Integrand size = 64, antiderivative size = 27 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=e^{\frac {-\frac {1}{5}+5 (5-x) x}{5+x}}-e^x x \] Output:
exp((5*x*(5-x)-1/5)/(5+x))-exp(x)*x
Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=e^{75-\frac {1251}{5 (5+x)}-5 (5+x)}-e^x x \] Input:
Integrate[(E^((-1 + 125*x - 25*x^2)/(25 + 5*x))*(626 - 250*x - 25*x^2) + E ^x*(-125 - 175*x - 55*x^2 - 5*x^3))/(125 + 50*x + 5*x^2),x]
Output:
E^(75 - 1251/(5*(5 + x)) - 5*(5 + x)) - E^x*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.
\(\displaystyle \int \frac {e^{\frac {-25 x^2+125 x-1}{5 x+25}} \left (-25 x^2-250 x+626\right )+e^x \left (-5 x^3-55 x^2-175 x-125\right )}{5 x^2+50 x+125} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{\frac {-25 x^2+125 x-1}{5 x+25}} \left (-25 x^2-250 x+626\right )+e^x \left (-5 x^3-55 x^2-175 x-125\right )}{\left (\sqrt {5} x+5 \sqrt {5}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {-25 x^2+125 x-1}{5 (x+5)}} \left (-25 x^2-250 x+626\right )}{5 (x+5)^2}-e^x (x+1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int e^{\frac {-25 x^2+125 x-1}{5 (x+5)}}dx+\frac {1251}{5} \int \frac {e^{\frac {-25 x^2+125 x-1}{5 (x+5)}}}{(x+5)^2}dx-e^x (x+1)+e^x\) |
Input:
Int[(E^((-1 + 125*x - 25*x^2)/(25 + 5*x))*(626 - 250*x - 25*x^2) + E^x*(-1 25 - 175*x - 55*x^2 - 5*x^3))/(125 + 50*x + 5*x^2),x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-{\mathrm e}^{x} x +{\mathrm e}^{-\frac {25 x^{2}-125 x +1}{5 \left (5+x \right )}}\) | \(25\) |
parallelrisch | \(-{\mathrm e}^{x} x +{\mathrm e}^{-\frac {25 x^{2}-125 x +1}{5 \left (5+x \right )}}\) | \(25\) |
parts | \(-{\mathrm e}^{x} x +\frac {x \,{\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}+5 \,{\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}}{5+x}\) | \(56\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}-5 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}+5 \,{\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}}{5+x}\) | \(62\) |
orering | \(\frac {\left (600 x^{5}+11950 x^{4}+26450 x^{3}-345510 x^{2}+288701 x -31250\right ) \left (\left (-5 x^{3}-55 x^{2}-175 x -125\right ) {\mathrm e}^{x}+\left (-25 x^{2}-250 x +626\right ) {\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}\right )}{\left (750 x^{5}+15875 x^{4}+61195 x^{3}-269370 x^{2}-74534 x +172826\right ) \left (5 x^{2}+50 x +125\right )}+\frac {5 \left (30 x^{3}+305 x^{2}-451 x +125\right ) \left (5+x \right )^{2} \left (\frac {\left (-15 x^{2}-110 x -175\right ) {\mathrm e}^{x}+\left (-5 x^{3}-55 x^{2}-175 x -125\right ) {\mathrm e}^{x}+\left (-50 x -250\right ) {\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}+\left (-25 x^{2}-250 x +626\right ) \left (\frac {-50 x +125}{25+5 x}-\frac {5 \left (-25 x^{2}+125 x -1\right )}{\left (25+5 x \right )^{2}}\right ) {\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}}{5 x^{2}+50 x +125}-\frac {\left (\left (-5 x^{3}-55 x^{2}-175 x -125\right ) {\mathrm e}^{x}+\left (-25 x^{2}-250 x +626\right ) {\mathrm e}^{\frac {-25 x^{2}+125 x -1}{25+5 x}}\right ) \left (10 x +50\right )}{\left (5 x^{2}+50 x +125\right )^{2}}\right )}{750 x^{5}+15875 x^{4}+61195 x^{3}-269370 x^{2}-74534 x +172826}\) | \(367\) |
