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

Optimal. Leaf size=24 \[ e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}}-x \]

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Rubi [F]  time = 2.06, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 26, normalized size = 1.08 \begin {gather*} e^{\frac {2 e^{1-x-\frac {1}{1+x}}}{\log (4)}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 21, normalized size = 0.88 \begin {gather*} -x + e^{\left (\frac {e^{\left (-\frac {x^{2}}{x + 1}\right )}}{\log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + e^(e^(-x^2/(x + 1))/log(2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.44, size = 22, normalized size = 0.92

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{x +1}}}{\ln \relax (2)}}-x\)	\(22\)
norman	\(\frac {\left ({\mathrm e}^{\frac {x^{2}}{x +1}}+{\mathrm e}^{\frac {x^{2}}{x +1}} {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{x +1}}}{\ln \relax (2)}}+x \,{\mathrm e}^{\frac {x^{2}}{x +1}} {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{x +1}}}{\ln \relax (2)}}-x^{2} {\mathrm e}^{\frac {x^{2}}{x +1}}\right ) {\mathrm e}^{-\frac {x^{2}}{x +1}}}{x +1}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.06, size = 75, normalized size = 3.12 \begin {gather*} -\frac {{\left (\frac {x^{2} + x - 1}{x + 1} - 2 \, \log \left (x + 1\right )\right )} \log \relax (2) + 2 \, {\left (\frac {1}{x + 1} + \log \left (x + 1\right )\right )} \log \relax (2) - e^{\left (\frac {e^{\left (-x - \frac {1}{x + 1} + 1\right )}}{\log \relax (2)}\right )} \log \relax (2) - \frac {\log \relax (2)}{x + 1}}{\log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.93, size = 21, normalized size = 0.88 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {x^2}{x+1}}}{\ln \relax (2)}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(-x^2/(x + 1))/log(2)) - x

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sympy [A]  time = 0.47, size = 15, normalized size = 0.62 \begin {gather*} - x + e^{\frac {e^{- \frac {x^{2}}{x + 1}}}{\log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + exp(exp(-x**2/(x + 1))/log(2))

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