\(\int \frac {1}{5} (5+4 e^{1+e^{\frac {1}{5} (1-20 x+4 x^2-\log (4))}-x} (5+e^{\frac {1}{5} (1-20 x+4 x^2-\log (4))} (20-8 x))) \, dx\) [4785]

Optimal result

Integrand size = 63, antiderivative size = 32 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=-4 e^{1+e^{\frac {1}{5} \left (1+4 \left (-5 x+x^2\right )-\log (4)\right )}-x}+x \]

[Out]

x-exp(exp(-2/5*ln(2)+4/5*x^2-4*x+1/5)+2*ln(2)-x+1)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {12, 6838} \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x-4 e^{\frac {e^{\frac {1}{5} \left (4 x^2-20 x+1\right )}}{2^{2/5}}-x+1} \]

[In]

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]

[Out]

-4*E^(1 + E^((1 - 20*x + 4*x^2)/5)/2^(2/5) - x) + x

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \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+\frac {4}{5} \int 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 ) \, dx \\ & = -4 e^{1+\frac {e^{\frac {1}{5} \left (1-20 x+4 x^2\right )}}{2^{2/5}}-x}+x \\ \end{align*}

Mathematica [F]

\[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=\int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx \]

[In]

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]

[Out]

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

Maple [A] (verified)

Time = 0.20 (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	\(x -{\mathrm e}^{{\mathrm e}^{-\frac {2 \ln \left (2\right )}{5}+\frac {4 x^{2}}{5}-4 x +\frac {1}{5}}+2 \ln \left (2\right )-x +1}\)	\(30\)

[In]

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,met
hod=_RETURNVERBOSE)

[Out]

x-4*exp(1/2*2^(3/5)*exp(1/5+4/5*x^2-4*x)+1-x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - e^{\left (-x + e^{\left (\frac {4}{5} \, x^{2} - 4 \, x - \frac {2}{5} \, \log \left (2\right ) + \frac {1}{5}\right )} + 2 \, \log \left (2\right ) + 1\right )} \]

[In]

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")

[Out]

x - e^(-x + e^(4/5*x^2 - 4*x - 2/5*log(2) + 1/5) + 2*log(2) + 1)

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - 4 e^{- x + \frac {2^{\frac {3}{5}} e^{\frac {4 x^{2}}{5} - 4 x + \frac {1}{5}}}{2} + 1} \]

[In]

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)

[Out]

x - 4*exp(-x + 2**(3/5)*exp(4*x**2/5 - 4*x + 1/5)/2 + 1)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - 4 \, e^{\left (\frac {1}{2} \cdot 2^{\frac {3}{5}} e^{\left (\frac {4}{5} \, x^{2} - 4 \, x + \frac {1}{5}\right )} - x + 1\right )} \]

[In]

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")

[Out]

x - 4*e^(1/2*2^(3/5)*e^(4/5*x^2 - 4*x + 1/5) - x + 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x - e^{\left (-x + e^{\left (\frac {4}{5} \, x^{2} - 4 \, x - \frac {2}{5} \, \log \left (2\right ) + \frac {1}{5}\right )} + 2 \, \log \left (2\right ) + 1\right )} \]

[In]

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")

[Out]

x - e^(-x + e^(4/5*x^2 - 4*x - 2/5*log(2) + 1/5) + 2*log(2) + 1)

Mupad [B] (verification not implemented)

Time = 10.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5} \left (5+4 e^{1+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )}-x} \left (5+e^{\frac {1}{5} \left (1-20 x+4 x^2-\log (4)\right )} (20-8 x)\right )\right ) \, dx=x-4\,{\mathrm {e}}^{\frac {2^{3/5}\,{\left ({\mathrm {e}}^{x^2}\right )}^{4/5}\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{1/5}}{2}-x+1} \]

[In]

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)

[Out]

x - 4*exp((2^(3/5)*exp(x^2)^(4/5)*exp(-4*x)*exp(1/5))/2 - x + 1)