Integrand size = 68, antiderivative size = 23 \[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=-x+10 e^{-\frac {4}{9} \left (1+e^{4+3 x}\right ) x} x \]
Time = 0.80 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=\frac {1}{9} \left (-9+90 e^{-\frac {4}{9} \left (1+e^{4+3 x}\right ) x}\right ) x \]
Integrate[(E^((-4*x - 4*E^(4 + 3*x)*x)/9)*(90 - 9*E^((4*x + 4*E^(4 + 3*x)* x)/9) - 40*x + E^(4 + 3*x)*(-40*x - 120*x^2)))/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.
\(\displaystyle \int \frac {1}{9} e^{\frac {1}{9} \left (-4 e^{3 x+4} x-4 x\right )} \left (e^{3 x+4} \left (-120 x^2-40 x\right )-40 x-9 e^{\frac {1}{9} \left (4 e^{3 x+4} x+4 x\right )}+90\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int e^{-\frac {4}{9} \left (e^{3 x+4} x+x\right )} \left (-40 x-9 e^{\frac {4}{9} \left (e^{3 x+4} x+x\right )}-40 e^{3 x+4} \left (3 x^2+x\right )+90\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{9} \int \left (-40 e^{-\frac {4}{9} \left (e^{3 x+4} x+x\right )} x-40 e^{3 x-\frac {4}{9} \left (e^{3 x+4} x+x\right )+4} (3 x+1) x+90 e^{-\frac {4}{9} \left (e^{3 x+4} x+x\right )}-9 \exp \left (\frac {4}{9} \left (1+e^{3 x+4}\right ) x-\frac {4}{9} \left (e^{3 x+4} x+x\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (-120 \int e^{\frac {1}{9} \left (-4 e^{3 x+4} x+23 x+36\right )} x^2dx+90 \int e^{-\frac {4}{9} \left (e^{3 x+4} x+x\right )}dx-40 \int e^{\frac {1}{9} \left (-4 e^{3 x+4} x+23 x+36\right )} xdx-40 \int e^{-\frac {4}{9} \left (e^{3 x+4} x+x\right )} xdx-9 x\right )\) |
Int[(E^((-4*x - 4*E^(4 + 3*x)*x)/9)*(90 - 9*E^((4*x + 4*E^(4 + 3*x)*x)/9) - 40*x + E^(4 + 3*x)*(-40*x - 120*x^2)))/9,x]
3.16.52.3.1 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 = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \(-x +10 x \,{\mathrm e}^{-\frac {4 x \left ({\mathrm e}^{4+3 x}+1\right )}{9}}\) | \(20\) |
norman | \(\left (10 x -x \,{\mathrm e}^{\frac {4 x \,{\mathrm e}^{4+3 x}}{9}+\frac {4 x}{9}}\right ) {\mathrm e}^{-\frac {4 x \,{\mathrm e}^{4+3 x}}{9}-\frac {4 x}{9}}\) | \(39\) |
parallelrisch | \(\frac {\left (-81 x \,{\mathrm e}^{\frac {4 x \,{\mathrm e}^{4+3 x}}{9}+\frac {4 x}{9}}+810 x \right ) {\mathrm e}^{-\frac {4 x \,{\mathrm e}^{4+3 x}}{9}-\frac {4 x}{9}}}{81}\) | \(40\) |
int(1/9*(-9*exp(4/9*x*exp(4+3*x)+4/9*x)+(-120*x^2-40*x)*exp(4+3*x)-40*x+90 )/exp(4/9*x*exp(4+3*x)+4/9*x),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=-{\left (x e^{\left (\frac {4}{9} \, x e^{\left (3 \, x + 4\right )} + \frac {4}{9} \, x\right )} - 10 \, x\right )} e^{\left (-\frac {4}{9} \, x e^{\left (3 \, x + 4\right )} - \frac {4}{9} \, x\right )} \]
integrate(1/9*(-9*exp(4/9*x*exp(4+3*x)+4/9*x)+(-120*x^2-40*x)*exp(4+3*x)-4 0*x+90)/exp(4/9*x*exp(4+3*x)+4/9*x),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=10 x e^{- \frac {4 x e^{3 x + 4}}{9} - \frac {4 x}{9}} - x \]
integrate(1/9*(-9*exp(4/9*x*exp(4+3*x)+4/9*x)+(-120*x**2-40*x)*exp(4+3*x)- 40*x+90)/exp(4/9*x*exp(4+3*x)+4/9*x),x)
\[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=\int { -\frac {1}{9} \, {\left (40 \, {\left (3 \, x^{2} + x\right )} e^{\left (3 \, x + 4\right )} + 40 \, x + 9 \, e^{\left (\frac {4}{9} \, x e^{\left (3 \, x + 4\right )} + \frac {4}{9} \, x\right )} - 90\right )} e^{\left (-\frac {4}{9} \, x e^{\left (3 \, x + 4\right )} - \frac {4}{9} \, x\right )} \,d x } \]
integrate(1/9*(-9*exp(4/9*x*exp(4+3*x)+4/9*x)+(-120*x^2-40*x)*exp(4+3*x)-4 0*x+90)/exp(4/9*x*exp(4+3*x)+4/9*x),x, algorithm=\
-x - 1/9*integrate(10*(4*(3*x^2*e^4 + x*e^4)*e^(3*x) + 4*x - 9)*e^(-4/9*x* e^(3*x + 4) - 4/9*x), x)
\[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=\int { -\frac {1}{9} \, {\left (40 \, {\left (3 \, x^{2} + x\right )} e^{\left (3 \, x + 4\right )} + 40 \, x + 9 \, e^{\left (\frac {4}{9} \, x e^{\left (3 \, x + 4\right )} + \frac {4}{9} \, x\right )} - 90\right )} e^{\left (-\frac {4}{9} \, x e^{\left (3 \, x + 4\right )} - \frac {4}{9} \, x\right )} \,d x } \]
integrate(1/9*(-9*exp(4/9*x*exp(4+3*x)+4/9*x)+(-120*x^2-40*x)*exp(4+3*x)-4 0*x+90)/exp(4/9*x*exp(4+3*x)+4/9*x),x, algorithm=\
integrate(-1/9*(40*(3*x^2 + x)*e^(3*x + 4) + 40*x + 9*e^(4/9*x*e^(3*x + 4) + 4/9*x) - 90)*e^(-4/9*x*e^(3*x + 4) - 4/9*x), x)
Time = 12.72 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {1}{9} e^{\frac {1}{9} \left (-4 x-4 e^{4+3 x} x\right )} \left (90-9 e^{\frac {1}{9} \left (4 x+4 e^{4+3 x} x\right )}-40 x+e^{4+3 x} \left (-40 x-120 x^2\right )\right ) \, dx=10\,x\,{\mathrm {e}}^{-\frac {4\,x}{9}-\frac {4\,x\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^4}{9}}-x \]