+\left (\log (3)+\frac {1}{x \log (3+x)}\right )^2 \] Output:
1/4*x+(ln(3)+1/ln(3+x)/x)^2
Time = 1.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {x}{4}+\frac {1}{x^2 \log ^2(3+x)}+\frac {2 \log (3)}{x \log (3+x)} \] Input:
Integrate[(-8*x + (-24 - 8*x - 8*x^2*Log[3])*Log[3 + x] + (-24*x - 8*x^2)* Log[3]*Log[3 + x]^2 + (3*x^3 + x^4)*Log[3 + x]^3)/((12*x^3 + 4*x^4)*Log[3 + x]^3),x]
Output:
x/4 + 1/(x^2*Log[3 + x]^2) + (2*Log[3])/(x*Log[3 + 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.
\(\displaystyle \int \frac {\left (-8 x^2-24 x\right ) \log (3) \log ^2(x+3)+\left (-8 x^2 \log (3)-8 x-24\right ) \log (x+3)+\left (x^4+3 x^3\right ) \log ^3(x+3)-8 x}{\left (4 x^4+12 x^3\right ) \log ^3(x+3)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-8 x^2-24 x\right ) \log (3) \log ^2(x+3)+\left (-8 x^2 \log (3)-8 x-24\right ) \log (x+3)+\left (x^4+3 x^3\right ) \log ^3(x+3)-8 x}{x^3 (4 x+12) \log ^3(x+3)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2}{x^2 (x+3) \log ^3(x+3)}-\frac {\log (9)}{x^2 \log (x+3)}-\frac {2 \left (x^2 \log (3)+x+3\right )}{x^3 (x+3) \log ^2(x+3)}+\frac {1}{4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \frac {1}{x^3 \log ^2(x+3)}dx-\frac {2}{3} \int \frac {1}{x^2 \log ^3(x+3)}dx-\log (9) \int \frac {1}{x^2 \log (x+3)}dx+\frac {2}{9} \int \frac {1}{x \log ^3(x+3)}dx-\frac {2}{3} \log (3) \int \frac {1}{x \log ^2(x+3)}dx+\frac {x}{4}+\frac {1}{9 \log ^2(x+3)}-\frac {2 \log (3)}{3 \log (x+3)}\) |
Input:
Int[(-8*x + (-24 - 8*x - 8*x^2*Log[3])*Log[3 + x] + (-24*x - 8*x^2)*Log[3] *Log[3 + x]^2 + (3*x^3 + x^4)*Log[3 + x]^3)/((12*x^3 + 4*x^4)*Log[3 + x]^3 ),x]
Output:
$Aborted
Time = 6.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24
method | result | size |
risch | \(\frac {x}{4}+\frac {2 x \ln \left (3\right ) \ln \left (3+x \right )+1}{x^{2} \ln \left (3+x \right )^{2}}\) | \(26\) |
norman | \(\frac {1+\frac {x^{3} \ln \left (3+x \right )^{2}}{4}+2 x \ln \left (3\right ) \ln \left (3+x \right )}{\ln \left (3+x \right )^{2} x^{2}}\) | \(33\) |
parallelrisch | \(\frac {x^{3} \ln \left (3+x \right )^{2}+4-6 \ln \left (3+x \right )^{2} x^{2}+8 x \ln \left (3\right ) \ln \left (3+x \right )}{4 x^{2} \ln \left (3+x \right )^{2}}\) | \(44\) |
derivativedivides | \(\frac {3}{4}+\frac {x}{4}+\frac {2 \ln \left (3\right )}{x \ln \left (3+x \right )}+\frac {3 \ln \left (3+x \right )-2 x}{3 \ln \left (3+x \right )^{3} x^{2}}+\frac {2}{3 \ln \left (3+x \right )^{3} x}\) | \(51\) |
default | \(\frac {3}{4}+\frac {x}{4}+\frac {2 \ln \left (3\right )}{x \ln \left (3+x \right )}+\frac {3 \ln \left (3+x \right )-2 x}{3 \ln \left (3+x \right )^{3} x^{2}}+\frac {2}{3 \ln \left (3+x \right )^{3} x}\) | \(51\) |
Input:
int(((x^4+3*x^3)*ln(3+x)^3+(-8*x^2-24*x)*ln(3)*ln(3+x)^2+(-8*x^2*ln(3)-8*x -24)*ln(3+x)-8*x)/(4*x^4+12*x^3)/ln(3+x)^3,x,method=_RETURNVERBOSE)
