Integrand size = 115, antiderivative size = 34 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=-x+\left (5 e^x+5 \left (x+e^{\frac {e^{2+x}}{5}} x-x^2\right )\right )^2 \] Output:
(5*exp(x)+5*x+5*exp(1/5*exp(2+x))*x-5*x^2)^2-x
Leaf count is larger than twice the leaf count of optimal. \(78\) vs. \(2(34)=68\).
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.29 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=25 e^{2 x}-x+25 x^2+25 e^{\frac {2 e^{2+x}}{5}} x^2-50 x^3+25 x^4+50 e^{\frac {e^{2+x}}{5}} x \left (e^x+x-x^2\right )-50 e^x \left (-x+x^2\right ) \] Input:
Integrate[-1 + 50*E^(2*x) + 50*x - 150*x^2 + 100*x^3 + E^x*(50 - 50*x - 50 *x^2) + E^((2*E^(2 + x))/5)*(50*x + 10*E^(2 + x)*x^2) + E^(E^(2 + x)/5)*(1 00*x - 150*x^2 + E^x*(50 + 50*x) + E^(2 + x)*(10*E^x*x + 10*x^2 - 10*x^3)) ,x]
Output:
25*E^(2*x) - x + 25*x^2 + 25*E^((2*E^(2 + x))/5)*x^2 - 50*x^3 + 25*x^4 + 5 0*E^(E^(2 + x)/5)*x*(E^x + x - x^2) - 50*E^x*(-x + x^2)
Leaf count is larger than twice the leaf count of optimal. \(83\) vs. \(2(34)=68\).
Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.44, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {2009}
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.
\(\displaystyle \int \left (100 x^3-150 x^2+e^x \left (-50 x^2-50 x+50\right )+e^{\frac {2 e^{x+2}}{5}} \left (10 e^{x+2} x^2+50 x\right )+e^{\frac {e^{x+2}}{5}} \left (-150 x^2+e^{x+2} \left (-10 x^3+10 x^2+10 e^x x\right )+100 x+e^x (50 x+50)\right )+50 x+50 e^{2 x}-1\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 25 x^4-50 x^3+25 e^{\frac {2 e^{x+2}}{5}} x^2-50 e^x x^2+25 x^2+50 e^{\frac {e^{x+2}}{5}} \left (-x^3+x^2+e^x x\right )+50 e^x x-x+25 e^{2 x}\) |
Input:
Int[-1 + 50*E^(2*x) + 50*x - 150*x^2 + 100*x^3 + E^x*(50 - 50*x - 50*x^2) + E^((2*E^(2 + x))/5)*(50*x + 10*E^(2 + x)*x^2) + E^(E^(2 + x)/5)*(100*x - 150*x^2 + E^x*(50 + 50*x) + E^(2 + x)*(10*E^x*x + 10*x^2 - 10*x^3)),x]
Output:
25*E^(2*x) - x + 50*E^x*x + 25*x^2 + 25*E^((2*E^(2 + x))/5)*x^2 - 50*E^x*x ^2 - 50*x^3 + 25*x^4 + 50*E^(E^(2 + x)/5)*(E^x*x + x^2 - x^3)
Leaf count of result is larger than twice the leaf count of optimal. \(70\) vs. \(2(29)=58\).
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.09
method | result | size |
risch | \(25 x^{2} {\mathrm e}^{\frac {2 \,{\mathrm e}^{2+x}}{5}}-50 x \left (x^{2}-x -{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}}+25 \,{\mathrm e}^{2 x}+\left (-50 x^{2}+50 x \right ) {\mathrm e}^{x}+25 x^{4}-50 x^{3}+25 x^{2}-x\) | \(71\) |
default | \(-x +25 x^{2} {\mathrm e}^{\frac {2 \,{\mathrm e}^{2+x}}{5}}-50 x^{3} {\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}} x^{2}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}} {\mathrm e}^{x} x +50 \,{\mathrm e}^{x} x -50 \,{\mathrm e}^{x} x^{2}+25 x^{2}-50 x^{3}+25 x^{4}+25 \,{\mathrm e}^{2 x}\) | \(88\) |
parallelrisch | \(-50 \,{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}} x^{3}+25 x^{4}+25 x^{2} {\mathrm e}^{\frac {2 \,{\mathrm e}^{2+x}}{5}}+50 \,{\mathrm e}^{x} {\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}} x -50 \,{\mathrm e}^{x} x^{2}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}} x^{2}-50 x^{3}+50 \,{\mathrm e}^{x} x +25 x^{2}+25 \,{\mathrm e}^{2 x}-x\) | \(88\) |
parts | \(-x +25 x^{2} {\mathrm e}^{\frac {2 \,{\mathrm e}^{2+x}}{5}}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}} x^{2}-50 \,{\mathrm e}^{\frac {{\mathrm e}^{2+x}}{5}} x^{3}+50 \,{\mathrm e}^{x} x -50 \,{\mathrm e}^{x} x^{2}+50 \,{\mathrm e}^{\frac {{\mathrm e}^{2} {\mathrm e}^{x}}{5}} {\mathrm e}^{x} x +25 x^{2}-50 x^{3}+25 x^{4}+25 \,{\mathrm e}^{2 x}\) | \(88\) |
Input:
int((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+10*x^2 )*exp(2+x)+(50*x+50)*exp(x)-150*x^2+100*x)*exp(1/5*exp(2+x))+50*exp(x)^2+( -50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x,method=_RETURNVERBOSE)
Output:
25*x^2*exp(2/5*exp(2+x))-50*x*(x^2-x-exp(x))*exp(1/5*exp(2+x))+25*exp(2*x) +(-50*x^2+50*x)*exp(x)+25*x^4-50*x^3+25*x^2-x
Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (29) = 58\).
