Integrand size = 103, antiderivative size = 29 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx=\frac {e^{\left (2+\frac {1}{3} (-2+x)\right )^2 x}+x}{-5-e^x+\log (3)} \]
Time = 1.71 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx=-\frac {e^{\frac {1}{9} x (4+x)^2}+x}{5+e^x-\log (3)} \]
Integrate[(-45 + E^x*(-9 + 9*x) + 9*Log[3] + E^((16*x + 8*x^2 + x^3)/9)*(- 80 - 80*x - 15*x^2 + E^x*(-7 - 16*x - 3*x^2) + (16 + 16*x + 3*x^2)*Log[3]) )/(225 + 9*E^(2*x) + E^x*(90 - 18*Log[3]) - 90*Log[3] + 9*Log[3]^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.
| \(\displaystyle \int \frac {e^{\frac {1}{9} \left (x^3+8 x^2+16 x\right )} \left (-15 x^2+e^x \left (-3 x^2-16 x-7\right )+\left (3 x^2+16 x+16\right ) \log (3)-80 x-80\right )+e^x (9 x-9)-45+9 \log (3)}{9 e^{2 x}+e^x (90-18 \log (3))+225+9 \log ^2(3)-90 \log (3)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{\frac {1}{9} \left (x^3+8 x^2+16 x\right )} \left (-15 x^2+e^x \left (-3 x^2-16 x-7\right )+\left (3 x^2+16 x+16\right ) \log (3)-80 x-80\right )+e^x (9 x-9)-45 \left (1-\frac {\log (3)}{5}\right )}{9 \left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \int -\frac {9 e^x (1-x)+e^{\frac {1}{9} \left (x^3+8 x^2+16 x\right )} \left (15 x^2+80 x+e^x \left (3 x^2+16 x+7\right )-\left (3 x^2+16 x+16\right ) \log (3)+80\right )+9 (5-\log (3))}{\left (5+e^x-\log (3)\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{9} \int \frac {9 e^x (1-x)+e^{\frac {1}{9} \left (x^3+8 x^2+16 x\right )} \left (15 x^2+80 x+e^x \left (3 x^2+16 x+7\right )-\left (3 x^2+16 x+16\right ) \log (3)+80\right )+9 (5-\log (3))}{\left (5+e^x-\log (3)\right )^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {1}{9} \int \frac {9 e^x (1-x)+e^{\frac {1}{9} \left (x^3+8 x^2+16 x\right )} \left (15 x^2+80 x+e^x \left (3 x^2+16 x+7\right )-\left (3 x^2+16 x+16\right ) \log (3)+80\right )+9 (5-\log (3))}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {1}{9} \int \left (\frac {9 \left (-e^x x+e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}+\frac {e^{\frac {1}{9} x (x+4)^2} \left (3 e^x x^2+15 \left (1-\frac {\log (3)}{5}\right ) x^2+16 e^x x+80 \left (1-\frac {\log (3)}{5}\right ) x+7 e^x+80 \left (1-\frac {\log (3)}{5}\right )\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{9} \left (-3 \int \frac {e^{\frac {1}{9} x (x+4)^2} x^2}{e^x+5 \left (1-\frac {\log (3)}{5}\right )}dx-9 (5-\log (3)) \int \frac {e^{\frac {1}{9} x (x+4)^2}}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}dx-7 \int \frac {e^{\frac {1}{9} x (x+4)^2}}{e^x+5 \left (1-\frac {\log (3)}{5}\right )}dx-16 \int \frac {e^{\frac {1}{9} x (x+4)^2} x}{e^x+5 \left (1-\frac {\log (3)}{5}\right )}dx-\frac {9 x^2}{2 (5-\log (3))}+\frac {9 (1-x)^2}{2 (5-\log (3))}+\frac {9 (1-x) \log \left (\frac {e^x}{5-\log (3)}+1\right )}{5-\log (3)}+\frac {9 x \log \left (\frac {e^x}{5-\log (3)}+1\right )}{5-\log (3)}-\frac {9 \log \left (e^x+5-\log (3)\right )}{5-\log (3)}-\frac {9 x}{e^x+5-\log (3)}+\frac {9 x}{5-\log (3)}\right )\) |
Int[(-45 + E^x*(-9 + 9*x) + 9*Log[3] + E^((16*x + 8*x^2 + x^3)/9)*(-80 - 8 0*x - 15*x^2 + E^x*(-7 - 16*x - 3*x^2) + (16 + 16*x + 3*x^2)*Log[3]))/(225 + 9*E^(2*x) + E^x*(90 - 18*Log[3]) - 90*Log[3] + 9*Log[3]^2),x]
3.26.33.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
| method | result | size |