Input:
int(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x^2+125 *x-1)/(25+5*x)))/(5*x^2+50*x+125),x,method=_RETURNVERBOSE)
Output:
-exp(x)*x+exp(-1/5*(25*x^2-125*x+1)/(5+x))
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=-x e^{x} + e^{\left (-\frac {25 \, x^{2} - 125 \, x + 1}{5 \, {\left (x + 5\right )}}\right )} \] Input:
integrate(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x ^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+125),x, algorithm="fricas")
Output:
-x*e^x + e^(-1/5*(25*x^2 - 125*x + 1)/(x + 5))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=- x e^{x} + e^{\frac {- 25 x^{2} + 125 x - 1}{5 x + 25}} \] Input:
integrate(((-5*x**3-55*x**2-175*x-125)*exp(x)+(-25*x**2-250*x+626)*exp((-2 5*x**2+125*x-1)/(25+5*x)))/(5*x**2+50*x+125),x)
Output:
-x*exp(x) + exp((-25*x**2 + 125*x - 1)/(5*x + 25))
\[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=\int { -\frac {5 \, {\left (x^{3} + 11 \, x^{2} + 35 \, x + 25\right )} e^{x} + {\left (25 \, x^{2} + 250 \, x - 626\right )} e^{\left (-\frac {25 \, x^{2} - 125 \, x + 1}{5 \, {\left (x + 5\right )}}\right )}}{5 \, {\left (x^{2} + 10 \, x + 25\right )}} \,d x } \] Input:
integrate(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x ^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+125),x, algorithm="maxima")
Output:
-(x*e^(6*x) - e^(-1251/5/(x + 5) + 50))*e^(-5*x) + 25*e^(-5)*exp_integral_ e(2, -x - 5)/(x + 5) + 25*integrate(e^x/(x^2 + 10*x + 25), x)
Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=-x e^{x} + e^{\left (-\frac {125 \, x^{2} - 626 \, x}{25 \, {\left (x + 5\right )}} - \frac {1}{25}\right )} \] Input:
integrate(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x ^2+125*x-1)/(25+5*x)))/(5*x^2+50*x+125),x, algorithm="giac")
Output:
-x*e^x + e^(-1/25*(125*x^2 - 626*x)/(x + 5) - 1/25)
Time = 4.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx={\mathrm {e}}^{-\frac {25\,x^2}{5\,x+25}}\,{\mathrm {e}}^{-\frac {1}{5\,x+25}}\,{\mathrm {e}}^{\frac {125\,x}{5\,x+25}}-x\,{\mathrm {e}}^x \] Input:
int(-(exp(-(25*x^2 - 125*x + 1)/(5*x + 25))*(250*x + 25*x^2 - 626) + exp(x )*(175*x + 55*x^2 + 5*x^3 + 125))/(50*x + 5*x^2 + 125),x)
Output:
exp(-(25*x^2)/(5*x + 25))*exp(-1/(5*x + 25))*exp((125*x)/(5*x + 25)) - x*e xp(x)
Time = 0.18 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {e^{\frac {-1+125 x-25 x^2}{25+5 x}} \left (626-250 x-25 x^2\right )+e^x \left (-125-175 x-55 x^2-5 x^3\right )}{125+50 x+5 x^2} \, dx=\frac {-e^{\frac {30 x^{2}+150 x +1251}{5 x +25}} x +e^{50}}{e^{\frac {25 x^{2}+125 x +1251}{5 x +25}}} \] Input:
int(((-5*x^3-55*x^2-175*x-125)*exp(x)+(-25*x^2-250*x+626)*exp((-25*x^2+125 *x-1)/(25+5*x)))/(5*x^2+50*x+125),x)
Output:
( - e**((30*x**2 + 150*x + 1251)/(5*x + 25))*x + e**50)/e**((25*x**2 + 125 *x + 1251)/(5*x + 25))