3.63.35 \(\int \frac {1}{3} (5+5 e^3+5 \log (4)) \, dx\)

Optimal. Leaf size=23 \[ 5+\frac {1}{3} x \left (-5-\frac {2}{x}+5 \left (2+e^3+\log (4)\right )\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 0.52, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {8} \begin {gather*} \frac {5}{3} x \left (1+e^3+\log (4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + 5*E^3 + 5*Log[4])/3,x]

[Out]

(5*x*(1 + E^3 + Log[4]))/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {5}{3} x \left (1+e^3+\log (4)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 0.91 \begin {gather*} \frac {5 x}{3}+\frac {5 e^3 x}{3}+\frac {5}{3} x \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 5*E^3 + 5*Log[4])/3,x]

[Out]

(5*x)/3 + (5*E^3*x)/3 + (5*x*Log[4])/3

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fricas [A]  time = 0.70, size = 14, normalized size = 0.61 \begin {gather*} \frac {5}{3} \, x e^{3} + \frac {10}{3} \, x \log \relax (2) + \frac {5}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/3*log(2)+5/3*exp(3)+5/3,x, algorithm="fricas")

[Out]

5/3*x*e^3 + 10/3*x*log(2) + 5/3*x

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giac [A]  time = 0.18, size = 11, normalized size = 0.48 \begin {gather*} \frac {5}{3} \, x {\left (e^{3} + 2 \, \log \relax (2) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/3*log(2)+5/3*exp(3)+5/3,x, algorithm="giac")

[Out]

5/3*x*(e^3 + 2*log(2) + 1)

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maple [A]  time = 0.02, size = 13, normalized size = 0.57

method	result	size
default	\(\left (\frac {10 \ln \relax (2)}{3}+\frac {5 \,{\mathrm e}^{3}}{3}+\frac {5}{3}\right ) x\)	\(13\)
norman	\(\left (\frac {10 \ln \relax (2)}{3}+\frac {5 \,{\mathrm e}^{3}}{3}+\frac {5}{3}\right ) x\)	\(13\)
risch	\(\frac {10 x \ln \relax (2)}{3}+\frac {5 x \,{\mathrm e}^{3}}{3}+\frac {5 x}{3}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10/3*ln(2)+5/3*exp(3)+5/3,x,method=_RETURNVERBOSE)

[Out]

(10/3*ln(2)+5/3*exp(3)+5/3)*x

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maxima [A]  time = 0.35, size = 11, normalized size = 0.48 \begin {gather*} \frac {5}{3} \, x {\left (e^{3} + 2 \, \log \relax (2) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/3*log(2)+5/3*exp(3)+5/3,x, algorithm="maxima")

[Out]

5/3*x*(e^3 + 2*log(2) + 1)

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mupad [B]  time = 0.00, size = 12, normalized size = 0.52 \begin {gather*} x\,\left (\frac {5\,{\mathrm {e}}^3}{3}+\frac {10\,\ln \relax (2)}{3}+\frac {5}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(3))/3 + (10*log(2))/3 + 5/3,x)

[Out]

x*((5*exp(3))/3 + (10*log(2))/3 + 5/3)

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sympy [A]  time = 0.05, size = 17, normalized size = 0.74 \begin {gather*} x \left (\frac {5}{3} + \frac {10 \log {\relax (2 )}}{3} + \frac {5 e^{3}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10/3*ln(2)+5/3*exp(3)+5/3,x)

[Out]

x*(5/3 + 10*log(2)/3 + 5*exp(3)/3)

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