| norman | \(\frac {x +{\mathrm e}^{\frac {1}{9} x^{3}+\frac {8}{9} x^{2}+\frac {16}{9} x}}{\ln \left (3\right )-{\mathrm e}^{x}-5}\) | \(29\) |
| parallelrisch | \(-\frac {-9 x -9 \,{\mathrm e}^{\frac {1}{9} x^{3}+\frac {8}{9} x^{2}+\frac {16}{9} x}}{9 \left (\ln \left (3\right )-{\mathrm e}^{x}-5\right )}\) | \(34\) |
int((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*ln(3)-15*x^2-80*x-80)*exp(1/9 *x^3+8/9*x^2+16/9*x)+(9*x-9)*exp(x)+9*ln(3)-45)/(9*exp(x)^2+(-18*ln(3)+90) *exp(x)+9*ln(3)^2-90*ln(3)+225),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx=-\frac {x + e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )}}{e^{x} - \log \left (3\right ) + 5} \]
integrate((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*log(3)-15*x^2-80*x-80)* exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*exp(x)+9*log(3)-45)/(9*exp(x)^2+(-18*l og(3)+90)*exp(x)+9*log(3)^2-90*log(3)+225),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx=- \frac {x}{e^{x} - \log {\left (3 \right )} + 5} - \frac {e^{\frac {x^{3}}{9} + \frac {8 x^{2}}{9} + \frac {16 x}{9}}}{e^{x} - \log {\left (3 \right )} + 5} \]
integrate((((-3*x**2-16*x-7)*exp(x)+(3*x**2+16*x+16)*ln(3)-15*x**2-80*x-80 )*exp(1/9*x**3+8/9*x**2+16/9*x)+(9*x-9)*exp(x)+9*ln(3)-45)/(9*exp(x)**2+(- 18*ln(3)+90)*exp(x)+9*ln(3)**2-90*ln(3)+225),x)
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (23) = 46\).
Time = 0.39 (sec) , antiderivative size = 205, normalized size of antiderivative = 7.07 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx={\left (\frac {x}{\log \left (3\right )^{2} - 10 \, \log \left (3\right ) + 25} - \frac {\log \left (e^{x} - \log \left (3\right ) + 5\right )}{\log \left (3\right )^{2} - 10 \, \log \left (3\right ) + 25} - \frac {1}{{\left (\log \left (3\right ) - 5\right )} e^{x} - \log \left (3\right )^{2} + 10 \, \log \left (3\right ) - 25}\right )} \log \left (3\right ) - \frac {x e^{x}}{{\left (\log \left (3\right ) - 5\right )} e^{x} - \log \left (3\right )^{2} + 10 \, \log \left (3\right ) - 25} - \frac {5 \, x}{\log \left (3\right )^{2} - 10 \, \log \left (3\right ) + 25} - \frac {e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )}}{e^{x} - \log \left (3\right ) + 5} + \frac {5 \, \log \left (e^{x} - \log \left (3\right ) + 5\right )}{\log \left (3\right )^{2} - 10 \, \log \left (3\right ) + 25} + \frac {\log \left (e^{x} - \log \left (3\right ) + 5\right )}{\log \left (3\right ) - 5} + \frac {5}{{\left (\log \left (3\right ) - 5\right )} e^{x} - \log \left (3\right )^{2} + 10 \, \log \left (3\right ) - 25} + \frac {1}{e^{x} - \log \left (3\right ) + 5} \]
integrate((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*log(3)-15*x^2-80*x-80)* exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*exp(x)+9*log(3)-45)/(9*exp(x)^2+(-18*l og(3)+90)*exp(x)+9*log(3)^2-90*log(3)+225),x, algorithm=\
(x/(log(3)^2 - 10*log(3) + 25) - log(e^x - log(3) + 5)/(log(3)^2 - 10*log( 3) + 25) - 1/((log(3) - 5)*e^x - log(3)^2 + 10*log(3) - 25))*log(3) - x*e^ x/((log(3) - 5)*e^x - log(3)^2 + 10*log(3) - 25) - 5*x/(log(3)^2 - 10*log( 3) + 25) - e^(1/9*x^3 + 8/9*x^2 + 16/9*x)/(e^x - log(3) + 5) + 5*log(e^x - log(3) + 5)/(log(3)^2 - 10*log(3) + 25) + log(e^x - log(3) + 5)/(log(3) - 5) + 5/((log(3) - 5)*e^x - log(3)^2 + 10*log(3) - 25) + 1/(e^x - log(3) + 5)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (23) = 46\).