Output:
1/4*x+1/x^2*(2*x*ln(3)*ln(3+x)+1)/ln(3+x)^2
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {x^{3} \log \left (x + 3\right )^{2} + 8 \, x \log \left (3\right ) \log \left (x + 3\right ) + 4}{4 \, x^{2} \log \left (x + 3\right )^{2}} \] Input:
integrate(((x^4+3*x^3)*log(3+x)^3+(-8*x^2-24*x)*log(3)*log(3+x)^2+(-8*x^2* log(3)-8*x-24)*log(3+x)-8*x)/(4*x^4+12*x^3)/log(3+x)^3,x, algorithm="frica s")
Output:
1/4*(x^3*log(x + 3)^2 + 8*x*log(3)*log(x + 3) + 4)/(x^2*log(x + 3)^2)
Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {x}{4} + \frac {2 x \log {\left (3 \right )} \log {\left (x + 3 \right )} + 1}{x^{2} \log {\left (x + 3 \right )}^{2}} \] Input:
integrate(((x**4+3*x**3)*ln(3+x)**3+(-8*x**2-24*x)*ln(3)*ln(3+x)**2+(-8*x* *2*ln(3)-8*x-24)*ln(3+x)-8*x)/(4*x**4+12*x**3)/ln(3+x)**3,x)
Output:
x/4 + (2*x*log(3)*log(x + 3) + 1)/(x**2*log(x + 3)**2)
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {x^{3} \log \left (x + 3\right )^{2} + 8 \, x \log \left (3\right ) \log \left (x + 3\right ) + 4}{4 \, x^{2} \log \left (x + 3\right )^{2}} \] Input:
integrate(((x^4+3*x^3)*log(3+x)^3+(-8*x^2-24*x)*log(3)*log(3+x)^2+(-8*x^2* log(3)-8*x-24)*log(3+x)-8*x)/(4*x^4+12*x^3)/log(3+x)^3,x, algorithm="maxim a")
Output:
1/4*(x^3*log(x + 3)^2 + 8*x*log(3)*log(x + 3) + 4)/(x^2*log(x + 3)^2)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {1}{4} \, x + \frac {2 \, x \log \left (3\right ) \log \left (x + 3\right ) + 1}{x^{2} \log \left (x + 3\right )^{2}} \] Input:
integrate(((x^4+3*x^3)*log(3+x)^3+(-8*x^2-24*x)*log(3)*log(3+x)^2+(-8*x^2* log(3)-8*x-24)*log(3+x)-8*x)/(4*x^4+12*x^3)/log(3+x)^3,x, algorithm="giac" )
Output:
1/4*x + (2*x*log(3)*log(x + 3) + 1)/(x^2*log(x + 3)^2)
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {x}{4}+\frac {2\,x\,\ln \left (x+3\right )\,\ln \left (3\right )+1}{x^2\,{\ln \left (x+3\right )}^2} \] Input:
int(-(8*x - log(x + 3)^3*(3*x^3 + x^4) + log(x + 3)*(8*x + 8*x^2*log(3) + 24) + log(x + 3)^2*log(3)*(24*x + 8*x^2))/(log(x + 3)^3*(12*x^3 + 4*x^4)), x)
Output:
x/4 + (2*x*log(x + 3)*log(3) + 1)/(x^2*log(x + 3)^2)
Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {-8 x+\left (-24-8 x-8 x^2 \log (3)\right ) \log (3+x)+\left (-24 x-8 x^2\right ) \log (3) \log ^2(3+x)+\left (3 x^3+x^4\right ) \log ^3(3+x)}{\left (12 x^3+4 x^4\right ) \log ^3(3+x)} \, dx=\frac {\mathrm {log}\left (x +3\right )^{2} x^{3}+8 \,\mathrm {log}\left (x +3\right ) \mathrm {log}\left (3\right ) x +4}{4 \mathrm {log}\left (x +3\right )^{2} x^{2}} \] Input:
int(((x^4+3*x^3)*log(3+x)^3+(-8*x^2-24*x)*log(3)*log(3+x)^2+(-8*x^2*log(3) -8*x-24)*log(3+x)-8*x)/(4*x^4+12*x^3)/log(3+x)^3,x)
Output:
(log(x + 3)**2*x**3 + 8*log(x + 3)*log(3)*x + 4)/(4*log(x + 3)**2*x**2)