3.8.57 \(\int \frac {1}{5} (5+(51-10 x) \log (4)) \, dx\)

Optimal. Leaf size=31 \[ x-\log (3)+\left (\left (2+e^3\right )^2-(5-x)^2+\frac {x}{5}\right ) \log (4) \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.48, number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {12} \begin {gather*} x-\frac {1}{100} (51-10 x)^2 \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + (51 - 10*x)*Log[4])/5,x]

[Out]

x - ((51 - 10*x)^2*Log[4])/100

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int (5+(51-10 x) \log (4)) \, dx\\ &=x-\frac {1}{100} (51-10 x)^2 \log (4)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.52 \begin {gather*} x+\frac {51}{5} x \log (4)-x^2 \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + (51 - 10*x)*Log[4])/5,x]

[Out]

x + (51*x*Log[4])/5 - x^2*Log[4]

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fricas [A]  time = 0.64, size = 15, normalized size = 0.48 \begin {gather*} -\frac {2}{5} \, {\left (5 \, x^{2} - 51 \, x\right )} \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*(-10*x+51)*log(2)+1,x, algorithm="fricas")

[Out]

-2/5*(5*x^2 - 51*x)*log(2) + x

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giac [A]  time = 0.31, size = 15, normalized size = 0.48 \begin {gather*} -\frac {2}{5} \, {\left (5 \, x^{2} - 51 \, x\right )} \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*(-10*x+51)*log(2)+1,x, algorithm="giac")

[Out]

-2/5*(5*x^2 - 51*x)*log(2) + x

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maple [A]  time = 0.03, size = 15, normalized size = 0.48

method	result	size
gosper	\(-\frac {x \left (10 x \ln \relax (2)-102 \ln \relax (2)-5\right )}{5}\)	\(15\)
risch	\(-2 x^{2} \ln \relax (2)+\frac {102 x \ln \relax (2)}{5}+x\)	\(15\)
default	\(\frac {2 \ln \relax (2) \left (-5 x^{2}+51 x \right )}{5}+x\)	\(16\)
norman	\(\left (\frac {102 \ln \relax (2)}{5}+1\right ) x -2 x^{2} \ln \relax (2)\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/5*(-10*x+51)*ln(2)+1,x,method=_RETURNVERBOSE)

[Out]

-1/5*x*(10*x*ln(2)-102*ln(2)-5)

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maxima [A]  time = 0.62, size = 15, normalized size = 0.48 \begin {gather*} -\frac {2}{5} \, {\left (5 \, x^{2} - 51 \, x\right )} \log \relax (2) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*(-10*x+51)*log(2)+1,x, algorithm="maxima")

[Out]

-2/5*(5*x^2 - 51*x)*log(2) + x

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mupad [B]  time = 0.56, size = 13, normalized size = 0.42 \begin {gather*} x-\frac {\ln \relax (2)\,{\left (10\,x-51\right )}^2}{50} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - (2*log(2)*(10*x - 51))/5,x)

[Out]

x - (log(2)*(10*x - 51)^2)/50

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sympy [A]  time = 0.05, size = 17, normalized size = 0.55 \begin {gather*} - 2 x^{2} \log {\relax (2 )} + x \left (1 + \frac {102 \log {\relax (2 )}}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/5*(-10*x+51)*ln(2)+1,x)

[Out]

-2*x**2*log(2) + x*(1 + 102*log(2)/5)

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