Integrand size = 118, antiderivative size = 29 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=3+x-\frac {x^3}{x+\frac {1}{3} \left (-5-e^x+e^{2 x}+x\right )} \] Output:
x-x^3/(4/3*x-1/3*exp(x)-5/3+1/3*exp(2*x))+3
Time = 2.92 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=x-\frac {3 x^3}{-5-e^x+e^{2 x}+4 x} \] Input:
Integrate[(25 + E^(2*x) + E^(4*x) - 40*x + 61*x^2 - 24*x^3 + E^x*(10 - 8*x + 9*x^2 - 3*x^3) + E^(2*x)*(-10 - 2*E^x + 8*x - 9*x^2 + 6*x^3))/(25 + E^( 2*x) + E^(4*x) + E^x*(10 - 8*x) - 40*x + 16*x^2 + E^(2*x)*(-10 - 2*E^x + 8 *x)),x]
Output:
x - (3*x^3)/(-5 - E^x + E^(2*x) + 4*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.
\(\displaystyle \int \frac {-24 x^3+61 x^2+e^x \left (-3 x^3+9 x^2-8 x+10\right )+e^{2 x} \left (6 x^3-9 x^2+8 x-2 e^x-10\right )-40 x+e^{2 x}+e^{4 x}+25}{16 x^2-40 x+e^{2 x}+e^{4 x}+e^x (10-8 x)+e^{2 x} \left (8 x-2 e^x-10\right )+25} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-24 x^3+61 x^2+e^x \left (-3 x^3+9 x^2-8 x+10\right )+e^{2 x} \left (6 x^3-9 x^2+8 x-2 e^x-10\right )-40 x+e^{2 x}+e^{4 x}+25}{\left (-4 x+e^x-e^{2 x}+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 \left (-8 x+e^x+14\right ) x^3}{\left (4 x-e^x+e^{2 x}-5\right )^2}+\frac {3 (2 x-3) x^2}{4 x-e^x+e^{2 x}-5}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -24 \int \frac {x^4}{\left (4 x-e^x+e^{2 x}-5\right )^2}dx+42 \int \frac {x^3}{\left (4 x-e^x+e^{2 x}-5\right )^2}dx+3 \int \frac {e^x x^3}{\left (4 x-e^x+e^{2 x}-5\right )^2}dx+6 \int \frac {x^3}{4 x-e^x+e^{2 x}-5}dx-9 \int \frac {x^2}{4 x-e^x+e^{2 x}-5}dx+x\) |
Input:
Int[(25 + E^(2*x) + E^(4*x) - 40*x + 61*x^2 - 24*x^3 + E^x*(10 - 8*x + 9*x ^2 - 3*x^3) + E^(2*x)*(-10 - 2*E^x + 8*x - 9*x^2 + 6*x^3))/(25 + E^(2*x) + E^(4*x) + E^x*(10 - 8*x) - 40*x + 16*x^2 + E^(2*x)*(-10 - 2*E^x + 8*x)),x ]
Output:
$Aborted
Time = 1.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79
method | result | size |
risch | \(x -\frac {3 x^{3}}{{\mathrm e}^{2 x}+4 x -{\mathrm e}^{x}-5}\) | \(23\) |
parallelrisch | \(-\frac {3 x^{3}-4 x^{2}+{\mathrm e}^{x} x -x \,{\mathrm e}^{2 x}+5 x}{{\mathrm e}^{2 x}+4 x -{\mathrm e}^{x}-5}\) | \(43\) |
norman | \(\frac {x \,{\mathrm e}^{2 x}-\frac {5 \,{\mathrm e}^{x}}{4}+\frac {5 \,{\mathrm e}^{2 x}}{4}+4 x^{2}-3 x^{3}-{\mathrm e}^{x} x -\frac {25}{4}}{{\mathrm e}^{2 x}+4 x -{\mathrm e}^{x}-5}\) | \(50\) |
Input:
int((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3*x^3+9 *x^2-8*x+10)*exp(x)-24*x^3+61*x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8*x-10)* exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x,method=_RETURNVERBOSE )
Output:
x-3*x^3/(exp(2*x)+4*x-exp(x)-5)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=-\frac {3 \, x^{3} - 4 \, x^{2} - x e^{\left (2 \, x\right )} + x e^{x} + 5 \, x}{4 \, x + e^{\left (2 \, x\right )} - e^{x} - 5} \] Input:
