Integrand size = 56, antiderivative size = 17 \[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=e^{-24+e^{(5+(5-x) x)^2}} \] Output:
exp(exp((x*(5-x)+5)^2)-24)
Time = 1.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=e^{-24+e^{\left (-5-5 x+x^2\right )^2}} \] Input:
Integrate[E^(1 + E^(25 + 50*x + 15*x^2 - 10*x^3 + x^4) + 50*x + 15*x^2 - 1 0*x^3 + x^4)*(50 + 30*x - 30*x^2 + 4*x^3),x]
Output:
E^(-24 + E^(-5 - 5*x + x^2)^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.
\(\displaystyle \int \left (4 x^3-30 x^2+30 x+50\right ) \exp \left (x^4-10 x^3+15 x^2+e^{x^4-10 x^3+15 x^2+50 x+25}+50 x+1\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (4 x^3 \exp \left (x^4-10 x^3+15 x^2+e^{x^4-10 x^3+15 x^2+50 x+25}+50 x+1\right )-30 x^2 \exp \left (x^4-10 x^3+15 x^2+e^{x^4-10 x^3+15 x^2+50 x+25}+50 x+1\right )+30 x \exp \left (x^4-10 x^3+15 x^2+e^{x^4-10 x^3+15 x^2+50 x+25}+50 x+1\right )+50 \exp \left (x^4-10 x^3+15 x^2+e^{x^4-10 x^3+15 x^2+50 x+25}+50 x+1\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 50 \int \exp \left (x^4-10 x^3+15 x^2+50 x+e^{x^4-10 x^3+15 x^2+50 x+25}+1\right )dx+30 \int \exp \left (x^4-10 x^3+15 x^2+50 x+e^{x^4-10 x^3+15 x^2+50 x+25}+1\right ) xdx-30 \int \exp \left (x^4-10 x^3+15 x^2+50 x+e^{x^4-10 x^3+15 x^2+50 x+25}+1\right ) x^2dx+4 \int \exp \left (x^4-10 x^3+15 x^2+50 x+e^{x^4-10 x^3+15 x^2+50 x+25}+1\right ) x^3dx\) |
Input:
Int[E^(1 + E^(25 + 50*x + 15*x^2 - 10*x^3 + x^4) + 50*x + 15*x^2 - 10*x^3 + x^4)*(50 + 30*x - 30*x^2 + 4*x^3),x]
Output:
$Aborted
Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{\left (x^{2}-5 x -5\right )^{2}}-24}\) | \(15\) |
derivativedivides | \({\mathrm e}^{{\mathrm e}^{x^{4}-10 x^{3}+15 x^{2}+50 x +25}-24}\) | \(23\) |
norman | \({\mathrm e}^{{\mathrm e}^{x^{4}-10 x^{3}+15 x^{2}+50 x +25}-24}\) | \(23\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{x^{4}-10 x^{3}+15 x^{2}+50 x +25}-24}\) | \(23\) |
Input:
int((4*x^3-30*x^2+30*x+50)*exp(x^4-10*x^3+15*x^2+50*x+25)*exp(exp(x^4-10*x ^3+15*x^2+50*x+25)-24),x,method=_RETURNVERBOSE)
Output:
exp(exp((x^2-5*x-5)^2)-24)
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=e^{\left (e^{\left (x^{4} - 10 \, x^{3} + 15 \, x^{2} + 50 \, x + 25\right )} - 24\right )} \] Input:
integrate((4*x^3-30*x^2+30*x+50)*exp(x^4-10*x^3+15*x^2+50*x+25)*exp(exp(x^ 4-10*x^3+15*x^2+50*x+25)-24),x, algorithm="fricas")
Output:
e^(e^(x^4 - 10*x^3 + 15*x^2 + 50*x + 25) - 24)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=e^{e^{x^{4} - 10 x^{3} + 15 x^{2} + 50 x + 25} - 24} \] Input:
integrate((4*x**3-30*x**2+30*x+50)*exp(x**4-10*x**3+15*x**2+50*x+25)*exp(e xp(x**4-10*x**3+15*x**2+50*x+25)-24),x)
Output:
exp(exp(x**4 - 10*x**3 + 15*x**2 + 50*x + 25) - 24)
