\(\int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} (-e^5+4 x^2+4 x \log (4))}{e^5} \, dx\) [6801]

Optimal result

Integrand size = 53, antiderivative size = 27 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=\left (5-e^{\frac {e^e-2 x^2}{e^5}}-x\right ) (x+\log (4)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 2326} \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=-\frac {e^{\frac {e^e-2 x^2}{e^5}} \left (x^2+x \log (4)\right )}{x}-\frac {1}{4} (5-2 x)^2-x \log (4) \]

[In]

Int[(E^5*(5 - 2*x) - E^5*Log[4] + E^((E^E - 2*x^2)/E^5)*(-E^5 + 4*x^2 + 4*x*Log[4]))/E^5,x]

[Out]

-1/4*(5 - 2*x)^2 - x*Log[4] - (E^((E^E - 2*x^2)/E^5)*(x^2 + x*Log[4]))/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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=-x (-5+x+\log (4))-\frac {1}{4} e^{\frac {e^e-2 x^2}{e^5}} (4 x+\log (256)) \]

[In]

Integrate[(E^5*(5 - 2*x) - E^5*Log[4] + E^((E^E - 2*x^2)/E^5)*(-E^5 + 4*x^2 + 4*x*Log[4]))/E^5,x]

[Out]

-(x*(-5 + x + Log[4])) - (E^((E^E - 2*x^2)/E^5)*(4*x + Log[256]))/4

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63

method	result	size
risch	\(-2 x \ln \left (2\right )-x^{2}+5 x +\left (-2 \,{\mathrm e}^{5} \ln \left (2\right )-x \,{\mathrm e}^{5}\right ) {\mathrm e}^{-2 x^{2} {\mathrm e}^{-5}+{\mathrm e}^{{\mathrm e}} {\mathrm e}^{-5}-5}\)	\(44\)
norman	\(\left (-2 \ln \left (2\right )+5\right ) x -x^{2}-x \,{\mathrm e}^{\left ({\mathrm e}^{{\mathrm e}}-2 x^{2}\right ) {\mathrm e}^{-5}}-2 \ln \left (2\right ) {\mathrm e}^{\left ({\mathrm e}^{{\mathrm e}}-2 x^{2}\right ) {\mathrm e}^{-5}}\)	\(52\)
parts	\(-x^{2}+5 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}} {\mathrm e}^{-5}} x \,{\mathrm e}^{-2 x^{2} {\mathrm e}^{-5}}-2 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}} {\mathrm e}^{-5}} \ln \left (2\right ) {\mathrm e}^{-2 x^{2} {\mathrm e}^{-5}}-2 x \ln \left (2\right )\)	\(61\)
parallelrisch	\({\mathrm e}^{-5} \left (-x^{2} {\mathrm e}^{5}-{\mathrm e}^{\left ({\mathrm e}^{{\mathrm e}}-2 x^{2}\right ) {\mathrm e}^{-5}} {\mathrm e}^{5} x +5 x \,{\mathrm e}^{5}-2 \,{\mathrm e}^{\left ({\mathrm e}^{{\mathrm e}}-2 x^{2}\right ) {\mathrm e}^{-5}} {\mathrm e}^{5} \ln \left (2\right )-2 x \,{\mathrm e}^{5} \ln \left (2\right )\right )\)	\(67\)
default	\({\mathrm e}^{-5} \left ({\mathrm e}^{5} \left (-x^{2}+5 x \right )-{\mathrm e}^{{\mathrm e}^{{\mathrm e}} {\mathrm e}^{-5}} {\mathrm e}^{5} x \,{\mathrm e}^{-2 x^{2} {\mathrm e}^{-5}}-2 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}} {\mathrm e}^{-5}} \ln \left (2\right ) {\mathrm e}^{-2 x^{2} {\mathrm e}^{-5}} {\mathrm e}^{5}-2 x \,{\mathrm e}^{5} \ln \left (2\right )\right )\)	\(76\)

