Integrand size = 90, antiderivative size = 31 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=16+\frac {4}{1+\frac {-x+\frac {x}{-5+e^x (5+2 x)}}{x^2}} \] Output:
4/((x/((5+2*x)*exp(x)-5)-x)/x^2+1)+16
Time = 3.63 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=-\frac {4 \left (6-e^x (5+2 x)\right )}{6-5 x+e^x \left (-5+3 x+2 x^2\right )} \] Input:
Integrate[(-120 + E^(2*x)*(-100 - 80*x - 16*x^2) + E^x*(220 + 116*x + 8*x^ 2))/(36 - 60*x + 25*x^2 + E^x*(-60 + 86*x - 6*x^2 - 20*x^3) + E^(2*x)*(25 - 30*x - 11*x^2 + 12*x^3 + 4*x^4)),x]
Output:
(-4*(6 - E^x*(5 + 2*x)))/(6 - 5*x + E^x*(-5 + 3*x + 2*x^2))
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.
\(\displaystyle \int \frac {e^{2 x} \left (-16 x^2-80 x-100\right )+e^x \left (8 x^2+116 x+220\right )-120}{25 x^2+e^x \left (-20 x^3-6 x^2+86 x-60\right )+e^{2 x} \left (4 x^4+12 x^3-11 x^2-30 x+25\right )-60 x+36} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 \left (e^x \left (2 x^2+29 x+55\right )-e^{2 x} (2 x+5)^2-30\right )}{\left (e^x \left (2 x^2+3 x-5\right )-5 x+6\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {e^{2 x} (2 x+5)^2-e^x \left (2 x^2+29 x+55\right )+30}{\left (-5 x-e^x \left (-2 x^2-3 x+5\right )+6\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {e^{2 x} (2 x+5)^2-e^x \left (2 x^2+29 x+55\right )+30}{\left (-5 x-e^x \left (-2 x^2-3 x+5\right )+6\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (-\frac {2 x^3+7 x^2+5}{(x-1)^2 (2 x+5) \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )}-\frac {x \left (10 x^3+13 x^2-67 x+37\right )}{(x-1)^2 (2 x+5) \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )^2}+\frac {1}{(x-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (-4 \int \frac {1}{\left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )^2}dx+\int \frac {1}{(x-1)^2 \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )^2}dx+\frac {16}{7} \int \frac {1}{(x-1) \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )^2}dx-5 \int \frac {x}{\left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )^2}dx+\frac {185}{7} \int \frac {1}{(2 x+5) \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )^2}dx-\int \frac {1}{2 e^x x^2+3 e^x x-5 x-5 e^x+6}dx-2 \int \frac {1}{(x-1)^2 \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )}dx-\frac {16}{7} \int \frac {1}{(x-1) \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )}dx-\frac {10}{7} \int \frac {1}{(2 x+5) \left (2 e^x x^2+3 e^x x-5 x-5 e^x+6\right )}dx+\frac {1}{1-x}\right )\) |
Input:
Int[(-120 + E^(2*x)*(-100 - 80*x - 16*x^2) + E^x*(220 + 116*x + 8*x^2))/(3 6 - 60*x + 25*x^2 + E^x*(-60 + 86*x - 6*x^2 - 20*x^3) + E^(2*x)*(25 - 30*x - 11*x^2 + 12*x^3 + 4*x^4)),x]
Output:
$Aborted
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16
method | result | size |
norman | \(\frac {20 \,{\mathrm e}^{x}+8 \,{\mathrm e}^{x} x -24}{2 \,{\mathrm e}^{x} x^{2}+3 \,{\mathrm e}^{x} x -5 \,{\mathrm e}^{x}-5 x +6}\) | \(36\) |
parallelrisch | \(\frac {-48+16 \,{\mathrm e}^{x} x +40 \,{\mathrm e}^{x}}{4 \,{\mathrm e}^{x} x^{2}+6 \,{\mathrm e}^{x} x -10 \,{\mathrm e}^{x}-10 x +12}\) | \(37\) |
