Integrand size = 63, antiderivative size = 32 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=-4 e^{1+e^{\frac {1}{5} \left (1+4 \left (-5 x+x^2\right )-\log (4)\right )}-x}+x \] Output:
x-exp(exp(-2/5*ln(2)+4/5*x^2-4*x+1/5)+2*ln(2)-x+1)
\[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=\int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx \] Input:
Integrate[(5 + 4*E^(1 + E^((1 - 20*x + 4*x^2 - Log[4])/5) - x)*(5 + E^((1 - 20*x + 4*x^2 - Log[4])/5)*(20 - 8*x)))/5,x]
Output:
Integrate[5 + 4*E^(1 + E^((1 - 20*x + 4*x^2 - Log[4])/5) - x)*(5 + E^((1 - 20*x + 4*x^2 - Log[4])/5)*(20 - 8*x)), x]/5
Time = 0.46 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, 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.
\(\displaystyle \int \frac {1}{5} \left (4 e^{e^{\frac {1}{5} \left (4 x^2-20 x+1-\log (4)\right )}-x+1} \left ((20-8 x) e^{\frac {1}{5} \left (4 x^2-20 x+1-\log (4)\right )}+5\right )+5\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \left (4 e^{-x+\frac {e^{\frac {1}{5} \left (4 x^2-20 x+1\right )}}{2^{2/5}}+1} \left (2\ 2^{3/5} e^{\frac {1}{5} \left (4 x^2-20 x+1\right )} (5-2 x)+5\right )+5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 x-20 e^{\frac {e^{\frac {1}{5} \left (4 x^2-20 x+1\right )}}{2^{2/5}}-x+1}\right )\) |
Input:
Int[(5 + 4*E^(1 + E^((1 - 20*x + 4*x^2 - Log[4])/5) - x)*(5 + E^((1 - 20*x + 4*x^2 - Log[4])/5)*(20 - 8*x)))/5,x]
Output:
(-20*E^(1 + E^((1 - 20*x + 4*x^2)/5)/2^(2/5) - x) + 5*x)/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.35 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
risch | \(x -4 \,{\mathrm e}^{\frac {2^{\frac {3}{5}} {\mathrm e}^{\frac {1}{5}+\frac {4}{5} x^{2}-4 x}}{2}+1-x}\) | \(27\) |
default | \(x -{\mathrm e}^{{\mathrm e}^{-\frac {2 \ln \left (2\right )}{5}+\frac {4 x^{2}}{5}-4 x +\frac {1}{5}}+2 \ln \left (2\right )-x +1}\) | \(30\) |
norman | \(x -{\mathrm e}^{{\mathrm e}^{-\frac {2 \ln \left (2\right )}{5}+\frac {4 x^{2}}{5}-4 x +\frac {1}{5}}+2 \ln \left (2\right )-x +1}\) | \(30\) |
parallelrisch | \(x -{\mathrm e}^{{\mathrm e}^{-\frac {2 \ln \left (2\right )}{5}+\frac {4 x^{2}}{5}-4 x +\frac {1}{5}}+2 \ln \left (2\right )-x +1}\) | \(30\) |
parts | \(x -{\mathrm e}^{{\mathrm e}^{-\frac {2 \ln \left (2\right )}{5}+\frac {4 x^{2}}{5}-4 x +\frac {1}{5}}+2 \ln \left (2\right )-x +1}\) | \(30\) |
Input:
int(1/5*((-8*x+20)*exp(-2/5*ln(2)+4/5*x^2-4*x+1/5)+5)*exp(exp(-2/5*ln(2)+4 /5*x^2-4*x+1/5)+2*ln(2)-x+1)+1,x,method=_RETURNVERBOSE)
Output:
x-4*exp(1/2*2^(3/5)*exp(1/5+4/5*x^2-4*x)+1-x)
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - e^{\left (-x + e^{\left (\frac {4}{5} \, x^{2} - 4 \, x - \frac {2}{5} \, \log \left (2\right ) + \frac {1}{5}\right )} + 2 \, \log \left (2\right ) + 1\right )} \] Input:
integrate(1/5*((-8*x+20)*exp(-2/5*log(2)+4/5*x^2-4*x+1/5)+5)*exp(exp(-2/5* log(2)+4/5*x^2-4*x+1/5)+2*log(2)-x+1)+1,x, algorithm="fricas")