[In]

int(((8*x*ln(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*ln(2)+(5-2*x)*exp(5))/exp(5),x,method=_
RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=-x^{2} - {\left (x + 2 \, \log \left (2\right )\right )} e^{\left (-{\left (2 \, x^{2} - e^{e}\right )} e^{\left (-5\right )}\right )} - 2 \, x \log \left (2\right ) + 5 \, x \]

[In]

integrate(((8*x*log(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*log(2)+(5-2*x)*exp(5))/exp(5),x,
algorithm="fricas")

[Out]

-x^2 - (x + 2*log(2))*e^(-(2*x^2 - e^e)*e^(-5)) - 2*x*log(2) + 5*x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=- x^{2} + x \left (5 - 2 \log {\left (2 \right )}\right ) + \left (- x - 2 \log {\left (2 \right )}\right ) e^{\frac {- 2 x^{2} + e^{e}}{e^{5}}} \]

[In]

integrate(((8*x*ln(2)-exp(5)+4*x**2)*exp((exp(exp(1))-2*x**2)/exp(5))-2*exp(5)*ln(2)+(5-2*x)*exp(5))/exp(5),x)

[Out]

-x**2 + x*(5 - 2*log(2)) + (-x - 2*log(2))*exp((-2*x**2 + exp(E))*exp(-5))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=-{\left (2 \, x e^{5} \log \left (2\right ) + {\left (x^{2} - 5 \, x\right )} e^{5} + {\left (x e^{\left (e^{\left (e - 5\right )} + 5\right )} + 2 \, e^{\left (e^{\left (e - 5\right )} + 5\right )} \log \left (2\right )\right )} e^{\left (-2 \, x^{2} e^{\left (-5\right )}\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((8*x*log(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*log(2)+(5-2*x)*exp(5))/exp(5),x,
algorithm="maxima")

[Out]

-(2*x*e^5*log(2) + (x^2 - 5*x)*e^5 + (x*e^(e^(e - 5) + 5) + 2*e^(e^(e - 5) + 5)*log(2))*e^(-2*x^2*e^(-5)))*e^(
-5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=-{\left (2 \, x e^{5} \log \left (2\right ) + {\left (x^{2} - 5 \, x\right )} e^{5} + {\left (x e^{5} + 2 \, e^{5} \log \left (2\right )\right )} e^{\left (-{\left (2 \, x^{2} - e^{e}\right )} e^{\left (-5\right )}\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((8*x*log(2)-exp(5)+4*x^2)*exp((exp(exp(1))-2*x^2)/exp(5))-2*exp(5)*log(2)+(5-2*x)*exp(5))/exp(5),x,
algorithm="giac")

[Out]

-(2*x*e^5*log(2) + (x^2 - 5*x)*e^5 + (x*e^5 + 2*e^5*log(2))*e^(-(2*x^2 - e^e)*e^(-5)))*e^(-5)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {e^5 (5-2 x)-e^5 \log (4)+e^{\frac {e^e-2 x^2}{e^5}} \left (-e^5+4 x^2+4 x \log (4)\right )}{e^5} \, dx=5\,x-2\,x\,\ln \left (2\right )-2\,{\mathrm {e}}^{{\mathrm {e}}^{-5}\,{\mathrm {e}}^{\mathrm {e}}-2\,x^2\,{\mathrm {e}}^{-5}}\,\ln \left (2\right )-x\,{\mathrm {e}}^{{\mathrm {e}}^{-5}\,{\mathrm {e}}^{\mathrm {e}}-2\,x^2\,{\mathrm {e}}^{-5}}-x^2 \]

[In]

int(-exp(-5)*(2*exp(5)*log(2) - exp(exp(-5)*(exp(exp(1)) - 2*x^2))*(8*x*log(2) - exp(5) + 4*x^2) + exp(5)*(2*x
- 5)),x)

[Out]

5*x - 2*x*log(2) - 2*exp(exp(-5)*exp(exp(1)) - 2*x^2*exp(-5))*log(2) - x*exp(exp(-5)*exp(exp(1)) - 2*x^2*exp(-
5)) - x^2