3.88.46 \(\int \frac {-2-x}{8 \log (\frac {e^4}{4})} \, dx\)

Optimal. Leaf size=26 \[ \frac {x-\frac {x^2 (4+x)}{16 \log \left (\frac {e^4}{4}\right )}}{x} \]

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Rubi [A]  time = 0.00, antiderivative size = 17, normalized size of antiderivative = 0.65, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {9} \begin {gather*} -\frac {(x+2)^2}{16 (4-\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - x)/(8*Log[E^4/4]),x]

[Out]

-1/16*(2 + x)^2/(4 - Log[4])

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {(2+x)^2}{16 (4-\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 0.81 \begin {gather*} \frac {2 x+\frac {x^2}{2}}{8 (-4+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - x)/(8*Log[E^4/4]),x]

[Out]

(2*x + x^2/2)/(8*(-4 + Log[4]))

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fricas [A]  time = 0.51, size = 15, normalized size = 0.58 \begin {gather*} \frac {x^{2} + 4 \, x}{32 \, {\left (\log \relax (2) - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(x^2 + 4*x)/(log(2) - 2)

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giac [A]  time = 0.17, size = 16, normalized size = 0.62 \begin {gather*} -\frac {x^{2} + 4 \, x}{16 \, \log \left (\frac {1}{4} \, e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(x^2 + 4*x)/log(1/4*e^4)

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maple [A]  time = 0.03, size = 14, normalized size = 0.54

method	result	size
gosper	\(-\frac {\left (4+x \right ) x}{16 \ln \left (\frac {{\mathrm e}^{4}}{4}\right )}\)	\(14\)
default	\(\frac {-\frac {1}{2} x^{2}-2 x}{8 \ln \left (\frac {{\mathrm e}^{4}}{4}\right )}\)	\(19\)
norman	\(\frac {x}{8 \ln \relax (2)-16}+\frac {x^{2}}{32 \ln \relax (2)-64}\)	\(22\)
risch	\(-\frac {x^{2}}{16 \left (4-2 \ln \relax (2)\right )}-\frac {x}{4 \left (4-2 \ln \relax (2)\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4+x)*x/ln(1/4*exp(4))

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maxima [A]  time = 0.35, size = 16, normalized size = 0.62 \begin {gather*} -\frac {x^{2} + 4 \, x}{16 \, \log \left (\frac {1}{4} \, e^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(x^2 + 4*x)/log(1/4*e^4)

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mupad [B]  time = 0.07, size = 15, normalized size = 0.58 \begin {gather*} \frac {4\,{\left (\frac {x}{8}+\frac {1}{4}\right )}^2}{\ln \relax (4)-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x/8 + 1/4)/log(exp(4)/4),x)

[Out]

(4*(x/8 + 1/4)^2)/(log(4) - 4)

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sympy [A]  time = 0.06, size = 17, normalized size = 0.65 \begin {gather*} \frac {x^{2}}{-64 + 32 \log {\relax (2 )}} + \frac {x}{-16 + 8 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/(-64 + 32*log(2)) + x/(-16 + 8*log(2))

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