Integrand size = 53, antiderivative size = 26 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=x \left (4-4 \left (2+e^{\frac {e^5 x}{25}}\right )+x\right ) \left (-x+x^2\right ) \] Output:
x*(x^2-x)*(x-4*exp(1/25*x*exp(5))-4)
Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=(-1+x) x^2 \left (-4-4 e^{\frac {e^5 x}{25}}+x\right ) \] Input:
Integrate[(200*x - 375*x^2 + 100*x^3 + E^((E^5*x)/25)*(200*x - 300*x^2 + E ^5*(4*x^2 - 4*x^3)))/25,x]
Output:
(-1 + x)*x^2*(-4 - 4*E^((E^5*x)/25) + x)
Time = 0.48 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {27, 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.
\(\displaystyle \int \frac {1}{25} \left (100 x^3-375 x^2+e^{\frac {e^5 x}{25}} \left (-300 x^2+e^5 \left (4 x^2-4 x^3\right )+200 x\right )+200 x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \left (100 x^3-375 x^2+200 x+4 e^{\frac {e^5 x}{25}} \left (-75 x^2+50 x+e^5 \left (x^2-x^3\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (25 x^4-100 e^{\frac {e^5 x}{25}} x^3-125 x^3+100 e^{\frac {e^5 x}{25}} x^2+100 x^2\right )\) |
Input:
Int[(200*x - 375*x^2 + 100*x^3 + E^((E^5*x)/25)*(200*x - 300*x^2 + E^5*(4* x^2 - 4*x^3)))/25,x]
Output:
(100*x^2 + 100*E^((E^5*x)/25)*x^2 - 125*x^3 - 100*E^((E^5*x)/25)*x^3 + 25* x^4)/25
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.80 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31
method | result | size |
risch | \(\frac {\left (-100 x^{3}+100 x^{2}\right ) {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}}{25}+x^{4}-5 x^{3}+4 x^{2}\) | \(34\) |
norman | \(x^{4}+4 x^{2}-5 x^{3}-4 x^{3} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}+4 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}\) | \(37\) |
parallelrisch | \(x^{4}+4 x^{2}-5 x^{3}-4 x^{3} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}+4 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}\) | \(37\) |
default | \({\mathrm e}^{-5} \left (5000 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )-187500 \,{\mathrm e}^{-10} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )+2500 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )-62500 \,{\mathrm e}^{-10} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{3} {\mathrm e}^{15}}{15625}-\frac {3 \,{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}+\frac {6 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}-6 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )\right )+4 x^{2}-5 x^{3}+x^{4}\) | \(185\) |
parts | \(2500 \,{\mathrm e}^{-5} \left ({\mathrm e}^{-5} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )+2 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )-75 \,{\mathrm e}^{-10} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )-25 \,{\mathrm e}^{-10} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{3} {\mathrm e}^{15}}{15625}-\frac {3 \,{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}+\frac {6 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}-6 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )\right )+4 x^{2}-5 x^{3}+x^{4}\) | \(185\) |
