3.5.54 \(\int \frac {1}{2} e^{-4+e^{\frac {1}{2} (1-2 x+\log (x))}} (2+e^{\frac {1}{2} (1-2 x+\log (x))} (1-2 x)) \, dx\)

Optimal. Leaf size=20 \[ e^{-4+e^{-x+\frac {1}{2} (1+\log (x))}} x \]

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Rubi [B]  time = 0.06, antiderivative size = 61, normalized size of antiderivative = 3.05, number of steps used = 2, number of rules used = 2, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 2288} \begin {gather*} -\frac {(1-2 x) \exp \left (e^{\frac {1}{2} (1-2 x)} \sqrt {x}+\frac {1}{2} (2 x-\log (x)-1)+\frac {1}{2} (-2 x+\log (x)+1)-4\right )}{2-\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-4 + E^((1 - 2*x + Log[x])/2))*(2 + E^((1 - 2*x + Log[x])/2)*(1 - 2*x)))/2,x]

[Out]

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

Rule 12

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

Rule 2288

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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int e^{-4+e^{\frac {1}{2} (1-2 x+\log (x))}} \left (2+e^{\frac {1}{2} (1-2 x+\log (x))} (1-2 x)\right ) \, dx\\ &=-\frac {\exp \left (-4+e^{\frac {1}{2} (1-2 x)} \sqrt {x}+\frac {1}{2} (-1+2 x-\log (x))+\frac {1}{2} (1-2 x+\log (x))\right ) (1-2 x)}{2-\frac {1}{x}}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.97, size = 45, normalized size = 2.25 \begin {gather*} \frac {1}{2} \int e^{-4+e^{\frac {1}{2} (1-2 x+\log (x))}} \left (2+e^{\frac {1}{2} (1-2 x+\log (x))} (1-2 x)\right ) \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 + E^((1 - 2*x + Log[x])/2))*(2 + E^((1 - 2*x + Log[x])/2)*(1 - 2*x)))/2,x]

[Out]

Integrate[E^(-4 + E^((1 - 2*x + Log[x])/2))*(2 + E^((1 - 2*x + Log[x])/2)*(1 - 2*x)), x]/2

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fricas [A]  time = 0.53, size = 15, normalized size = 0.75 \begin {gather*} x e^{\left (e^{\left (-x + \frac {1}{2} \, \log \relax (x) + \frac {1}{2}\right )} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(e^(-x + 1/2*log(x) + 1/2) - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{2} \, {\left ({\left (2 \, x - 1\right )} e^{\left (-x + \frac {1}{2} \, \log \relax (x) + \frac {1}{2}\right )} - 2\right )} e^{\left (e^{\left (-x + \frac {1}{2} \, \log \relax (x) + \frac {1}{2}\right )} - 4\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 16, normalized size = 0.80

method	result	size
norman	\({\mathrm e}^{{\mathrm e}^{\frac {\ln \relax (x )}{2}+\frac {1}{2}-x}-4} x\)	\(16\)
risch	\({\mathrm e}^{\sqrt {x}\, {\mathrm e}^{\frac {1}{2}-x}-4} x\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((1-2*x)*exp(1/2*ln(x)+1/2-x)+2)*exp(exp(1/2*ln(x)+1/2-x)-4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{4} \, e^{\left (-5\right )} \int \frac {{\left ({\left (2 \, x^{2} e^{\frac {1}{2}} - x e^{\frac {1}{2}}\right )} e^{x} - 2 \, {\left (2 \, x^{3} e - x^{2} e - 2 \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )}\right )} \sqrt {x}\right )} e^{\left (\sqrt {x} e^{\left (-x + \frac {1}{2}\right )}\right )}}{x^{\frac {7}{2}}}\,{d x} + \frac {{\left (2 \, x^{3} e^{\frac {3}{2}} + 2 \, x^{\frac {3}{2}} e^{\left (x + 1\right )} + {\left (2 \, x e^{\frac {1}{2}} - e^{\frac {1}{2}}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (\sqrt {x} e^{\left (-x + \frac {1}{2}\right )} - \frac {11}{2}\right )}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*e^(-5)*integrate(((2*x^2*e^(1/2) - x*e^(1/2))*e^x - 2*(2*x^3*e - x^2*e - 2*(2*x^2 - 2*x + 1)*e^(2*x))*sqr
t(x))*e^(sqrt(x)*e^(-x + 1/2))/x^(7/2), x) + 1/2*(2*x^3*e^(3/2) + 2*x^(3/2)*e^(x + 1) + (2*x*e^(1/2) - e^(1/2)
)*e^(2*x))*e^(sqrt(x)*e^(-x + 1/2) - 11/2)/x^2

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mupad [B]  time = 0.50, size = 15, normalized size = 0.75 \begin {gather*} x\,{\mathrm {e}}^{\sqrt {x}\,{\mathrm {e}}^{-x}\,\sqrt {\mathrm {e}}}\,{\mathrm {e}}^{-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(log(x)/2 - x + 1/2) - 4)*(exp(log(x)/2 - x + 1/2)*(2*x - 1) - 2))/2,x)

[Out]

x*exp(x^(1/2)*exp(-x)*exp(1/2))*exp(-4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \left (- \sqrt {x} e^{\frac {1}{2}} e^{- x} e^{\sqrt {x} e^{\frac {1}{2}} e^{- x}}\right )\, dx + \int 2 x^{\frac {3}{2}} e^{\frac {1}{2}} e^{- x} e^{\sqrt {x} e^{\frac {1}{2}} e^{- x}}\, dx + \int \left (- 2 e^{\sqrt {x} e^{\frac {1}{2}} e^{- x}}\right )\, dx}{2 e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(-sqrt(x)*exp(1/2)*exp(-x)*exp(sqrt(x)*exp(1/2)*exp(-x)), x) + Integral(2*x**(3/2)*exp(1/2)*exp(-x)*
exp(sqrt(x)*exp(1/2)*exp(-x)), x) + Integral(-2*exp(sqrt(x)*exp(1/2)*exp(-x)), x))*exp(-4)/2

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