Time = 0.49 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=e^{\left (e^{\left (x^{4} - 10 \, x^{3} + 15 \, x^{2} + 50 \, x + 25\right )} - 24\right )} \] Input:
integrate((4*x^3-30*x^2+30*x+50)*exp(x^4-10*x^3+15*x^2+50*x+25)*exp(exp(x^ 4-10*x^3+15*x^2+50*x+25)-24),x, algorithm="maxima")
Output:
e^(e^(x^4 - 10*x^3 + 15*x^2 + 50*x + 25) - 24)
\[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=\int { 2 \, {\left (2 \, x^{3} - 15 \, x^{2} + 15 \, x + 25\right )} e^{\left (x^{4} - 10 \, x^{3} + 15 \, x^{2} + 50 \, x + e^{\left (x^{4} - 10 \, x^{3} + 15 \, x^{2} + 50 \, x + 25\right )} + 1\right )} \,d x } \] Input:
integrate((4*x^3-30*x^2+30*x+50)*exp(x^4-10*x^3+15*x^2+50*x+25)*exp(exp(x^ 4-10*x^3+15*x^2+50*x+25)-24),x, algorithm="giac")
Output:
integrate(2*(2*x^3 - 15*x^2 + 15*x + 25)*e^(x^4 - 10*x^3 + 15*x^2 + 50*x + e^(x^4 - 10*x^3 + 15*x^2 + 50*x + 25) + 1), x)
Time = 3.43 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx={\mathrm {e}}^{-24}\,{\mathrm {e}}^{{\mathrm {e}}^{50\,x}\,{\mathrm {e}}^{x^4}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^{-10\,x^3}\,{\mathrm {e}}^{15\,x^2}} \] Input:
int(exp(exp(50*x + 15*x^2 - 10*x^3 + x^4 + 25) - 24)*exp(50*x + 15*x^2 - 1 0*x^3 + x^4 + 25)*(30*x - 30*x^2 + 4*x^3 + 50),x)
Output:
exp(-24)*exp(exp(50*x)*exp(x^4)*exp(25)*exp(-10*x^3)*exp(15*x^2))
\[ \int e^{1+e^{25+50 x+15 x^2-10 x^3+x^4}+50 x+15 x^2-10 x^3+x^4} \left (50+30 x-30 x^2+4 x^3\right ) \, dx=2 e \left (25 \left (\int \frac {e^{\frac {e^{x^{4}+15 x^{2}+50 x} e^{25}+e^{10 x^{3}} x^{4}+15 e^{10 x^{3}} x^{2}+50 e^{10 x^{3}} x}{e^{10 x^{3}}}}}{e^{10 x^{3}}}d x \right )+2 \left (\int \frac {e^{\frac {e^{x^{4}+15 x^{2}+50 x} e^{25}+e^{10 x^{3}} x^{4}+15 e^{10 x^{3}} x^{2}+50 e^{10 x^{3}} x}{e^{10 x^{3}}}} x^{3}}{e^{10 x^{3}}}d x \right )-15 \left (\int \frac {e^{\frac {e^{x^{4}+15 x^{2}+50 x} e^{25}+e^{10 x^{3}} x^{4}+15 e^{10 x^{3}} x^{2}+50 e^{10 x^{3}} x}{e^{10 x^{3}}}} x^{2}}{e^{10 x^{3}}}d x \right )+15 \left (\int \frac {e^{\frac {e^{x^{4}+15 x^{2}+50 x} e^{25}+e^{10 x^{3}} x^{4}+15 e^{10 x^{3}} x^{2}+50 e^{10 x^{3}} x}{e^{10 x^{3}}}} x}{e^{10 x^{3}}}d x \right )\right ) \] Input:
int((4*x^3-30*x^2+30*x+50)*exp(x^4-10*x^3+15*x^2+50*x+25)*exp(exp(x^4-10*x ^3+15*x^2+50*x+25)-24),x)
Output:
2*e*(25*int(e**((e**(x**4 + 15*x**2 + 50*x)*e**25 + e**(10*x**3)*x**4 + 15 *e**(10*x**3)*x**2 + 50*e**(10*x**3)*x)/e**(10*x**3))/e**(10*x**3),x) + 2* int((e**((e**(x**4 + 15*x**2 + 50*x)*e**25 + e**(10*x**3)*x**4 + 15*e**(10 *x**3)*x**2 + 50*e**(10*x**3)*x)/e**(10*x**3))*x**3)/e**(10*x**3),x) - 15* int((e**((e**(x**4 + 15*x**2 + 50*x)*e**25 + e**(10*x**3)*x**4 + 15*e**(10 *x**3)*x**2 + 50*e**(10*x**3)*x)/e**(10*x**3))*x**2)/e**(10*x**3),x) + 15* int((e**((e**(x**4 + 15*x**2 + 50*x)*e**25 + e**(10*x**3)*x**4 + 15*e**(10 *x**3)*x**2 + 50*e**(10*x**3)*x)/e**(10*x**3))*x)/e**(10*x**3),x))