3.17.70 \(\int \frac {1}{25} (40-2 x-10 \log (2)) \, dx\)

Optimal. Leaf size=17 \[ \frac {11}{3}-\left (-4+\frac {x}{5}+\log (2)\right )^2 \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 0.94, 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 {1}{25} (x-5 (4-\log (2)))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40 - 2*x - 10*Log[2])/25,x]

[Out]

-1/25*(x - 5*(4 - Log[2]))^2

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 {1}{25} (x-5 (4-\log (2)))^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.12 \begin {gather*} -\frac {2}{25} \left (-20 x+\frac {x^2}{2}+x \log (32)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40 - 2*x - 10*Log[2])/25,x]

[Out]

(-2*(-20*x + x^2/2 + x*Log[32]))/25

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fricas [A]  time = 0.77, size = 14, normalized size = 0.82 \begin {gather*} -\frac {1}{25} \, x^{2} - \frac {2}{5} \, x \log \relax (2) + \frac {8}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*log(2)-2/25*x+8/5,x, algorithm="fricas")

[Out]

-1/25*x^2 - 2/5*x*log(2) + 8/5*x

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giac [A]  time = 0.29, size = 14, normalized size = 0.82 \begin {gather*} -\frac {1}{25} \, x^{2} - \frac {2}{5} \, x \log \relax (2) + \frac {8}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*log(2)-2/25*x+8/5,x, algorithm="giac")

[Out]

-1/25*x^2 - 2/5*x*log(2) + 8/5*x

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maple [A]  time = 0.02, size = 11, normalized size = 0.65

method	result	size
gosper	\(-\frac {x \left (x +10 \ln \relax (2)-40\right )}{25}\)	\(11\)
default	\(-\frac {2 x \ln \relax (2)}{5}-\frac {x^{2}}{25}+\frac {8 x}{5}\)	\(15\)
norman	\(\left (-\frac {2 \ln \relax (2)}{5}+\frac {8}{5}\right ) x -\frac {x^{2}}{25}\)	\(15\)
risch	\(-\frac {2 x \ln \relax (2)}{5}-\frac {x^{2}}{25}+\frac {8 x}{5}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2/5*ln(2)-2/25*x+8/5,x,method=_RETURNVERBOSE)

[Out]

-1/25*x*(x+10*ln(2)-40)

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maxima [A]  time = 0.37, size = 14, normalized size = 0.82 \begin {gather*} -\frac {1}{25} \, x^{2} - \frac {2}{5} \, x \log \relax (2) + \frac {8}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*log(2)-2/25*x+8/5,x, algorithm="maxima")

[Out]

-1/25*x^2 - 2/5*x*log(2) + 8/5*x

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mupad [B]  time = 0.04, size = 10, normalized size = 0.59 \begin {gather*} -\frac {x\,\left (x+10\,\ln \relax (2)-40\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8/5 - (2*log(2))/5 - (2*x)/25,x)

[Out]

-(x*(x + 10*log(2) - 40))/25

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sympy [A]  time = 0.06, size = 15, normalized size = 0.88 \begin {gather*} - \frac {x^{2}}{25} + x \left (\frac {8}{5} - \frac {2 \log {\relax (2 )}}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2/5*ln(2)-2/25*x+8/5,x)

[Out]

-x**2/25 + x*(8/5 - 2*log(2)/5)

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