Integrand size = 155, antiderivative size = 30 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=e^3 x+\frac {e^3 x}{(-5+x) \left (-x^2+\log (3 (3+x))\right )} \] Output:
x*exp(3)+x/(-5+x)/exp(ln(ln(3*x+9)-x^2)-3)
Time = 0.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=e^3 \left (x+\frac {x}{(-5+x) \left (-x^2+\log (3 (3+x))\right )}\right ) \] Input:
Integrate[(E^3*(5*x - 16*x^2 + x^3 + 2*x^4 + (-15 - 5*x)*Log[9 + 3*x] + (( -x^2 + Log[9 + 3*x])*(E^3*(-75*x^2 + 5*x^3 + 7*x^4 - x^5) + E^3*(75 - 5*x - 7*x^2 + x^3)*Log[9 + 3*x]))/E^3))/((-x^2 + Log[9 + 3*x])*(-75*x^2 + 5*x^ 3 + 7*x^4 - x^5 + (75 - 5*x - 7*x^2 + x^3)*Log[9 + 3*x])),x]
Output:
E^3*(x + x/((-5 + x)*(-x^2 + Log[3*(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.
| \(\displaystyle \int \frac {e^3 \left (2 x^4+x^3-16 x^2+\frac {\left (\log (3 x+9)-x^2\right ) \left (e^3 \left (x^3-7 x^2-5 x+75\right ) \log (3 x+9)+e^3 \left (-x^5+7 x^4+5 x^3-75 x^2\right )\right )}{e^3}+5 x+(-5 x-15) \log (3 x+9)\right )}{\left (\log (3 x+9)-x^2\right ) \left (-x^5+7 x^4+5 x^3-75 x^2+\left (x^3-7 x^2-5 x+75\right ) \log (3 x+9)\right )} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle e^3 \int \frac {2 x^4+x^3-16 x^2+5 x-5 (x+3) \log (3 x+9)+\left (x^2-\log (3 x+9)\right ) \left (x^5-7 x^4-5 x^3+75 x^2-\left (x^3-7 x^2-5 x+75\right ) \log (3 x+9)\right )}{\left (x^2-\log (3 x+9)\right ) \left (x^5-7 x^4-5 x^3+75 x^2-\left (x^3-7 x^2-5 x+75\right ) \log (3 x+9)\right )}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle e^3 \int \frac {2 x^4+x^3-16 x^2+5 x-5 (x+3) \log (3 x+9)+\left (x^2-\log (3 x+9)\right ) \left (x^5-7 x^4-5 x^3+75 x^2-\left (x^3-7 x^2-5 x+75\right ) \log (3 x+9)\right )}{(5-x)^2 (x+3) \left (x^2-\log (3 x+9)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle e^3 \int \left (\frac {x \left (2 x^2+6 x-1\right )}{(x-5) (x+3) \left (x^2-\log (3 x+9)\right )^2}+\frac {5}{(x-5)^2 \left (x^2-\log (3 x+9)\right )}+1\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle e^3 \left (10 \int \frac {1}{\left (x^2-\log (3 x+9)\right )^2}dx+\frac {395}{8} \int \frac {1}{(x-5) \left (x^2-\log (3 x+9)\right )^2}dx+2 \int \frac {x}{\left (x^2-\log (3 x+9)\right )^2}dx-\frac {3}{8} \int \frac {1}{(x+3) \left (x^2-\log (3 x+9)\right )^2}dx+5 \int \frac {1}{(x-5)^2 \left (x^2-\log (3 x+9)\right )}dx+x\right )\) |
Input:
