3.61.97 \(\int \frac {-3+x}{16+8 \log (\log (16))} \, dx\)

Optimal. Leaf size=22 \[ \frac {4-\frac {1}{16} (1-x) (-5+x)}{2+\log (\log (16))} \]

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Rubi [A]  time = 0.00, antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {9} \begin {gather*} \frac {(3-x)^2}{16 (2+\log (\log (16)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 + x)/(16 + 8*Log[Log[16]]),x]

[Out]

(3 - x)^2/(16*(2 + Log[Log[16]]))

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 {(3-x)^2}{16 (2+\log (\log (16)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 22, normalized size = 1.00 \begin {gather*} \frac {-3 x+\frac {x^2}{2}}{8 (2+\log (\log (16)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + x)/(16 + 8*Log[Log[16]]),x]

[Out]

(-3*x + x^2/2)/(8*(2 + Log[Log[16]]))

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fricas [A]  time = 1.11, size = 18, normalized size = 0.82 \begin {gather*} \frac {x^{2} - 6 \, x}{16 \, {\left (\log \left (4 \, \log \relax (2)\right ) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-3)/(8*log(4*log(2))+16),x, algorithm="fricas")

[Out]

1/16*(x^2 - 6*x)/(log(4*log(2)) + 2)

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giac [A]  time = 0.12, size = 18, normalized size = 0.82 \begin {gather*} \frac {x^{2} - 6 \, x}{16 \, {\left (\log \left (4 \, \log \relax (2)\right ) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-3)/(8*log(4*log(2))+16),x, algorithm="giac")

[Out]

1/16*(x^2 - 6*x)/(log(4*log(2)) + 2)

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maple [A]  time = 0.04, size = 16, normalized size = 0.73

method	result	size
gosper	\(\frac {x \left (x -6\right )}{32+16 \ln \left (4 \ln \relax (2)\right )}\)	\(16\)
default	\(\frac {\frac {1}{2} x^{2}-3 x}{8 \ln \left (4 \ln \relax (2)\right )+16}\)	\(22\)
norman	\(-\frac {3 x}{8 \left (2+2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )}+\frac {x^{2}}{32 \ln \relax (2)+16 \ln \left (\ln \relax (2)\right )+32}\)	\(32\)
risch	\(\frac {x^{2}}{32 \ln \relax (2)+16 \ln \left (\ln \relax (2)\right )+32}-\frac {3 x}{16 \ln \relax (2)+8 \ln \left (\ln \relax (2)\right )+16}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x-3)/(8*ln(4*ln(2))+16),x,method=_RETURNVERBOSE)

[Out]

1/16*x*(x-6)/(2+ln(4*ln(2)))

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maxima [A]  time = 0.33, size = 18, normalized size = 0.82 \begin {gather*} \frac {x^{2} - 6 \, x}{16 \, {\left (\log \left (4 \, \log \relax (2)\right ) + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-3)/(8*log(4*log(2))+16),x, algorithm="maxima")

[Out]

1/16*(x^2 - 6*x)/(log(4*log(2)) + 2)

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mupad [B]  time = 0.06, size = 15, normalized size = 0.68 \begin {gather*} \frac {{\left (x-3\right )}^2}{16\,\ln \left (\ln \left (16\right )\right )+32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 3)/(8*log(4*log(2)) + 16),x)

[Out]

(x - 3)^2/(16*log(log(16)) + 32)

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sympy [A]  time = 0.06, size = 32, normalized size = 1.45 \begin {gather*} \frac {x^{2}}{16 \log {\left (\log {\relax (2 )} \right )} + 32 \log {\relax (2 )} + 32} - \frac {3 x}{8 \log {\left (\log {\relax (2 )} \right )} + 16 \log {\relax (2 )} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x-3)/(8*ln(4*ln(2))+16),x)

[Out]

x**2/(16*log(log(2)) + 32*log(2) + 32) - 3*x/(8*log(log(2)) + 16*log(2) + 16)

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