Integrand size = 135, antiderivative size = 26 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=4-\frac {1}{x}+x-\frac {x}{3 (5-x-\log (8+x))} \] Output:
4-1/3*x/(5-ln(x+8)-x)+x-1/x
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {1}{3} \left (-\frac {3}{x}+3 x+\frac {x}{-5+x+\log (8+x)}\right ) \] Input:
Integrate[(600 - 165*x + 554*x^2 - 168*x^3 - 6*x^4 + 3*x^5 + (-240 + 18*x - 226*x^2 + 19*x^3 + 6*x^4)*Log[8 + x] + (24 + 3*x + 24*x^2 + 3*x^3)*Log[8 + x]^2)/(600*x^2 - 165*x^3 - 6*x^4 + 3*x^5 + (-240*x^2 + 18*x^3 + 6*x^4)* Log[8 + x] + (24*x^2 + 3*x^3)*Log[8 + x]^2),x]
Output:
(-3/x + 3*x + x/(-5 + x + Log[8 + x]))/3
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.
\(\displaystyle \int \frac {3 x^5-6 x^4-168 x^3+554 x^2+\left (3 x^3+24 x^2+3 x+24\right ) \log ^2(x+8)+\left (6 x^4+19 x^3-226 x^2+18 x-240\right ) \log (x+8)-165 x+600}{3 x^5-6 x^4-165 x^3+600 x^2+\left (3 x^3+24 x^2\right ) \log ^2(x+8)+\left (6 x^4+18 x^3-240 x^2\right ) \log (x+8)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 x^5-6 x^4-168 x^3+554 x^2+\left (3 x^3+24 x^2+3 x+24\right ) \log ^2(x+8)+\left (6 x^4+19 x^3-226 x^2+18 x-240\right ) \log (x+8)-165 x+600}{3 x^2 (x+8) (-x-\log (x+8)+5)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {3 x^5-6 x^4-168 x^3+554 x^2-165 x+3 \left (x^3+8 x^2+x+8\right ) \log ^2(x+8)-\left (-6 x^4-19 x^3+226 x^2-18 x+240\right ) \log (x+8)+600}{x^2 (x+8) (-x-\log (x+8)+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (-\frac {x (x+9)}{(x+8) (x+\log (x+8)-5)^2}+\frac {3 \left (x^2+1\right )}{x^2}+\frac {1}{x+\log (x+8)-5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\int \frac {1}{(x+\log (x+8)-5)^2}dx-\int \frac {x}{(x+\log (x+8)-5)^2}dx+8 \int \frac {1}{(x+8) (x+\log (x+8)-5)^2}dx+\int \frac {1}{x+\log (x+8)-5}dx+3 x-\frac {3}{x}\right )\) |
Input:
Int[(600 - 165*x + 554*x^2 - 168*x^3 - 6*x^4 + 3*x^5 + (-240 + 18*x - 226* x^2 + 19*x^3 + 6*x^4)*Log[8 + x] + (24 + 3*x + 24*x^2 + 3*x^3)*Log[8 + x]^ 2)/(600*x^2 - 165*x^3 - 6*x^4 + 3*x^5 + (-240*x^2 + 18*x^3 + 6*x^4)*Log[8 + x] + (24*x^2 + 3*x^3)*Log[8 + x]^2),x]
Output:
$Aborted
Time = 1.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x^{2}-1}{x}+\frac {x}{3 \ln \left (x +8\right )+3 x -15}\) | \(23\) |
norman | \(\frac {5+x^{3}-\frac {73 x}{3}+x^{2} \ln \left (x +8\right )+\frac {14 x \ln \left (x +8\right )}{3}-\ln \left (x +8\right )}{x \left (\ln \left (x +8\right )+x -5\right )}\) | \(43\) |
derivativedivides | \(\frac {-211 x +911-195 \ln \left (x +8\right )-38 \left (x +8\right )^{2}+3 \left (x +8\right )^{3}+3 \ln \left (x +8\right ) \left (x +8\right )^{2}}{3 x \left (\ln \left (x +8\right )+x -5\right )}\) | \(51\) |
default | \(\frac {-211 x +911-195 \ln \left (x +8\right )-38 \left (x +8\right )^{2}+3 \left (x +8\right )^{3}+3 \ln \left (x +8\right ) \left (x +8\right )^{2}}{3 x \left (\ln \left (x +8\right )+x -5\right )}\) | \(51\) |
parallelrisch | \(\frac {15+3 x^{3}+3 x^{2} \ln \left (x +8\right )-62 x^{2}-48 x \ln \left (x +8\right )+237 x -3 \ln \left (x +8\right )}{3 x \left (\ln \left (x +8\right )+x -5\right )}\) | \(52\) |
Input:
int(((3*x^3+24*x^2+3*x+24)*ln(x+8)^2+(6*x^4+19*x^3-226*x^2+18*x-240)*ln(x+ 8)+3*x^5-6*x^4-168*x^3+554*x^2-165*x+600)/((3*x^3+24*x^2)*ln(x+8)^2+(6*x^4 +18*x^3-240*x^2)*ln(x+8)+3*x^5-6*x^4-165*x^3+600*x^2),x,method=_RETURNVERB OSE)
Output:
(x^2-1)/x+1/3*x/(ln(x+8)+x-5)