risch | \(\frac {4}{-1+x}-\frac {4 x}{\left (-1+x \right ) \left (2 \,{\mathrm e}^{x} x^{2}+3 \,{\mathrm e}^{x} x -5 \,{\mathrm e}^{x}-5 x +6\right )}\) | \(40\) |
Input:
int(((-16*x^2-80*x-100)*exp(x)^2+(8*x^2+116*x+220)*exp(x)-120)/((4*x^4+12* x^3-11*x^2-30*x+25)*exp(x)^2+(-20*x^3-6*x^2+86*x-60)*exp(x)+25*x^2-60*x+36 ),x,method=_RETURNVERBOSE)
Output:
(20*exp(x)+8*exp(x)*x-24)/(2*exp(x)*x^2+3*exp(x)*x-5*exp(x)-5*x+6)
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=\frac {4 \, {\left ({\left (2 \, x + 5\right )} e^{x} - 6\right )}}{{\left (2 \, x^{2} + 3 \, x - 5\right )} e^{x} - 5 \, x + 6} \] Input:
integrate(((-16*x^2-80*x-100)*exp(x)^2+(8*x^2+116*x+220)*exp(x)-120)/((4*x ^4+12*x^3-11*x^2-30*x+25)*exp(x)^2+(-20*x^3-6*x^2+86*x-60)*exp(x)+25*x^2-6 0*x+36),x, algorithm="fricas")
Output:
4*((2*x + 5)*e^x - 6)/((2*x^2 + 3*x - 5)*e^x - 5*x + 6)
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=- \frac {4 x}{- 5 x^{2} + 11 x + \left (2 x^{3} + x^{2} - 8 x + 5\right ) e^{x} - 6} + \frac {4}{x - 1} \] Input:
integrate(((-16*x**2-80*x-100)*exp(x)**2+(8*x**2+116*x+220)*exp(x)-120)/(( 4*x**4+12*x**3-11*x**2-30*x+25)*exp(x)**2+(-20*x**3-6*x**2+86*x-60)*exp(x) +25*x**2-60*x+36),x)
Output:
-4*x/(-5*x**2 + 11*x + (2*x**3 + x**2 - 8*x + 5)*exp(x) - 6) + 4/(x - 1)
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=\frac {4 \, {\left ({\left (2 \, x + 5\right )} e^{x} - 6\right )}}{{\left (2 \, x^{2} + 3 \, x - 5\right )} e^{x} - 5 \, x + 6} \] Input:
integrate(((-16*x^2-80*x-100)*exp(x)^2+(8*x^2+116*x+220)*exp(x)-120)/((4*x ^4+12*x^3-11*x^2-30*x+25)*exp(x)^2+(-20*x^3-6*x^2+86*x-60)*exp(x)+25*x^2-6 0*x+36),x, algorithm="maxima")
Output:
4*((2*x + 5)*e^x - 6)/((2*x^2 + 3*x - 5)*e^x - 5*x + 6)
Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=\frac {4 \, {\left (2 \, x e^{x} + 5 \, e^{x} - 6\right )}}{2 \, x^{2} e^{x} + 3 \, x e^{x} - 5 \, x - 5 \, e^{x} + 6} \] Input:
integrate(((-16*x^2-80*x-100)*exp(x)^2+(8*x^2+116*x+220)*exp(x)-120)/((4*x ^4+12*x^3-11*x^2-30*x+25)*exp(x)^2+(-20*x^3-6*x^2+86*x-60)*exp(x)+25*x^2-6 0*x+36),x, algorithm="giac")
Output:
4*(2*x*e^x + 5*e^x - 6)/(2*x^2*e^x + 3*x*e^x - 5*x - 5*e^x + 6)
Time = 7.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=\frac {4\,\left (5\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^x-6\right )}{2\,x^2\,{\mathrm {e}}^x-5\,{\mathrm {e}}^x-5\,x+3\,x\,{\mathrm {e}}^x+6} \] Input:
int(-(exp(2*x)*(80*x + 16*x^2 + 100) - exp(x)*(116*x + 8*x^2 + 220) + 120) /(exp(2*x)*(12*x^3 - 11*x^2 - 30*x + 4*x^4 + 25) - 60*x + 25*x^2 - exp(x)* (6*x^2 - 86*x + 20*x^3 + 60) + 36),x)
Output:
(4*(5*exp(x) + 2*x*exp(x) - 6))/(2*x^2*exp(x) - 5*exp(x) - 5*x + 3*x*exp(x ) + 6)
Time = 0.18 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {-120+e^{2 x} \left (-100-80 x-16 x^2\right )+e^x \left (220+116 x+8 x^2\right )}{36-60 x+25 x^2+e^x \left (-60+86 x-6 x^2-20 x^3\right )+e^{2 x} \left (25-30 x-11 x^2+12 x^3+4 x^4\right )} \, dx=\frac {8 e^{x} x +20 e^{x}-24}{2 e^{x} x^{2}+3 e^{x} x -5 e^{x}-5 x +6} \] Input:
int(((-16*x^2-80*x-100)*exp(x)^2+(8*x^2+116*x+220)*exp(x)-120)/((4*x^4+12* x^3-11*x^2-30*x+25)*exp(x)^2+(-20*x^3-6*x^2+86*x-60)*exp(x)+25*x^2-60*x+36 ),x)
Output:
(4*(2*e**x*x + 5*e**x - 6))/(2*e**x*x**2 + 3*e**x*x - 5*e**x - 5*x + 6)