derivativedivides | \({\mathrm e}^{-5} \left (5000 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}-{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )-187500 \,{\mathrm e}^{-10} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )+2500 \,{\mathrm e}^{-5} \left (\frac {{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}+2 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )-62500 \,{\mathrm e}^{-10} \left (\frac {{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{3} {\mathrm e}^{15}}{15625}-\frac {3 \,{\mathrm e}^{10} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x^{2}}{625}+\frac {6 \,{\mathrm e}^{5} {\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}} x}{25}-6 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{5}}{25}}\right )+4 x^{2} {\mathrm e}^{5}-5 x^{3} {\mathrm e}^{5}+x^{4} {\mathrm e}^{5}\right )\) | \(191\) |
Input:
int(1/25*((-4*x^3+4*x^2)*exp(5)-300*x^2+200*x)*exp(1/25*x*exp(5))+4*x^3-15 *x^2+8*x,x,method=_RETURNVERBOSE)
Output:
1/25*(-100*x^3+100*x^2)*exp(1/25*x*exp(5))+x^4-5*x^3+4*x^2
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=x^{4} - 5 \, x^{3} + 4 \, x^{2} - 4 \, {\left (x^{3} - x^{2}\right )} e^{\left (\frac {1}{25} \, x e^{5}\right )} \] Input:
integrate(1/25*((-4*x^3+4*x^2)*exp(5)-300*x^2+200*x)*exp(1/25*x*exp(5))+4* x^3-15*x^2+8*x,x, algorithm="fricas")
Output:
x^4 - 5*x^3 + 4*x^2 - 4*(x^3 - x^2)*e^(1/25*x*e^5)
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=x^{4} - 5 x^{3} + 4 x^{2} + \left (- 4 x^{3} + 4 x^{2}\right ) e^{\frac {x e^{5}}{25}} \] Input:
integrate(1/25*((-4*x**3+4*x**2)*exp(5)-300*x**2+200*x)*exp(1/25*x*exp(5)) +4*x**3-15*x**2+8*x,x)
Output:
x**4 - 5*x**3 + 4*x**2 + (-4*x**3 + 4*x**2)*exp(x*exp(5)/25)
Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=x^{4} - 5 \, x^{3} + 4 \, x^{2} - 4 \, {\left (x^{3} - x^{2}\right )} e^{\left (\frac {1}{25} \, x e^{5}\right )} \] Input:
integrate(1/25*((-4*x^3+4*x^2)*exp(5)-300*x^2+200*x)*exp(1/25*x*exp(5))+4* x^3-15*x^2+8*x,x, algorithm="maxima")
Output:
x^4 - 5*x^3 + 4*x^2 - 4*(x^3 - x^2)*e^(1/25*x*e^5)
Leaf count of result is larger than twice the leaf count of optimal. 93 vs. \(2 (20) = 40\).
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.58 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=x^{4} - 5 \, x^{3} + 4 \, x^{2} - 4 \, {\left (x^{3} e^{15} - x^{2} e^{15} - 75 \, x^{2} e^{10} + 50 \, x e^{10} + 3750 \, x e^{5} - 1250 \, e^{5} - 93750\right )} e^{\left (\frac {1}{25} \, x e^{5} - 15\right )} - 100 \, {\left (3 \, x^{2} e^{10} - 2 \, x e^{10} - 150 \, x e^{5} + 50 \, e^{5} + 3750\right )} e^{\left (\frac {1}{25} \, x e^{5} - 15\right )} \] Input:
integrate(1/25*((-4*x^3+4*x^2)*exp(5)-300*x^2+200*x)*exp(1/25*x*exp(5))+4* x^3-15*x^2+8*x,x, algorithm="giac")
Output:
x^4 - 5*x^3 + 4*x^2 - 4*(x^3*e^15 - x^2*e^15 - 75*x^2*e^10 + 50*x*e^10 + 3 750*x*e^5 - 1250*e^5 - 93750)*e^(1/25*x*e^5 - 15) - 100*(3*x^2*e^10 - 2*x* e^10 - 150*x*e^5 + 50*e^5 + 3750)*e^(1/25*x*e^5 - 15)
Time = 1.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=-x^2\,\left (x-1\right )\,\left (4\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^5}{25}}-x+4\right ) \] Input:
int(8*x + (exp((x*exp(5))/25)*(200*x + exp(5)*(4*x^2 - 4*x^3) - 300*x^2))/ 25 - 15*x^2 + 4*x^3,x)
Output:
-x^2*(x - 1)*(4*exp((x*exp(5))/25) - x + 4)
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {1}{25} \left (200 x-375 x^2+100 x^3+e^{\frac {e^5 x}{25}} \left (200 x-300 x^2+e^5 \left (4 x^2-4 x^3\right )\right )\right ) \, dx=x^{2} \left (-4 e^{\frac {e^{5} x}{25}} x +4 e^{\frac {e^{5} x}{25}}+x^{2}-5 x +4\right ) \] Input:
int(1/25*((-4*x^3+4*x^2)*exp(5)-300*x^2+200*x)*exp(1/25*x*exp(5))+4*x^3-15 *x^2+8*x,x)
Output:
x**2*( - 4*e**((e**5*x)/25)*x + 4*e**((e**5*x)/25) + x**2 - 5*x + 4)