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {3 \, x^{3} - 14 \, x^{2} + 3 \, {\left (x^{2} - 1\right )} \log \left (x + 8\right ) - 3 \, x + 15}{3 \, {\left (x^{2} + x \log \left (x + 8\right ) - 5 \, x\right )}} \] Input:
integrate(((3*x^3+24*x^2+3*x+24)*log(x+8)^2+(6*x^4+19*x^3-226*x^2+18*x-240 )*log(x+8)+3*x^5-6*x^4-168*x^3+554*x^2-165*x+600)/((3*x^3+24*x^2)*log(x+8) ^2+(6*x^4+18*x^3-240*x^2)*log(x+8)+3*x^5-6*x^4-165*x^3+600*x^2),x, algorit hm="fricas")
Output:
1/3*(3*x^3 - 14*x^2 + 3*(x^2 - 1)*log(x + 8) - 3*x + 15)/(x^2 + x*log(x + 8) - 5*x)
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=x + \frac {x}{3 x + 3 \log {\left (x + 8 \right )} - 15} - \frac {1}{x} \] Input:
integrate(((3*x**3+24*x**2+3*x+24)*ln(x+8)**2+(6*x**4+19*x**3-226*x**2+18* x-240)*ln(x+8)+3*x**5-6*x**4-168*x**3+554*x**2-165*x+600)/((3*x**3+24*x**2 )*ln(x+8)**2+(6*x**4+18*x**3-240*x**2)*ln(x+8)+3*x**5-6*x**4-165*x**3+600* x**2),x)
Output:
x + x/(3*x + 3*log(x + 8) - 15) - 1/x
Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {3 \, x^{3} - 14 \, x^{2} + 3 \, {\left (x^{2} - 1\right )} \log \left (x + 8\right ) - 3 \, x + 15}{3 \, {\left (x^{2} + x \log \left (x + 8\right ) - 5 \, x\right )}} \] Input:
integrate(((3*x^3+24*x^2+3*x+24)*log(x+8)^2+(6*x^4+19*x^3-226*x^2+18*x-240 )*log(x+8)+3*x^5-6*x^4-168*x^3+554*x^2-165*x+600)/((3*x^3+24*x^2)*log(x+8) ^2+(6*x^4+18*x^3-240*x^2)*log(x+8)+3*x^5-6*x^4-165*x^3+600*x^2),x, algorit hm="maxima")
Output:
1/3*(3*x^3 - 14*x^2 + 3*(x^2 - 1)*log(x + 8) - 3*x + 15)/(x^2 + x*log(x + 8) - 5*x)
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=x + \frac {x}{3 \, {\left (x + \log \left (x + 8\right ) - 5\right )}} - \frac {1}{x} \] Input:
integrate(((3*x^3+24*x^2+3*x+24)*log(x+8)^2+(6*x^4+19*x^3-226*x^2+18*x-240 )*log(x+8)+3*x^5-6*x^4-168*x^3+554*x^2-165*x+600)/((3*x^3+24*x^2)*log(x+8) ^2+(6*x^4+18*x^3-240*x^2)*log(x+8)+3*x^5-6*x^4-165*x^3+600*x^2),x, algorit hm="giac")
Output:
x + 1/3*x/(x + log(x + 8) - 5) - 1/x
Time = 3.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=x-\frac {\ln \left (x+8\right )+x\,\left (\frac {\ln \left (x+8\right )}{3}-\frac {2}{3}\right )-5}{x\,\left (x+\ln \left (x+8\right )-5\right )} \] Input:
int((log(x + 8)*(18*x - 226*x^2 + 19*x^3 + 6*x^4 - 240) - 165*x + log(x + 8)^2*(3*x + 24*x^2 + 3*x^3 + 24) + 554*x^2 - 168*x^3 - 6*x^4 + 3*x^5 + 600 )/(log(x + 8)*(18*x^3 - 240*x^2 + 6*x^4) + log(x + 8)^2*(24*x^2 + 3*x^3) + 600*x^2 - 165*x^3 - 6*x^4 + 3*x^5),x)
Output:
x - (log(x + 8) + x*(log(x + 8)/3 - 2/3) - 5)/(x*(x + log(x + 8) - 5))
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.77 \[ \int \frac {600-165 x+554 x^2-168 x^3-6 x^4+3 x^5+\left (-240+18 x-226 x^2+19 x^3+6 x^4\right ) \log (8+x)+\left (24+3 x+24 x^2+3 x^3\right ) \log ^2(8+x)}{600 x^2-165 x^3-6 x^4+3 x^5+\left (-240 x^2+18 x^3+6 x^4\right ) \log (8+x)+\left (24 x^2+3 x^3\right ) \log ^2(8+x)} \, dx=\frac {3 \,\mathrm {log}\left (x +8\right ) x^{2}+14 \,\mathrm {log}\left (x +8\right ) x -3 \,\mathrm {log}\left (x +8\right )+3 x^{3}-73 x +15}{3 x \left (\mathrm {log}\left (x +8\right )+x -5\right )} \] Input:
int(((3*x^3+24*x^2+3*x+24)*log(x+8)^2+(6*x^4+19*x^3-226*x^2+18*x-240)*log( x+8)+3*x^5-6*x^4-168*x^3+554*x^2-165*x+600)/((3*x^3+24*x^2)*log(x+8)^2+(6* x^4+18*x^3-240*x^2)*log(x+8)+3*x^5-6*x^4-165*x^3+600*x^2),x)
Output:
(3*log(x + 8)*x**2 + 14*log(x + 8)*x - 3*log(x + 8) + 3*x**3 - 73*x + 15)/ (3*x*(log(x + 8) + x - 5))