3.54.46 \(\int \frac {e^{-2 e^{\frac {x^2+\log (x)}{4 \log (2)}}} (e^{\frac {x^2+\log (x)}{4 \log (2)}} (-x-2 x^3) \log (4)+4 x \log (2) \log (4))}{2 \log (2)} \, dx\)

Optimal. Leaf size=28 \[ 4+e^{-2 e^{\frac {x^2+\log (x)}{4 \log (2)}}} x^2 \log (4) \]

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Rubi [A]  time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.50, number of steps used = 2, number of rules used = 2, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {12, 2288} \begin {gather*} \frac {\left (2 x^3+x\right ) \log (4) e^{-2 e^{\frac {x^2}{4 \log (2)}} x^{\frac {1}{\log (16)}}}}{2 x+\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x^2 + Log[x])/(4*Log[2]))*(-x - 2*x^3)*Log[4] + 4*x*Log[2]*Log[4])/(2*E^(2*E^((x^2 + Log[x])/(4*Log[2
])))*Log[2]),x]

[Out]

((x + 2*x^3)*Log[4])/(E^(2*E^(x^2/(4*Log[2]))*x^Log[16]^(-1))*(x^(-1) + 2*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 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 {\int e^{-2 e^{\frac {x^2+\log (x)}{4 \log (2)}}} \left (e^{\frac {x^2+\log (x)}{4 \log (2)}} \left (-x-2 x^3\right ) \log (4)+4 x \log (2) \log (4)\right ) \, dx}{2 \log (2)}\\ &=\frac {e^{-2 e^{\frac {x^2}{4 \log (2)}} x^{\frac {1}{\log (16)}}} \left (x+2 x^3\right ) \log (4)}{\frac {1}{x}+2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 26, normalized size = 0.93 \begin {gather*} e^{-2 e^{\frac {x^2}{\log (16)}} x^{\frac {1}{\log (16)}}} x^2 \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^2 + Log[x])/(4*Log[2]))*(-x - 2*x^3)*Log[4] + 4*x*Log[2]*Log[4])/(2*E^(2*E^((x^2 + Log[x])/(4
*Log[2])))*Log[2]),x]

[Out]

(x^2*Log[4])/E^(2*E^(x^2/Log[16])*x^Log[16]^(-1))

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fricas [A]  time = 0.61, size = 23, normalized size = 0.82 \begin {gather*} 2 \, x^{2} e^{\left (-2 \, e^{\left (\frac {x^{2} + \log \relax (x)}{4 \, \log \relax (2)}\right )}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x^3-x)*log(2)*exp(1/4*(log(x)+x^2)/log(2))+8*x*log(2)^2)/log(2)/exp(exp(1/4*(log(x)+x^2)/
log(2)))^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (2 \, x^{3} + x\right )} e^{\left (\frac {x^{2} + \log \relax (x)}{4 \, \log \relax (2)}\right )} \log \relax (2) - 4 \, x \log \relax (2)^{2}\right )} e^{\left (-2 \, e^{\left (\frac {x^{2} + \log \relax (x)}{4 \, \log \relax (2)}\right )}\right )}}{\log \relax (2)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x^3-x)*log(2)*exp(1/4*(log(x)+x^2)/log(2))+8*x*log(2)^2)/log(2)/exp(exp(1/4*(log(x)+x^2)/
log(2)))^2,x, algorithm="giac")

[Out]

integrate(-((2*x^3 + x)*e^(1/4*(x^2 + log(x))/log(2))*log(2) - 4*x*log(2)^2)*e^(-2*e^(1/4*(x^2 + log(x))/log(2
)))/log(2), x)

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maple [A]  time = 0.16, size = 24, normalized size = 0.86

method	result	size
risch	\(2 x^{2} {\mathrm e}^{-2 \,{\mathrm e}^{\frac {\ln \relax (x )+x^{2}}{4 \ln \relax (2)}}} \ln \relax (2)\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(2*(-2*x^3-x)*ln(2)*exp(1/4*(ln(x)+x^2)/ln(2))+8*x*ln(2)^2)/ln(2)/exp(exp(1/4*(ln(x)+x^2)/ln(2)))^2,x,
method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 29, normalized size = 1.04 \begin {gather*} 2 \, x^{2} e^{\left (-2 \, e^{\left (\frac {x^{2}}{4 \, \log \relax (2)} + \frac {\log \relax (x)}{4 \, \log \relax (2)}\right )}\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x^3-x)*log(2)*exp(1/4*(log(x)+x^2)/log(2))+8*x*log(2)^2)/log(2)/exp(exp(1/4*(log(x)+x^2)/
log(2)))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.78, size = 28, normalized size = 1.00 \begin {gather*} 2\,x^2\,{\mathrm {e}}^{-2\,x^{\frac {1}{4\,\ln \relax (2)}}\,{\mathrm {e}}^{\frac {x^2}{4\,\ln \relax (2)}}}\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*exp((log(x)/4 + x^2/4)/log(2)))*(4*x*log(2)^2 - exp((log(x)/4 + x^2/4)/log(2))*log(2)*(x + 2*x^3))
)/log(2),x)

[Out]

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

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sympy [A]  time = 2.01, size = 26, normalized size = 0.93 \begin {gather*} 2 x^{2} e^{- 2 e^{\frac {\frac {x^{2}}{4} + \frac {\log {\relax (x )}}{4}}{\log {\relax (2 )}}}} \log {\relax (2 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(2*(-2*x**3-x)*ln(2)*exp(1/4*(ln(x)+x**2)/ln(2))+8*x*ln(2)**2)/ln(2)/exp(exp(1/4*(ln(x)+x**2)/ln
(2)))**2,x)

[Out]

2*x**2*exp(-2*exp((x**2/4 + log(x)/4)/log(2)))*log(2)

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