Int[(E^3*(5*x - 16*x^2 + x^3 + 2*x^4 + (-15 - 5*x)*Log[9 + 3*x] + ((-x^2 + Log[9 + 3*x])*(E^3*(-75*x^2 + 5*x^3 + 7*x^4 - x^5) + E^3*(75 - 5*x - 7*x^ 2 + x^3)*Log[9 + 3*x]))/E^3))/((-x^2 + Log[9 + 3*x])*(-75*x^2 + 5*x^3 + 7* x^4 - x^5 + (75 - 5*x - 7*x^2 + x^3)*Log[9 + 3*x])),x]
Output:
$Aborted
Time = 23.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(x \,{\mathrm e}^{3}-\frac {{\mathrm e}^{3} x}{\left (-5+x \right ) \left (x^{2}-\ln \left (3 x +9\right )\right )}\) | \(30\) |
| default | \(x \,{\mathrm e}^{3}+\frac {{\mathrm e}^{3} x}{\left (-5+x \right ) \left (-\left (3+x \right )^{2}+\ln \left (3\right )+\ln \left (3+x \right )+6 x +9\right )}\) | \(35\) |
| parts | \(x \,{\mathrm e}^{3}+\frac {{\mathrm e}^{3} x}{\left (-5+x \right ) \left (-\left (3+x \right )^{2}+\ln \left (3\right )+\ln \left (3+x \right )+6 x +9\right )}\) | \(35\) |
| norman | \(\frac {x^{4} {\mathrm e}^{3}-25 x^{2} {\mathrm e}^{3}+25 \,{\mathrm e}^{3} \ln \left (3 x +9\right )-{\mathrm e}^{3} x^{2} \ln \left (3 x +9\right )-x \,{\mathrm e}^{3}}{\left (-5+x \right ) \left (x^{2}-\ln \left (3 x +9\right )\right )}\) | \(63\) |
| parallelrisch | \(\frac {\left (12 \,{\mathrm e}^{3} \ln \left (3 x +9\right ) {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3} x^{3}-10 \,{\mathrm e}^{3} \ln \left (3 x +9\right ) {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3} x^{2}+x^{5}-6 \,{\mathrm e}^{3} {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3} x^{5}+5 \,{\mathrm e}^{3} {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3} x^{4}-2 \ln \left (3 x +9\right ) x^{3}+\ln \left (3 x +9\right )^{2} x +{\mathrm e}^{3} x^{6} {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3}+5 \ln \left (3 x +9\right )^{2} {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3} {\mathrm e}^{3}-6 \,{\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3} \ln \left (3 x +9\right )^{2} x \,{\mathrm e}^{3}-2 \,{\mathrm e}^{3} x^{4} \ln \left (3 x +9\right ) {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3}+{\mathrm e}^{3} x^{2} \ln \left (3 x +9\right )^{2} {\mathrm e}^{\ln \left (\ln \left (3 x +9\right )-x^{2}\right )-3}\right ) {\mathrm e}^{3}}{\left (-5+x \right ) \left (\ln \left (3 x +9\right )-x^{2}\right ) \left (x^{2}-\ln \left (3 x +9\right )\right )^{2}}\) | \(306\) |
Input:
int((((x^3-7*x^2-5*x+75)*exp(3)*ln(3*x+9)+(-x^5+7*x^4+5*x^3-75*x^2)*exp(3) )*exp(ln(ln(3*x+9)-x^2)-3)+(-5*x-15)*ln(3*x+9)+2*x^4+x^3-16*x^2+5*x)/((x^3 -7*x^2-5*x+75)*ln(3*x+9)-x^5+7*x^4+5*x^3-75*x^2)/exp(ln(ln(3*x+9)-x^2)-3), x,method=_RETURNVERBOSE)
Output:
x*exp(3)-exp(3)*x/(-5+x)/(x^2-ln(3*x+9))
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.90 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=-\frac {{\left (x^{2} - 5 \, x\right )} e^{3} \log \left (3 \, x + 9\right ) - {\left (x^{4} - 5 \, x^{3} - x\right )} e^{3}}{x^{3} - 5 \, x^{2} - {\left (x - 5\right )} \log \left (3 \, x + 9\right )} \] Input:
integrate((((x^3-7*x^2-5*x+75)*exp(3)*log(3*x+9)+(-x^5+7*x^4+5*x^3-75*x^2) *exp(3))*exp(log(log(3*x+9)-x^2)-3)+(-5*x-15)*log(3*x+9)+2*x^4+x^3-16*x^2+ 5*x)/((x^3-7*x^2-5*x+75)*log(3*x+9)-x^5+7*x^4+5*x^3-75*x^2)/exp(log(log(3* x+9)-x^2)-3),x, algorithm="fricas")
Output:
-((x^2 - 5*x)*e^3*log(3*x + 9) - (x^4 - 5*x^3 - x)*e^3)/(x^3 - 5*x^2 - (x - 5)*log(3*x + 9))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=x e^{3} + \frac {x e^{3}}{- x^{3} + 5 x^{2} + \left (x - 5\right ) \log {\left (3 x + 9 \right )}} \] Input:
integrate((((x**3-7*x**2-5*x+75)*exp(3)*ln(3*x+9)+(-x**5+7*x**4+5*x**3-75* x**2)*exp(3))*exp(ln(ln(3*x+9)-x**2)-3)+(-5*x-15)*ln(3*x+9)+2*x**4+x**3-16 *x**2+5*x)/((x**3-7*x**2-5*x+75)*ln(3*x+9)-x**5+7*x**4+5*x**3-75*x**2)/exp (ln(ln(3*x+9)-x**2)-3),x)
Output:
x*exp(3) + x*exp(3)/(-x**3 + 5*x**2 + (x - 5)*log(3*x + 9))
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (30) = 60\).
Time = 0.16 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=\frac {{\left (x^{4} - 5 \, x^{3} - x^{2} \log \left (3\right ) + x {\left (5 \, \log \left (3\right ) - 1\right )} - {\left (x^{2} - 5 \, x\right )} \log \left (x + 3\right )\right )} e^{3}}{x^{3} - 5 \, x^{2} - x \log \left (3\right ) - {\left (x - 5\right )} \log \left (x + 3\right ) + 5 \, \log \left (3\right )} \] Input:
integrate((((x^3-7*x^2-5*x+75)*exp(3)*log(3*x+9)+(-x^5+7*x^4+5*x^3-75*x^2) *exp(3))*exp(log(log(3*x+9)-x^2)-3)+(-5*x-15)*log(3*x+9)+2*x^4+x^3-16*x^2+ 5*x)/((x^3-7*x^2-5*x+75)*log(3*x+9)-x^5+7*x^4+5*x^3-75*x^2)/exp(log(log(3* x+9)-x^2)-3),x, algorithm="maxima")
Output:
(x^4 - 5*x^3 - x^2*log(3) + x*(5*log(3) - 1) - (x^2 - 5*x)*log(x + 3))*e^3 /(x^3 - 5*x^2 - x*log(3) - (x - 5)*log(x + 3) + 5*log(3))
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (30) = 60\).
