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

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

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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 = {} $$\begin{array}{c}\int \frac{e^{\frac{1+2x+x^2}{4x^2}}(-1-x-4x^2)+4x^2\log (2)+(-2e^{\frac{1+2x+x^2}{4x^2}}{x}^2+2x^2\log (2))\log \left(\frac{e^{e^2}}{e^{\frac{1+2x+x^2}{4x^2}}-\log (2)}\right)}{2e^{\frac{1+2x+x^2}{4x^2}}{x}^2-2x^2\log (2)}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{-e^{\frac{(1+x{)}^2}{4x^2}}(1+x+4x^2)+x^2\log (16)}{2x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}-\log \left(\frac{e^{e^2}}{e^{\frac{(1+x{)}^2}{4x^2}}-\log (2)}\right))dx\\ & =\frac{1}{2}\int \frac{-e^{\frac{(1+x{)}^2}{4x^2}}(1+x+4x^2)+x^2\log (16)}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx-\int \log \left(\frac{e^{e^2}}{e^{\frac{(1+x{)}^2}{4x^2}}-\log (2)}\right)dx\\ & =-x\log \left(\frac{e^{e^2}}{e^{\frac{(1+x{)}^2}{4x^2}}-\log (2)}\right)+\frac{1}{2}\int (\frac{-1-x-4x^2}{x^2}-\frac{(1+x)\log (2)}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))})dx+\int \frac{e^{\frac{(1+x{)}^2}{4x^2}}(1+x)}{2x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx\\ & =-x\log \left(\frac{e^{e^2}}{e^{\frac{(1+x{)}^2}{4x^2}}-\log (2)}\right)+\frac{1}{2}\int \frac{-1-x-4x^2}{x^2}dx+\frac{1}{2}\int \frac{e^{\frac{(1+x{)}^2}{4x^2}}(1+x)}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx-\frac{1}{2}\log (2)\int \frac{1+x}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx\\ & =-x\log \left(\frac{e^{e^2}}{e^{\frac{(1+x{)}^2}{4x^2}}-\log (2)}\right)+\frac{1}{2}\int (-4-\frac{1}{x^2}-\frac{1}{x})dx+\frac{1}{2}\int (\frac{e^{\frac{(1+x{)}^2}{4x^2}}}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}+\frac{e^{\frac{(1+x{)}^2}{4x^2}}}{x(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))})dx-\frac{1}{2}\log (2)\int (\frac{1}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}+\frac{1}{x(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))})dx\\ & =\frac{1}{2x}-2x-\frac{\log (x)}{2}-x\log \left(\frac{e^{e^2}}{e^{\frac{(1+x{)}^2}{4x^2}}-\log (2)}\right)+\frac{1}{2}\int \frac{e^{\frac{(1+x{)}^2}{4x^2}}}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx+\frac{1}{2}\int \frac{e^{\frac{(1+x{)}^2}{4x^2}}}{x(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx-\frac{1}{2}\log (2)\int \frac{1}{x^2(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx-\frac{1}{2}\log (2)\int \frac{1}{x(e^{\frac{(1+x{)}^2}{4x^2}}-\log (2))}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}-x\log \left(\frac{e^{\left(e^2\right)}}{e^{\left(\frac{x^2+2x+1}{4x^2}\right)}-\log (2)}\right)-2x\end{array}$$

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 $$\begin{array}{c}-x{e}^2+x\log (e^{\left(\frac{x^2+2x+1}{4x^2}\right)}-\log (2))-2x\end{array}$$

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	$\frac{-2x^2-x^2\ln \left(\frac{{\mathrm{e}}^{{\mathrm{e}}^2}}{{\mathrm{e}}^{\frac{x^2+2x+1}{4x^2}}-\ln (2)}\right)}{x}$	$42$
default	$\frac{(-2\ln \left(\frac{{\mathrm{e}}^{{\mathrm{e}}^2}}{{\mathrm{e}}^{\frac{x^2+2x+1}{4x^2}}-\ln (2)}\right)-2\ln ({\mathrm{e}}^{\frac{x^2+2x+1}{4x^2}}-\ln (2))-4)x^2+2x^2\ln ({\mathrm{e}}^{\frac{x^2+2x+1}{4x^2}}-\ln (2))}{2x}$	$88$
risch	$x\ln (-{\mathrm{e}}^{\frac{{(x+1)}^2}{4x^2}}+\ln (2))-\frac{(-2i\pi \mathrm{csgn}{\left(\frac{i}{-{\mathrm{e}}^{\frac{{(x+1)}^2}{4x^2}}+\ln (2)}\right)}^2+2i\pi \mathrm{csgn}{\left(\frac{i}{-{\mathrm{e}}^{\frac{{(x+1)}^2}{4x^2}}+\ln (2)}\right)}^3+2i\pi +2{\mathrm{e}}^2+4)x}{2}$	$90$

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 $$\begin{array}{c}-x(e^2+2)+x\log (e^{(\frac{1}{2x}+\frac{1}{4x^2}+\frac{1}{4})}-\log (2))\end{array}$$

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 $$\begin{array}{c}-2x-x{\mathrm{e}}^2-x\ln (-\frac{1}{\ln (2)-\sqrt{{\mathrm{e}}^{1/x}}{\left({\mathrm{e}}^{\frac{1}{x^2}}\right)}^{1/4}{\mathrm{e}}^{1/4}})\end{array}$$

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 $$\begin{array}{c}-x\log \left(\frac{e^{e^2}}{e^{\frac{\frac{x^2}{4}+\frac{x}{2}+\frac{1}{4}}{x^2}}-\log (2)}\right)-2x\end{array}$$

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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