Output:
x - e^(-x + e^(4/5*x^2 - 4*x - 2/5*log(2) + 1/5) + 2*log(2) + 1)
Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - 4 e^{- x + \frac {2^{\frac {3}{5}} e^{\frac {4 x^{2}}{5} - 4 x + \frac {1}{5}}}{2} + 1} \] Input:
integrate(1/5*((-8*x+20)*exp(-2/5*ln(2)+4/5*x**2-4*x+1/5)+5)*exp(exp(-2/5* ln(2)+4/5*x**2-4*x+1/5)+2*ln(2)-x+1)+1,x)
Output:
x - 4*exp(-x + 2**(3/5)*exp(4*x**2/5 - 4*x + 1/5)/2 + 1)
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - 4 \, e^{\left (\frac {1}{2} \cdot 2^{\frac {3}{5}} e^{\left (\frac {4}{5} \, x^{2} - 4 \, x + \frac {1}{5}\right )} - x + 1\right )} \] Input:
integrate(1/5*((-8*x+20)*exp(-2/5*log(2)+4/5*x^2-4*x+1/5)+5)*exp(exp(-2/5* log(2)+4/5*x^2-4*x+1/5)+2*log(2)-x+1)+1,x, algorithm="maxima")
Output:
x - 4*e^(1/2*2^(3/5)*e^(4/5*x^2 - 4*x + 1/5) - x + 1)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - e^{\left (-x + e^{\left (\frac {4}{5} \, x^{2} - 4 \, x - \frac {2}{5} \, \log \left (2\right ) + \frac {1}{5}\right )} + 2 \, \log \left (2\right ) + 1\right )} \] Input:
integrate(1/5*((-8*x+20)*exp(-2/5*log(2)+4/5*x^2-4*x+1/5)+5)*exp(exp(-2/5* log(2)+4/5*x^2-4*x+1/5)+2*log(2)-x+1)+1,x, algorithm="giac")
Output:
x - e^(-x + e^(4/5*x^2 - 4*x - 2/5*log(2) + 1/5) + 2*log(2) + 1)
Time = 4.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x-4\,{\mathrm {e}}^{\frac {2^{3/5}\,{\left ({\mathrm {e}}^{x^2}\right )}^{4/5}\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{1/5}}{2}-x+1} \] Input:
int(1 - (exp(2*log(2) - x + exp((4*x^2)/5 - (2*log(2))/5 - 4*x + 1/5) + 1) *(exp((4*x^2)/5 - (2*log(2))/5 - 4*x + 1/5)*(8*x - 20) - 5))/5,x)
Output:
x - 4*exp((2^(3/5)*exp(x^2)^(4/5)*exp(-4*x)*exp(1/5))/2 - x + 1)
\[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=\frac {\left (20 \,2^{\frac {2}{5}} \left (\int \frac {e^{\frac {e^{\frac {4 x^{2}}{5}+\frac {1}{5}} 2^{\frac {3}{5}}}{2 e^{4 x}}}}{e^{x}}d x \right ) e +5 \,2^{\frac {2}{5}} x +80 \left (\int \frac {e^{\frac {\left (5 e^{\frac {4 x^{2}}{5}+\frac {1}{5}}+4 e^{4 x} 2^{\frac {2}{5}} x^{2}+e^{4 x} 2^{\frac {2}{5}}\right ) 2^{\frac {3}{5}}}{10 e^{4 x}}}}{e^{5 x}}d x \right ) e -32 \left (\int \frac {e^{\frac {\left (5 e^{\frac {4 x^{2}}{5}+\frac {1}{5}}+4 e^{4 x} 2^{\frac {2}{5}} x^{2}+e^{4 x} 2^{\frac {2}{5}}\right ) 2^{\frac {3}{5}}}{10 e^{4 x}}} x}{e^{5 x}}d x \right ) e \right ) 2^{\frac {3}{5}}}{10} \] Input:
int(1/5*((-8*x+20)*exp(-2/5*log(2)+4/5*x^2-4*x+1/5)+5)*exp(exp(-2/5*log(2) +4/5*x^2-4*x+1/5)+2*log(2)-x+1)+1,x)
Output:
(20*2**(2/5)*int(e**(e**((4*x**2 + 1)/5)/(e**(4*x)*2**(2/5)))/e**x,x)*e + 5*2**(2/5)*x + 80*int(e**((5*e**((4*x**2 + 1)/5) + 4*e**(4*x)*2**(2/5)*x** 2 + e**(4*x)*2**(2/5))/(5*e**(4*x)*2**(2/5)))/e**(5*x),x)*e - 32*int((e**( (5*e**((4*x**2 + 1)/5) + 4*e**(4*x)*2**(2/5)*x**2 + e**(4*x)*2**(2/5))/(5* e**(4*x)*2**(2/5)))*x)/e**(5*x),x)*e)/(5*2**(2/5))