Integrand size = 79, antiderivative size = 30 \[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=5 e^{\frac {1}{9} \left (-1-e^{x (2+\log (2+x))}-x\right )-x} x \] Output:
5*x/exp(10/9*x+1/9+1/9*exp((ln(2+x)+2)*x))
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=5 e^{-\frac {1}{9}-\frac {10 x}{9}-\frac {1}{9} e^{2 x} (2+x)^x} x \] Input:
Integrate[(E^((-1 - E^(2*x + x*Log[2 + x]) - 10*x)/9)*(90 - 55*x - 50*x^2 + E^(2*x + x*Log[2 + x])*(-20*x - 15*x^2 + (-10*x - 5*x^2)*Log[2 + x])))/( 18 + 9*x),x]
Output:
5*E^(-1/9 - (10*x)/9 - (E^(2*x)*(2 + x)^x)/9)*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.
\(\displaystyle \int \frac {e^{\frac {1}{9} \left (-10 x-e^{2 x+x \log (x+2)}-1\right )} \left (-50 x^2+e^{2 x+x \log (x+2)} \left (-15 x^2+\left (-5 x^2-10 x\right ) \log (x+2)-20 x\right )-55 x+90\right )}{9 x+18} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5}{9} x (x+2)^{x-1} (3 x+x \log (x+2)+2 \log (x+2)+4) \exp \left (2 x+\frac {1}{9} \left (-10 x-e^{2 x+x \log (x+2)}-1\right )\right )-\frac {5}{9} (10 x-9) e^{\frac {1}{9} \left (-10 x-e^{2 x+x \log (x+2)}-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{3} \int e^{\frac {1}{9} \left (-e^{2 x} (x+2)^x+8 x-1\right )} x^2 (x+2)^{x-1}dx-\frac {20}{9} \int e^{\frac {1}{9} \left (-e^{2 x} (x+2)^x+8 x-1\right )} x (x+2)^{x-1}dx+\frac {5}{9} \int \frac {\int e^{\frac {1}{9} \left (-e^{2 x} (x+2)^x+8 x-1\right )} x (x+2)^xdx}{x+2}dx+5 \int e^{\frac {1}{9} \left (-10 x-e^{\log (x+2) x+2 x}-1\right )}dx-\frac {50}{9} \int e^{\frac {1}{9} \left (-10 x-e^{\log (x+2) x+2 x}-1\right )} xdx-\frac {5}{9} \log (x+2) \int e^{\frac {1}{9} \left (-e^{2 x} (x+2)^x+8 x-1\right )} x (x+2)^xdx\) |
Input:
Int[(E^((-1 - E^(2*x + x*Log[2 + x]) - 10*x)/9)*(90 - 55*x - 50*x^2 + E^(2 *x + x*Log[2 + x])*(-20*x - 15*x^2 + (-10*x - 5*x^2)*Log[2 + x])))/(18 + 9 *x),x]
Output:
$Aborted
Time = 0.74 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70
method | result | size |
risch | \(5 x \,{\mathrm e}^{-\frac {\left (2+x \right )^{x} {\mathrm e}^{2 x}}{9}-\frac {10 x}{9}-\frac {1}{9}}\) | \(21\) |
parallelrisch | \(5 x \,{\mathrm e}^{-\frac {10 x}{9}-\frac {1}{9}-\frac {{\mathrm e}^{\left (\ln \left (2+x \right )+2\right ) x}}{9}}\) | \(23\) |
Input:
int((((-5*x^2-10*x)*ln(2+x)-15*x^2-20*x)*exp(x*ln(2+x)+2*x)-50*x^2-55*x+90 )/(9*x+18)/exp(1/9*exp(x*ln(2+x)+2*x)+10/9*x+1/9),x,method=_RETURNVERBOSE)
Output:
5*x*exp(-1/9*(2+x)^x*exp(2*x)-10/9*x-1/9)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=5 \, x e^{\left (-\frac {10}{9} \, x - \frac {1}{9} \, e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} - \frac {1}{9}\right )} \] Input:
integrate((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2 -55*x+90)/(9*x+18)/exp(1/9*exp(x*log(2+x)+2*x)+10/9*x+1/9),x, algorithm="f ricas")
Output:
5*x*e^(-10/9*x - 1/9*e^(x*log(x + 2) + 2*x) - 1/9)
Time = 31.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=5 x e^{- \frac {10 x}{9} - \frac {e^{x \log {\left (x + 2 \right )} + 2 x}}{9} - \frac {1}{9}} \] Input:
integrate((((-5*x**2-10*x)*ln(2+x)-15*x**2-20*x)*exp(x*ln(2+x)+2*x)-50*x** 2-55*x+90)/(9*x+18)/exp(1/9*exp(x*ln(2+x)+2*x)+10/9*x+1/9),x)
Output:
5*x*exp(-10*x/9 - exp(x*log(x + 2) + 2*x)/9 - 1/9)
\[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=\int { -\frac {5 \, {\left (10 \, x^{2} + {\left (3 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (x + 2\right ) + 4 \, x\right )} e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} + 11 \, x - 18\right )} e^{\left (-\frac {10}{9} \, x - \frac {1}{9} \, e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} - \frac {1}{9}\right )}}{9 \, {\left (x + 2\right )}} \,d x } \] Input:
integrate((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2 -55*x+90)/(9*x+18)/exp(1/9*exp(x*log(2+x)+2*x)+10/9*x+1/9),x, algorithm="m axima")
Output:
5*x*e^(-10/9*x - 1/9*e^(x*log(x + 2) + 2*x) - 1/9) + 5/9*integrate(0, x)
\[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=\int { -\frac {5 \, {\left (10 \, x^{2} + {\left (3 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (x + 2\right ) + 4 \, x\right )} e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} + 11 \, x - 18\right )} e^{\left (-\frac {10}{9} \, x - \frac {1}{9} \, e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} - \frac {1}{9}\right )}}{9 \, {\left (x + 2\right )}} \,d x } \] Input:
integrate((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2 -55*x+90)/(9*x+18)/exp(1/9*exp(x*log(2+x)+2*x)+10/9*x+1/9),x, algorithm="g iac")
Output:
integrate(-5/9*(10*x^2 + (3*x^2 + (x^2 + 2*x)*log(x + 2) + 4*x)*e^(x*log(x + 2) + 2*x) + 11*x - 18)*e^(-10/9*x - 1/9*e^(x*log(x + 2) + 2*x) - 1/9)/( x + 2), x)
Time = 4.48 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=5\,x\,{\mathrm {e}}^{-\frac {10\,x}{9}}\,{\mathrm {e}}^{-\frac {1}{9}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}\,{\left (x+2\right )}^x}{9}} \] Input:
int(-(exp(- (10*x)/9 - exp(2*x + x*log(x + 2))/9 - 1/9)*(55*x + exp(2*x + x*log(x + 2))*(20*x + log(x + 2)*(10*x + 5*x^2) + 15*x^2) + 50*x^2 - 90))/ (9*x + 18),x)
Output:
5*x*exp(-(10*x)/9)*exp(-1/9)*exp(-(exp(2*x)*(x + 2)^x)/9)
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx=\frac {5 x}{e^{\frac {e^{2 x} \left (x +2\right )^{x}}{9}+\frac {10 x}{9}+\frac {1}{9}}} \] Input:
int((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2-55*x+ 90)/(9*x+18)/exp(1/9*exp(x*log(2+x)+2*x)+10/9*x+1/9),x)
Output:
(5*x)/e**((e**(2*x)*(x + 2)**x + 10*x + 1)/9)