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

Optimal. Leaf size=36 \[ 1-2 x-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right ) \]

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Rubi [F]  time = 2.90, 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 {1+2 x+x^2}{4 x^2}} \left (-1-x-4 x^2\right )+4 x^2 \log (2)+\left (-2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2+2 x^2 \log (2)\right ) \log \left (\frac {e^{e^2}}{e^{\frac {1+2 x+x^2}{4 x^2}}-\log (2)}\right )}{2 e^{\frac {1+2 x+x^2}{4 x^2}} x^2-2 x^2 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-e^{\frac {(1+x)^2}{4 x^2}} \left (1+x+4 x^2\right )+x^2 \log (16)}{2 x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}-\log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )\right ) \, dx\\ &=\frac {1}{2} \int \frac {-e^{\frac {(1+x)^2}{4 x^2}} \left (1+x+4 x^2\right )+x^2 \log (16)}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\int \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right ) \, dx\\ &=-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \left (\frac {-1-x-4 x^2}{x^2}-\frac {(1+x) \log (2)}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}\right ) \, dx+\int \frac {e^{\frac {(1+x)^2}{4 x^2}} (1+x)}{2 x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx\\ &=-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \frac {-1-x-4 x^2}{x^2} \, dx+\frac {1}{2} \int \frac {e^{\frac {(1+x)^2}{4 x^2}} (1+x)}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\frac {1}{2} \log (2) \int \frac {1+x}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx\\ &=-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \left (-4-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx+\frac {1}{2} \int \left (\frac {e^{\frac {(1+x)^2}{4 x^2}}}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}+\frac {e^{\frac {(1+x)^2}{4 x^2}}}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}\right ) \, dx-\frac {1}{2} \log (2) \int \left (\frac {1}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}+\frac {1}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )}\right ) \, dx\\ &=\frac {1}{2 x}-2 x-\frac {\log (x)}{2}-x \log \left (\frac {e^{e^2}}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )+\frac {1}{2} \int \frac {e^{\frac {(1+x)^2}{4 x^2}}}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx+\frac {1}{2} \int \frac {e^{\frac {(1+x)^2}{4 x^2}}}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\frac {1}{2} \log (2) \int \frac {1}{x^2 \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx-\frac {1}{2} \log (2) \int \frac {1}{x \left (e^{\frac {(1+x)^2}{4 x^2}}-\log (2)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.54, size = 30, normalized size = 0.83 \begin {gather*} -x \left (2+e^2+\log \left (\frac {1}{e^{\frac {(1+x)^2}{4 x^2}}-\log (2)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 33, normalized size = 0.92 \begin {gather*} -x \log \left (\frac {e^{\left (e^{2}\right )}}{e^{\left (\frac {x^{2} + 2 \, x + 1}{4 \, x^{2}}\right )} - \log \relax (2)}\right ) - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.52, size = 31, normalized size = 0.86 \begin {gather*} -x e^{2} + x \log \left (e^{\left (\frac {x^{2} + 2 \, x + 1}{4 \, x^{2}}\right )} - \log \relax (2)\right ) - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 42, normalized size = 1.17

method	result	size
norman	\(\frac {-2 x^{2}-x^{2} \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{2}}}{{\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \relax (2)}\right )}{x}\)	\(42\)
default	\(\frac {\left (-2 \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{2}}}{{\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \relax (2)}\right )-2 \ln \left ({\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \relax (2)\right )-4\right ) x^{2}+2 x^{2} \ln \left ({\mathrm e}^{\frac {x^{2}+2 x +1}{4 x^{2}}}-\ln \relax (2)\right )}{2 x}\)	\(88\)
risch	\(x \ln \left (-{\mathrm e}^{\frac {\left (x +1\right )^{2}}{4 x^{2}}}+\ln \relax (2)\right )-\frac {\left (-2 i \pi \mathrm {csgn}\left (\frac {i}{-{\mathrm e}^{\frac {\left (x +1\right )^{2}}{4 x^{2}}}+\ln \relax (2)}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i}{-{\mathrm e}^{\frac {\left (x +1\right )^{2}}{4 x^{2}}}+\ln \relax (2)}\right )^{3}+2 i \pi +2 \,{\mathrm e}^{2}+4\right ) x}{2}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.07, size = 29, normalized size = 0.81 \begin {gather*} -x {\left (e^{2} + 2\right )} + x \log \left (e^{\left (\frac {1}{2 \, x} + \frac {1}{4 \, x^{2}} + \frac {1}{4}\right )} - \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.63, size = 36, normalized size = 1.00 \begin {gather*} -2\,x-x\,{\mathrm {e}}^2-x\,\ln \left (-\frac {1}{\ln \relax (2)-\sqrt {{\mathrm {e}}^{1/x}}\,{\left ({\mathrm {e}}^{\frac {1}{x^2}}\right )}^{1/4}\,{\mathrm {e}}^{1/4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.55, size = 32, normalized size = 0.89 \begin {gather*} - x \log {\left (\frac {e^{e^{2}}}{e^{\frac {\frac {x^{2}}{4} + \frac {x}{2} + \frac {1}{4}}{x^{2}}} - \log {\relax (2 )}} \right )} - 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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