3.98.66 \(\int \frac {-5+2 x+5 \log (4)}{5 \log (4)} \, dx\)

Optimal. Leaf size=15 \[ 1+x+\frac {(-5+x) x}{5 \log (4)} \]

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Rubi [A]  time = 0.00, antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 1, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {9} \begin {gather*} \frac {(2 x-5 (1-\log (4)))^2}{20 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 2*x + 5*Log[4])/(5*Log[4]),x]

[Out]

(2*x - 5*(1 - Log[4]))^2/(20*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-5 (1-\log (4)))^2}{20 \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.33 \begin {gather*} \frac {-5 x+x^2+5 x \log (4)}{5 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 2*x + 5*Log[4])/(5*Log[4]),x]

[Out]

(-5*x + x^2 + 5*x*Log[4])/(5*Log[4])

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fricas [A]  time = 0.82, size = 18, normalized size = 1.20 \begin {gather*} \frac {x^{2} + 10 \, x \log \relax (2) - 5 \, x}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(10*log(2)+2*x-5)/log(2),x, algorithm="fricas")

[Out]

1/10*(x^2 + 10*x*log(2) - 5*x)/log(2)

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giac [A]  time = 0.13, size = 18, normalized size = 1.20 \begin {gather*} \frac {x^{2} + 10 \, x \log \relax (2) - 5 \, x}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(10*log(2)+2*x-5)/log(2),x, algorithm="giac")

[Out]

1/10*(x^2 + 10*x*log(2) - 5*x)/log(2)

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maple [A]  time = 0.02, size = 15, normalized size = 1.00

method	result	size
gosper	\(\frac {x \left (x +10 \ln \relax (2)-5\right )}{10 \ln \relax (2)}\)	\(15\)
default	\(\frac {10 x \ln \relax (2)+x^{2}-5 x}{10 \ln \relax (2)}\)	\(19\)
risch	\(x +\frac {x^{2}}{10 \ln \relax (2)}-\frac {x}{2 \ln \relax (2)}\)	\(19\)
norman	\(\frac {x^{2}}{10 \ln \relax (2)}+\frac {\left (2 \ln \relax (2)-1\right ) x}{2 \ln \relax (2)}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(10*ln(2)+2*x-5)/ln(2),x,method=_RETURNVERBOSE)

[Out]

1/10*x*(x+10*ln(2)-5)/ln(2)

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maxima [A]  time = 0.35, size = 18, normalized size = 1.20 \begin {gather*} \frac {x^{2} + 10 \, x \log \relax (2) - 5 \, x}{10 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(10*log(2)+2*x-5)/log(2),x, algorithm="maxima")

[Out]

1/10*(x^2 + 10*x*log(2) - 5*x)/log(2)

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mupad [B]  time = 6.17, size = 15, normalized size = 1.00 \begin {gather*} \frac {5\,{\left (\frac {x}{5}+\ln \relax (2)-\frac {1}{2}\right )}^2}{2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x/5 + log(2) - 1/2)/log(2),x)

[Out]

(5*(x/5 + log(2) - 1/2)^2)/(2*log(2))

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sympy [A]  time = 0.06, size = 20, normalized size = 1.33 \begin {gather*} \frac {x^{2}}{10 \log {\relax (2 )}} + \frac {x \left (-1 + 2 \log {\relax (2 )}\right )}{2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(10*ln(2)+2*x-5)/ln(2),x)

[Out]

x**2/(10*log(2)) + x*(-1 + 2*log(2))/(2*log(2))

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