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.65 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx={\left (25 \, x^{2} e^{\left (\frac {2}{5} \, e^{\left (x + 2\right )} + 4\right )} + {\left (25 \, x^{4} - 50 \, x^{3} + 25 \, x^{2} - x\right )} e^{4} - 50 \, {\left (x^{2} - x\right )} e^{\left (x + 4\right )} - 50 \, {\left ({\left (x^{3} - x^{2}\right )} e^{4} - x e^{\left (x + 4\right )}\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 2\right )}\right )} + 25 \, e^{\left (2 \, x + 4\right )}\right )} e^{\left (-4\right )} \] Input:
integrate((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+ 10*x^2)*exp(2+x)+(50*x+50)*exp(x)-150*x^2+100*x)*exp(1/5*exp(2+x))+50*exp( x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x, algorithm="fricas" )
Output:
(25*x^2*e^(2/5*e^(x + 2) + 4) + (25*x^4 - 50*x^3 + 25*x^2 - x)*e^4 - 50*(x ^2 - x)*e^(x + 4) - 50*((x^3 - x^2)*e^4 - x*e^(x + 4))*e^(1/5*e^(x + 2)) + 25*e^(2*x + 4))*e^(-4)
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (27) = 54\).
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.29 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=25 x^{4} - 50 x^{3} + 25 x^{2} e^{\frac {2 e^{2} e^{x}}{5}} + 25 x^{2} - x + \left (- 50 x^{2} + 50 x\right ) e^{x} + \left (- 50 x^{3} + 50 x^{2} + 50 x e^{x}\right ) e^{\frac {e^{2} e^{x}}{5}} + 25 e^{2 x} \] Input:
integrate((10*x**2*exp(2+x)+50*x)*exp(1/5*exp(2+x))**2+((10*exp(x)*x-10*x* *3+10*x**2)*exp(2+x)+(50*x+50)*exp(x)-150*x**2+100*x)*exp(1/5*exp(2+x))+50 *exp(x)**2+(-50*x**2-50*x+50)*exp(x)+100*x**3-150*x**2+50*x-1,x)
Output:
25*x**4 - 50*x**3 + 25*x**2*exp(2*exp(2)*exp(x)/5) + 25*x**2 - x + (-50*x* *2 + 50*x)*exp(x) + (-50*x**3 + 50*x**2 + 50*x*exp(x))*exp(exp(2)*exp(x)/5 ) + 25*exp(2*x)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (29) = 58\).
Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.09 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=25 \, x^{4} - 50 \, x^{3} + 25 \, x^{2} e^{\left (\frac {2}{5} \, e^{\left (x + 2\right )}\right )} + 25 \, x^{2} - 50 \, {\left (x^{2} - x\right )} e^{x} - 50 \, {\left (x^{3} - x^{2} - x e^{x}\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 2\right )}\right )} - x + 25 \, e^{\left (2 \, x\right )} \] Input:
integrate((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+ 10*x^2)*exp(2+x)+(50*x+50)*exp(x)-150*x^2+100*x)*exp(1/5*exp(2+x))+50*exp( x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x, algorithm="maxima" )
Output:
25*x^4 - 50*x^3 + 25*x^2*e^(2/5*e^(x + 2)) + 25*x^2 - 50*(x^2 - x)*e^x - 5 0*(x^3 - x^2 - x*e^x)*e^(1/5*e^(x + 2)) - x + 25*e^(2*x)
\[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=\int { 100 \, x^{3} - 150 \, x^{2} - 50 \, {\left (x^{2} + x - 1\right )} e^{x} + 10 \, {\left (x^{2} e^{\left (x + 2\right )} + 5 \, x\right )} e^{\left (\frac {2}{5} \, e^{\left (x + 2\right )}\right )} - 10 \, {\left (15 \, x^{2} + {\left (x^{3} - x^{2} - x e^{x}\right )} e^{\left (x + 2\right )} - 5 \, {\left (x + 1\right )} e^{x} - 10 \, x\right )} e^{\left (\frac {1}{5} \, e^{\left (x + 2\right )}\right )} + 50 \, x + 50 \, e^{\left (2 \, x\right )} - 1 \,d x } \] Input:
integrate((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+ 10*x^2)*exp(2+x)+(50*x+50)*exp(x)-150*x^2+100*x)*exp(1/5*exp(2+x))+50*exp( x)^2+(-50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x, algorithm="giac")
Output:
integrate(100*x^3 - 150*x^2 - 50*(x^2 + x - 1)*e^x + 10*(x^2*e^(x + 2) + 5 *x)*e^(2/5*e^(x + 2)) - 10*(15*x^2 + (x^3 - x^2 - x*e^x)*e^(x + 2) - 5*(x + 1)*e^x - 10*x)*e^(1/5*e^(x + 2)) + 50*x + 50*e^(2*x) - 1, x)
Time = 2.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.50 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=25\,{\mathrm {e}}^{2\,x}-x-50\,x^2\,{\mathrm {e}}^x+50\,x^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}+25\,x^2\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}-50\,x^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}+50\,x\,{\mathrm {e}}^x+50\,x\,{\mathrm {e}}^{x+\frac {{\mathrm {e}}^2\,{\mathrm {e}}^x}{5}}+25\,x^2-50\,x^3+25\,x^4 \] Input:
int(50*x + 50*exp(2*x) + exp((2*exp(x + 2))/5)*(50*x + 10*x^2*exp(x + 2)) + exp(exp(x + 2)/5)*(100*x + exp(x + 2)*(10*x*exp(x) + 10*x^2 - 10*x^3) + exp(x)*(50*x + 50) - 150*x^2) - exp(x)*(50*x + 50*x^2 - 50) - 150*x^2 + 10 0*x^3 - 1,x)
Output:
25*exp(2*x) - x - 50*x^2*exp(x) + 50*x^2*exp((exp(2)*exp(x))/5) + 25*x^2*e xp((2*exp(2)*exp(x))/5) - 50*x^3*exp((exp(2)*exp(x))/5) + 50*x*exp(x) + 50 *x*exp(x + (exp(2)*exp(x))/5) + 25*x^2 - 50*x^3 + 25*x^4
Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.94 \[ \int \left (-1+50 e^{2 x}+50 x-150 x^2+100 x^3+e^x \left (50-50 x-50 x^2\right )+e^{\frac {2 e^{2+x}}{5}} \left (50 x+10 e^{2+x} x^2\right )+e^{\frac {e^{2+x}}{5}} \left (100 x-150 x^2+e^x (50+50 x)+e^{2+x} \left (10 e^x x+10 x^2-10 x^3\right )\right )\right ) \, dx=25 e^{\frac {2 e^{x} e^{2}}{5}} x^{2}+50 e^{\frac {e^{x} e^{2}}{5}+x} x -50 e^{\frac {e^{x} e^{2}}{5}} x^{3}+50 e^{\frac {e^{x} e^{2}}{5}} x^{2}+25 e^{2 x}-50 e^{x} x^{2}+50 e^{x} x +25 x^{4}-50 x^{3}+25 x^{2}-x \] Input:
int((10*x^2*exp(2+x)+50*x)*exp(1/5*exp(2+x))^2+((10*exp(x)*x-10*x^3+10*x^2 )*exp(2+x)+(50*x+50)*exp(x)-150*x^2+100*x)*exp(1/5*exp(2+x))+50*exp(x)^2+( -50*x^2-50*x+50)*exp(x)+100*x^3-150*x^2+50*x-1,x)
Output:
25*e**((2*e**x*e**2)/5)*x**2 + 50*e**((e**x*e**2 + 5*x)/5)*x - 50*e**((e** x*e**2)/5)*x**3 + 50*e**((e**x*e**2)/5)*x**2 + 25*e**(2*x) - 50*e**x*x**2 + 50*e**x*x + 25*x**4 - 50*x**3 + 25*x**2 - x