Integrand size = 67, antiderivative size = 26 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=\log \left (\frac {x}{e^2+\frac {x+5 (-4+x-\log (x))^2}{x}}\right ) \] Output:
ln(x/((5*(-ln(x)+x-4)^2+x)/x+exp(2)))
Time = 0.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=2 \log (x)-\log \left (80-39 x+e^2 x+5 x^2+40 \log (x)-10 x \log (x)+5 \log ^2(x)\right ) \] Input:
Integrate[(120 - 29*x + E^2*x + (70 - 10*x)*Log[x] + 10*Log[x]^2)/(80*x - 39*x^2 + E^2*x^2 + 5*x^3 + (40*x - 10*x^2)*Log[x] + 5*x*Log[x]^2),x]
Output:
2*Log[x] - Log[80 - 39*x + E^2*x + 5*x^2 + 40*Log[x] - 10*x*Log[x] + 5*Log [x]^2]
Time = 0.77 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {6, 6, 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.
\(\displaystyle \int \frac {e^2 x-29 x+10 \log ^2(x)+(70-10 x) \log (x)+120}{5 x^3+e^2 x^2-39 x^2+\left (40 x-10 x^2\right ) \log (x)+80 x+5 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (e^2-29\right ) x+10 \log ^2(x)+(70-10 x) \log (x)+120}{5 x^3+e^2 x^2-39 x^2+\left (40 x-10 x^2\right ) \log (x)+80 x+5 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {\left (e^2-29\right ) x+10 \log ^2(x)+(70-10 x) \log (x)+120}{5 x^3+\left (e^2-39\right ) x^2+\left (40 x-10 x^2\right ) \log (x)+80 x+5 x \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-10 x^2+49 \left (1-\frac {e^2}{49}\right ) x+10 x \log (x)-10 \log (x)-40}{x \left (5 x^2-39 \left (1-\frac {e^2}{39}\right ) x+5 \log ^2(x)-10 x \log (x)+40 \log (x)+80\right )}+\frac {2}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log (x)-\log \left (5 x^2-\left (39-e^2\right ) x+5 \log ^2(x)-10 x \log (x)+40 \log (x)+80\right )\) |
Input:
Int[(120 - 29*x + E^2*x + (70 - 10*x)*Log[x] + 10*Log[x]^2)/(80*x - 39*x^2 + E^2*x^2 + 5*x^3 + (40*x - 10*x^2)*Log[x] + 5*x*Log[x]^2),x]
Output:
2*Log[x] - Log[80 - (39 - E^2)*x + 5*x^2 + 40*Log[x] - 10*x*Log[x] + 5*Log [x]^2]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 0.32 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \(2 \ln \left (x \right )-\ln \left (\ln \left (x \right )^{2}+\left (-2 x +8\right ) \ln \left (x \right )+\frac {{\mathrm e}^{2} x}{5}+x^{2}-\frac {39 x}{5}+16\right )\) | \(34\) |
parallelrisch | \(-\ln \left (\ln \left (x \right )^{2}-2 x \ln \left (x \right )+8 \ln \left (x \right )+\frac {{\mathrm e}^{2} x}{5}+x^{2}-\frac {39 x}{5}+16\right )+2 \ln \left (x \right )\) | \(35\) |
default | \(2 \ln \left (x \right )-\ln \left (5 \ln \left (x \right )^{2}-10 x \ln \left (x \right )+{\mathrm e}^{2} x +5 x^{2}+40 \ln \left (x \right )-39 x +80\right )\) | \(38\) |
norman | \(2 \ln \left (x \right )-\ln \left (5 \ln \left (x \right )^{2}-10 x \ln \left (x \right )+{\mathrm e}^{2} x +5 x^{2}+40 \ln \left (x \right )-39 x +80\right )\) | \(38\) |
Input:
int((10*ln(x)^2+(-10*x+70)*ln(x)+exp(2)*x-29*x+120)/(5*x*ln(x)^2+(-10*x^2+ 40*x)*ln(x)+x^2*exp(2)+5*x^3-39*x^2+80*x),x,method=_RETURNVERBOSE)
Output:
2*ln(x)-ln(ln(x)^2+(-2*x+8)*ln(x)+1/5*exp(2)*x+x^2-39/5*x+16)
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=-\log \left (5 \, x^{2} + x e^{2} - 10 \, {\left (x - 4\right )} \log \left (x\right ) + 5 \, \log \left (x\right )^{2} - 39 \, x + 80\right ) + 2 \, \log \left (x\right ) \] Input:
integrate((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+ (-10*x^2+40*x)*log(x)+x^2*exp(2)+5*x^3-39*x^2+80*x),x, algorithm="fricas")
Output:
-log(5*x^2 + x*e^2 - 10*(x - 4)*log(x) + 5*log(x)^2 - 39*x + 80) + 2*log(x )
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=2 \log {\left (x \right )} - \log {\left (x^{2} - \frac {39 x}{5} + \frac {x e^{2}}{5} + \left (8 - 2 x\right ) \log {\left (x \right )} + \log {\left (x \right )}^{2} + 16 \right )} \] Input:
integrate((10*ln(x)**2+(-10*x+70)*ln(x)+exp(2)*x-29*x+120)/(5*x*ln(x)**2+( -10*x**2+40*x)*ln(x)+x**2*exp(2)+5*x**3-39*x**2+80*x),x)
Output:
2*log(x) - log(x**2 - 39*x/5 + x*exp(2)/5 + (8 - 2*x)*log(x) + log(x)**2 + 16)
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=-\log \left (x^{2} + \frac {1}{5} \, x {\left (e^{2} - 39\right )} - 2 \, {\left (x - 4\right )} \log \left (x\right ) + \log \left (x\right )^{2} + 16\right ) + 2 \, \log \left (x\right ) \] Input:
integrate((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+ (-10*x^2+40*x)*log(x)+x^2*exp(2)+5*x^3-39*x^2+80*x),x, algorithm="maxima")
Output:
-log(x^2 + 1/5*x*(e^2 - 39) - 2*(x - 4)*log(x) + log(x)^2 + 16) + 2*log(x)
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=-\log \left (-5 \, x^{2} - x e^{2} + 10 \, x \log \left (x\right ) - 5 \, \log \left (x\right )^{2} + 39 \, x - 40 \, \log \left (x\right ) - 80\right ) + 2 \, \log \left (x\right ) \] Input:
integrate((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+ (-10*x^2+40*x)*log(x)+x^2*exp(2)+5*x^3-39*x^2+80*x),x, algorithm="giac")
Output:
-log(-5*x^2 - x*e^2 + 10*x*log(x) - 5*log(x)^2 + 39*x - 40*log(x) - 80) + 2*log(x)
Time = 2.39 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=2\,\ln \left (x\right )-\ln \left (8\,\ln \left (x\right )-\frac {39\,x}{5}+\frac {x\,{\mathrm {e}}^2}{5}+{\ln \left (x\right )}^2-2\,x\,\ln \left (x\right )+x^2+16\right ) \] Input:
int((x*exp(2) - 29*x + 10*log(x)^2 - log(x)*(10*x - 70) + 120)/(80*x + 5*x *log(x)^2 + x^2*exp(2) + log(x)*(40*x - 10*x^2) - 39*x^2 + 5*x^3),x)
Output:
2*log(x) - log(8*log(x) - (39*x)/5 + (x*exp(2))/5 + log(x)^2 - 2*x*log(x) + x^2 + 16)
\[ \int \frac {120-29 x+e^2 x+(70-10 x) \log (x)+10 \log ^2(x)}{80 x-39 x^2+e^2 x^2+5 x^3+\left (40 x-10 x^2\right ) \log (x)+5 x \log ^2(x)} \, dx=10 \left (\int \frac {\mathrm {log}\left (x \right )^{2}}{5 \mathrm {log}\left (x \right )^{2} x -10 \,\mathrm {log}\left (x \right ) x^{2}+40 \,\mathrm {log}\left (x \right ) x +e^{2} x^{2}+5 x^{3}-39 x^{2}+80 x}d x \right )+70 \left (\int \frac {\mathrm {log}\left (x \right )}{5 \mathrm {log}\left (x \right )^{2} x -10 \,\mathrm {log}\left (x \right ) x^{2}+40 \,\mathrm {log}\left (x \right ) x +e^{2} x^{2}+5 x^{3}-39 x^{2}+80 x}d x \right )-10 \left (\int \frac {\mathrm {log}\left (x \right )}{5 \mathrm {log}\left (x \right )^{2}-10 \,\mathrm {log}\left (x \right ) x +40 \,\mathrm {log}\left (x \right )+e^{2} x +5 x^{2}-39 x +80}d x \right )+120 \left (\int \frac {1}{5 \mathrm {log}\left (x \right )^{2} x -10 \,\mathrm {log}\left (x \right ) x^{2}+40 \,\mathrm {log}\left (x \right ) x +e^{2} x^{2}+5 x^{3}-39 x^{2}+80 x}d x \right )+\left (\int \frac {1}{5 \mathrm {log}\left (x \right )^{2}-10 \,\mathrm {log}\left (x \right ) x +40 \,\mathrm {log}\left (x \right )+e^{2} x +5 x^{2}-39 x +80}d x \right ) e^{2}-29 \left (\int \frac {1}{5 \mathrm {log}\left (x \right )^{2}-10 \,\mathrm {log}\left (x \right ) x +40 \,\mathrm {log}\left (x \right )+e^{2} x +5 x^{2}-39 x +80}d x \right ) \] Input:
int((10*log(x)^2+(-10*x+70)*log(x)+exp(2)*x-29*x+120)/(5*x*log(x)^2+(-10*x ^2+40*x)*log(x)+x^2*exp(2)+5*x^3-39*x^2+80*x),x)
Output:
10*int(log(x)**2/(5*log(x)**2*x - 10*log(x)*x**2 + 40*log(x)*x + e**2*x**2 + 5*x**3 - 39*x**2 + 80*x),x) + 70*int(log(x)/(5*log(x)**2*x - 10*log(x)* x**2 + 40*log(x)*x + e**2*x**2 + 5*x**3 - 39*x**2 + 80*x),x) - 10*int(log( x)/(5*log(x)**2 - 10*log(x)*x + 40*log(x) + e**2*x + 5*x**2 - 39*x + 80),x ) + 120*int(1/(5*log(x)**2*x - 10*log(x)*x**2 + 40*log(x)*x + e**2*x**2 + 5*x**3 - 39*x**2 + 80*x),x) + int(1/(5*log(x)**2 - 10*log(x)*x + 40*log(x) + e**2*x + 5*x**2 - 39*x + 80),x)*e**2 - 29*int(1/(5*log(x)**2 - 10*log(x )*x + 40*log(x) + e**2*x + 5*x**2 - 39*x + 80),x)