3.88.53 \(\int \frac {1}{2} (2-\log (5 e^x)) \, dx\) [8753]

Optimal. Leaf size=23 \[ -1+x+\frac {1}{4} \left (5+x^2\right )-\frac {1}{2} x \log \left (5 e^x\right ) \]

[Out]

1/4+x-1/2*ln(5*exp(x))*x+1/4*x^2

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Rubi [A]
time = 0.00, antiderivative size = 14, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 2188, 30} \begin {gather*} x-\frac {1}{4} \log ^2\left (5 e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - Log[5*E^x])/2,x]

[Out]

x - Log[5*E^x]^2/4

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (2-\log \left (5 e^x\right )\right ) \, dx\\ &=x-\frac {1}{2} \int \log \left (5 e^x\right ) \, dx\\ &=x-\frac {1}{2} \text {Subst}\left (\int x \, dx,x,\log \left (5 e^x\right )\right )\\ &=x-\frac {1}{4} \log ^2\left (5 e^x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 14, normalized size = 0.61 \begin {gather*} x-\frac {1}{4} \log ^2\left (5 e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - Log[5*E^x])/2,x]

[Out]

x - Log[5*E^x]^2/4

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Maple [A]
time = 1.09, size = 12, normalized size = 0.52

method	result	size
default	\(x -\frac {\ln \left (5 \,{\mathrm e}^{x}\right )^{2}}{4}\)	\(12\)
derivativedivides	\(\ln \left (5 \,{\mathrm e}^{x}\right )-\frac {\ln \left (5 \,{\mathrm e}^{x}\right )^{2}}{4}\)	\(16\)
norman	\(\ln \left (5 \,{\mathrm e}^{x}\right )-\frac {\ln \left (5 \,{\mathrm e}^{x}\right )^{2}}{4}\)	\(16\)
risch	\(-\frac {x \ln \left ({\mathrm e}^{x}\right )}{2}+\frac {x \left (x -2 \ln \left (5\right )\right )}{4}+x\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x-1/4*ln(5*exp(x))^2

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Maxima [A]
time = 0.25, size = 11, normalized size = 0.48 \begin {gather*} -\frac {1}{4} \, \log \left (5 \, e^{x}\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(5*e^x)^2 + x

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Fricas [A]
time = 0.36, size = 12, normalized size = 0.52 \begin {gather*} -\frac {1}{4} \, x^{2} - \frac {1}{2} \, x \log \left (5\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2 - 1/2*x*log(5) + x

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Sympy [A]
time = 0.02, size = 12, normalized size = 0.52 \begin {gather*} - \frac {x^{2}}{4} + x \left (1 - \frac {\log {\left (5 \right )}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2/4 + x*(1 - log(5)/2)

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Giac [A]
time = 0.41, size = 12, normalized size = 0.52 \begin {gather*} -\frac {1}{4} \, x^{2} - \frac {1}{2} \, x \log \left (5\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2 - 1/2*x*log(5) + x

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Mupad [B]
time = 5.44, size = 8, normalized size = 0.35 \begin {gather*} -\frac {x\,\left (x+\ln \left (25\right )-4\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1 - log(5*exp(x))/2,x)

[Out]

-(x*(x + log(25) - 4))/4

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