integrate((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3 *x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8* x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x, algorithm="fri cas")
Output:
-(3*x^3 - 4*x^2 - x*e^(2*x) + x*e^x + 5*x)/(4*x + e^(2*x) - e^x - 5)
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=- \frac {3 x^{3}}{4 x + e^{2 x} - e^{x} - 5} + x \] Input:
integrate((exp(2*x)**2+(-2*exp(x)+6*x**3-9*x**2+8*x-10)*exp(2*x)+exp(x)**2 +(-3*x**3+9*x**2-8*x+10)*exp(x)-24*x**3+61*x**2-40*x+25)/(exp(2*x)**2+(-2* exp(x)+8*x-10)*exp(2*x)+exp(x)**2+(-8*x+10)*exp(x)+16*x**2-40*x+25),x)
Output:
-3*x**3/(4*x + exp(2*x) - exp(x) - 5) + x
Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=-\frac {3 \, x^{3} - 4 \, x^{2} - x e^{\left (2 \, x\right )} + x e^{x} + 5 \, x}{4 \, x + e^{\left (2 \, x\right )} - e^{x} - 5} \] Input:
integrate((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3 *x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8* x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x, algorithm="max ima")
Output:
-(3*x^3 - 4*x^2 - x*e^(2*x) + x*e^x + 5*x)/(4*x + e^(2*x) - e^x - 5)
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=-\frac {6 \, x^{3} - 4 \, x^{2} - x e^{\left (2 \, x\right )} + x e^{x} + 5 \, x}{4 \, x + e^{\left (2 \, x\right )} - e^{x} - 5} \] Input:
integrate((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3 *x^3+9*x^2-8*x+10)*exp(x)-24*x^3+61*x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8* x-10)*exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x, algorithm="gia c")
Output:
-(6*x^3 - 4*x^2 - x*e^(2*x) + x*e^x + 5*x)/(4*x + e^(2*x) - e^x - 5)
Time = 2.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=x-\frac {3\,x^3}{4\,x+{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x-5} \] Input:
int(-(40*x - exp(2*x) - exp(4*x) + exp(2*x)*(2*exp(x) - 8*x + 9*x^2 - 6*x^ 3 + 10) - 61*x^2 + 24*x^3 + exp(x)*(8*x - 9*x^2 + 3*x^3 - 10) - 25)/(exp(2 *x) - 40*x + exp(4*x) - exp(2*x)*(2*exp(x) - 8*x + 10) - exp(x)*(8*x - 10) + 16*x^2 + 25),x)
Output:
x - (3*x^3)/(4*x + exp(2*x) - exp(x) - 5)
Time = 0.18 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {25+e^{2 x}+e^{4 x}-40 x+61 x^2-24 x^3+e^x \left (10-8 x+9 x^2-3 x^3\right )+e^{2 x} \left (-10-2 e^x+8 x-9 x^2+6 x^3\right )}{25+e^{2 x}+e^{4 x}+e^x (10-8 x)-40 x+16 x^2+e^{2 x} \left (-10-2 e^x+8 x\right )} \, dx=\frac {x \left (e^{2 x}-e^{x}-3 x^{2}+4 x -5\right )}{e^{2 x}-e^{x}+4 x -5} \] Input:
int((exp(2*x)^2+(-2*exp(x)+6*x^3-9*x^2+8*x-10)*exp(2*x)+exp(x)^2+(-3*x^3+9 *x^2-8*x+10)*exp(x)-24*x^3+61*x^2-40*x+25)/(exp(2*x)^2+(-2*exp(x)+8*x-10)* exp(2*x)+exp(x)^2+(-8*x+10)*exp(x)+16*x^2-40*x+25),x)
Output:
(x*(e**(2*x) - e**x - 3*x**2 + 4*x - 5))/(e**(2*x) - e**x + 4*x - 5)