3.100.49 \(\int \frac {5+4 e^8-e^8 \log (x)}{5 e^8} \, dx\)

Optimal. Leaf size=18 \[ x+\frac {x}{e^8}+\frac {1}{5} (5-x \log (x)) \]

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.56, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {12, 2295} \begin {gather*} \frac {\left (5+4 e^8\right ) x}{5 e^8}+\frac {x}{5}-\frac {1}{5} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + 4*E^8 - E^8*Log[x])/(5*E^8),x]

[Out]

x/5 + ((5 + 4*E^8)*x)/(5*E^8) - (x*Log[x])/5

Rule 12

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (5+4 e^8-e^8 \log (x)\right ) \, dx}{5 e^8}\\ &=\frac {\left (5+4 e^8\right ) x}{5 e^8}-\frac {1}{5} \int \log (x) \, dx\\ &=\frac {x}{5}+\frac {\left (5+4 e^8\right ) x}{5 e^8}-\frac {1}{5} x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.78 \begin {gather*} x+\frac {x}{e^8}-\frac {1}{5} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 4*E^8 - E^8*Log[x])/(5*E^8),x]

[Out]

x + x/E^8 - (x*Log[x])/5

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fricas [A]  time = 0.76, size = 19, normalized size = 1.06 \begin {gather*} -\frac {1}{5} \, {\left (x e^{8} \log \relax (x) - 5 \, x e^{8} - 5 \, x\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(4)^2*log(x)+4*exp(4)^2+5)/exp(4)^2,x, algorithm="fricas")

[Out]

-1/5*(x*e^8*log(x) - 5*x*e^8 - 5*x)*e^(-8)

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giac [A]  time = 0.17, size = 24, normalized size = 1.33 \begin {gather*} -\frac {1}{5} \, {\left ({\left (x \log \relax (x) - x\right )} e^{8} - 4 \, x e^{8} - 5 \, x\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(4)^2*log(x)+4*exp(4)^2+5)/exp(4)^2,x, algorithm="giac")

[Out]

-1/5*((x*log(x) - x)*e^8 - 4*x*e^8 - 5*x)*e^(-8)

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maple [A]  time = 0.02, size = 17, normalized size = 0.94

method	result	size
risch	\(-\frac {x \ln \relax (x )}{5}+{\mathrm e}^{-8} x \,{\mathrm e}^{8}+{\mathrm e}^{-8} x\)	\(17\)
norman	\(\left (\left ({\mathrm e}^{8}+1\right ) {\mathrm e}^{-4} x -\frac {x \,{\mathrm e}^{4} \ln \relax (x )}{5}\right ) {\mathrm e}^{-4}\)	\(26\)
default	\(\frac {{\mathrm e}^{-8} \left (5 x +5 x \,{\mathrm e}^{8}-{\mathrm e}^{8} \ln \relax (x ) x \right )}{5}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-exp(4)^2*ln(x)+4*exp(4)^2+5)/exp(4)^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*x*ln(x)+exp(-8)*x*exp(8)+exp(-8)*x

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maxima [A]  time = 0.36, size = 24, normalized size = 1.33 \begin {gather*} -\frac {1}{5} \, {\left ({\left (x \log \relax (x) - x\right )} e^{8} - 4 \, x e^{8} - 5 \, x\right )} e^{\left (-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(4)^2*log(x)+4*exp(4)^2+5)/exp(4)^2,x, algorithm="maxima")

[Out]

-1/5*((x*log(x) - x)*e^8 - 4*x*e^8 - 5*x)*e^(-8)

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mupad [B]  time = 7.41, size = 13, normalized size = 0.72 \begin {gather*} \frac {x\,\left (5\,{\mathrm {e}}^{-8}-\ln \relax (x)+5\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-8)*((4*exp(8))/5 - (exp(8)*log(x))/5 + 1),x)

[Out]

(x*(5*exp(-8) - log(x) + 5))/5

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sympy [A]  time = 0.10, size = 15, normalized size = 0.83 \begin {gather*} - \frac {x \log {\relax (x )}}{5} + \frac {x \left (1 + e^{8}\right )}{e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(x)/5 + x*(1 + exp(8))*exp(-8)

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