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.97 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=\frac {x^{4} e^{3} - 5 \, x^{3} e^{3} - x^{2} e^{3} \log \left (3\right ) - x^{2} e^{3} \log \left (x + 3\right ) + 5 \, x e^{3} \log \left (3\right ) + 5 \, x e^{3} \log \left (x + 3\right ) - x e^{3}}{x^{3} - 5 \, x^{2} - x \log \left (3\right ) - x \log \left (x + 3\right ) + 5 \, \log \left (3\right ) + 5 \, \log \left (x + 3\right )} \] Input:
integrate((((x^3-7*x^2-5*x+75)*exp(3)*log(3*x+9)+(-x^5+7*x^4+5*x^3-75*x^2) *exp(3))*exp(log(log(3*x+9)-x^2)-3)+(-5*x-15)*log(3*x+9)+2*x^4+x^3-16*x^2+ 5*x)/((x^3-7*x^2-5*x+75)*log(3*x+9)-x^5+7*x^4+5*x^3-75*x^2)/exp(log(log(3* x+9)-x^2)-3),x, algorithm="giac")
Output:
(x^4*e^3 - 5*x^3*e^3 - x^2*e^3*log(3) - x^2*e^3*log(x + 3) + 5*x*e^3*log(3 ) + 5*x*e^3*log(x + 3) - x*e^3)/(x^3 - 5*x^2 - x*log(3) - x*log(x + 3) + 5 *log(3) + 5*log(x + 3))
Time = 3.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.70 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=\frac {x\,{\mathrm {e}}^3\,\left (x\,\ln \left (3\,x+9\right )-5\,\ln \left (3\,x+9\right )+5\,x^2-x^3+1\right )}{\left (\ln \left (3\,x+9\right )-x^2\right )\,\left (x-5\right )} \] Input:
int(-(exp(3 - log(log(3*x + 9) - x^2))*(5*x - log(3*x + 9)*(5*x + 15) - ex p(log(log(3*x + 9) - x^2) - 3)*(exp(3)*(75*x^2 - 5*x^3 - 7*x^4 + x^5) + ex p(3)*log(3*x + 9)*(5*x + 7*x^2 - x^3 - 75)) - 16*x^2 + x^3 + 2*x^4))/(log( 3*x + 9)*(5*x + 7*x^2 - x^3 - 75) + 75*x^2 - 5*x^3 - 7*x^4 + x^5),x)
Output:
(x*exp(3)*(x*log(3*x + 9) - 5*log(3*x + 9) + 5*x^2 - x^3 + 1))/((log(3*x + 9) - x^2)*(x - 5))
Time = 0.15 (sec) , antiderivative size = 130, normalized size of antiderivative = 4.33 \[ \int \frac {e^3 \left (5 x-16 x^2+x^3+2 x^4+(-15-5 x) \log (9+3 x)+\frac {\left (-x^2+\log (9+3 x)\right ) \left (e^3 \left (-75 x^2+5 x^3+7 x^4-x^5\right )+e^3 \left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )}{e^3}\right )}{\left (-x^2+\log (9+3 x)\right ) \left (-75 x^2+5 x^3+7 x^4-x^5+\left (75-5 x-7 x^2+x^3\right ) \log (9+3 x)\right )} \, dx=\frac {e^{3} \left (-\mathrm {log}\left (3 x +9\right )^{2} x +5 \mathrm {log}\left (3 x +9\right )^{2}+\mathrm {log}\left (3 x +9\right ) \mathrm {log}\left (x +3\right ) x -5 \,\mathrm {log}\left (3 x +9\right ) \mathrm {log}\left (x +3\right )+\mathrm {log}\left (3 x +9\right ) x^{3}-25 \,\mathrm {log}\left (3 x +9\right ) x -\mathrm {log}\left (x +3\right ) x^{3}+5 \,\mathrm {log}\left (x +3\right ) x^{2}-5 x^{4}+25 x^{3}+5 x \right )}{5 \,\mathrm {log}\left (3 x +9\right ) x -25 \,\mathrm {log}\left (3 x +9\right )-5 x^{3}+25 x^{2}} \] Input:
int((((x^3-7*x^2-5*x+75)*exp(3)*log(3*x+9)+(-x^5+7*x^4+5*x^3-75*x^2)*exp(3 ))*exp(log(log(3*x+9)-x^2)-3)+(-5*x-15)*log(3*x+9)+2*x^4+x^3-16*x^2+5*x)/( (x^3-7*x^2-5*x+75)*log(3*x+9)-x^5+7*x^4+5*x^3-75*x^2)/exp(log(log(3*x+9)-x ^2)-3),x)
Output:
(e**3*( - log(3*x + 9)**2*x + 5*log(3*x + 9)**2 + log(3*x + 9)*log(x + 3)* x - 5*log(3*x + 9)*log(x + 3) + log(3*x + 9)*x**3 - 25*log(3*x + 9)*x - lo g(x + 3)*x**3 + 5*log(x + 3)*x**2 - 5*x**4 + 25*x**3 + 5*x))/(5*(log(3*x + 9)*x - 5*log(3*x + 9) - x**3 + 5*x**2))