Time = 0.35 (sec) , antiderivative size = 140, normalized size of antiderivative = 4.83 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx=-\frac {x \log \left (3\right ) + e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )} \log \left (3\right ) - e^{x} \log \left (e^{x} - \log \left (3\right ) + 5\right ) + \log \left (3\right ) \log \left (e^{x} - \log \left (3\right ) + 5\right ) + e^{x} \log \left (-e^{x} + \log \left (3\right ) - 5\right ) - \log \left (3\right ) \log \left (-e^{x} + \log \left (3\right ) - 5\right ) - 5 \, x - 5 \, e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )} - 5 \, \log \left (e^{x} - \log \left (3\right ) + 5\right ) + 5 \, \log \left (-e^{x} + \log \left (3\right ) - 5\right )}{e^{x} \log \left (3\right ) - \log \left (3\right )^{2} - 5 \, e^{x} + 10 \, \log \left (3\right ) - 25} \]
integrate((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*log(3)-15*x^2-80*x-80)* exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*exp(x)+9*log(3)-45)/(9*exp(x)^2+(-18*l og(3)+90)*exp(x)+9*log(3)^2-90*log(3)+225),x, algorithm=\
-(x*log(3) + e^(1/9*x^3 + 8/9*x^2 + 16/9*x)*log(3) - e^x*log(e^x - log(3) + 5) + log(3)*log(e^x - log(3) + 5) + e^x*log(-e^x + log(3) - 5) - log(3)* log(-e^x + log(3) - 5) - 5*x - 5*e^(1/9*x^3 + 8/9*x^2 + 16/9*x) - 5*log(e^ x - log(3) + 5) + 5*log(-e^x + log(3) - 5))/(e^x*log(3) - log(3)^2 - 5*e^x + 10*log(3) - 25)
Time = 15.67 (sec) , antiderivative size = 225, normalized size of antiderivative = 7.76 \[ \int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx=\frac {5\,\ln \left ({\mathrm {e}}^x-\ln \left (3\right )+5\right )}{{\ln \left (3\right )}^2-10\,\ln \left (3\right )+25}+\frac {\frac {{\mathrm {e}}^x}{\ln \left (3\right )-5}-\frac {x\,{\mathrm {e}}^x}{\ln \left (3\right )-5}}{{\mathrm {e}}^x-\ln \left (3\right )+5}+\frac {\frac {5\,x}{\ln \left (3\right )-5}+\frac {5\,{\mathrm {e}}^x}{{\left (\ln \left (3\right )-5\right )}^2}-\frac {5\,x\,{\mathrm {e}}^x}{{\ln \left (3\right )}^2-10\,\ln \left (3\right )+25}}{{\mathrm {e}}^x-\ln \left (3\right )+5}-\frac {{\mathrm {e}}^{\frac {x^3}{9}+\frac {8\,x^2}{9}+\frac {16\,x}{9}}}{{\mathrm {e}}^x-\ln \left (3\right )+5}-\frac {\frac {x\,\ln \left (3\right )}{\ln \left (3\right )-5}+\frac {{\mathrm {e}}^x\,\ln \left (3\right )}{{\left (\ln \left (3\right )-5\right )}^2}-\frac {x\,{\mathrm {e}}^x\,\ln \left (3\right )}{{\ln \left (3\right )}^2-10\,\ln \left (3\right )+25}}{{\mathrm {e}}^x-\ln \left (3\right )+5}+\frac {\ln \left ({\mathrm {e}}^x-\ln \left (3\right )+5\right )}{\ln \left (3\right )-5}-\frac {\ln \left (3\right )\,\ln \left ({\mathrm {e}}^x-\ln \left (3\right )+5\right )}{{\ln \left (3\right )}^2-10\,\ln \left (3\right )+25} \]
int((9*log(3) + exp(x)*(9*x - 9) - exp((16*x)/9 + (8*x^2)/9 + x^3/9)*(80*x - log(3)*(16*x + 3*x^2 + 16) + exp(x)*(16*x + 3*x^2 + 7) + 15*x^2 + 80) - 45)/(9*exp(2*x) - 90*log(3) - exp(x)*(18*log(3) - 90) + 9*log(3)^2 + 225) ,x)
(5*log(exp(x) - log(3) + 5))/(log(3)^2 - 10*log(3) + 25) + (exp(x)/(log(3) - 5) - (x*exp(x))/(log(3) - 5))/(exp(x) - log(3) + 5) + ((5*x)/(log(3) - 5) + (5*exp(x))/(log(3) - 5)^2 - (5*x*exp(x))/(log(3)^2 - 10*log(3) + 25)) /(exp(x) - log(3) + 5) - exp((16*x)/9 + (8*x^2)/9 + x^3/9)/(exp(x) - log(3 ) + 5) - ((x*log(3))/(log(3) - 5) + (exp(x)*log(3))/(log(3) - 5)^2 - (x*ex p(x)*log(3))/(log(3)^2 - 10*log(3) + 25))/(exp(x) - log(3) + 5) + log(exp( x) - log(3) + 5)/(log(3) - 5) - (log(3)*log(exp(x) - log(3) + 5))/(log(3)^ 2 - 10*log(3) + 25)