3.71.95 \(\int \frac {-6 x+3 x^2+(3-2 x) \log (3)}{2 e^5} \, dx\)

Optimal. Leaf size=21 \[ -3+\frac {(3-x) x (-x+\log (3))}{2 e^5} \]

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Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.76, number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {12} \begin {gather*} \frac {x^3}{2 e^5}-\frac {3 x^2}{2 e^5}-\frac {(3-2 x)^2 \log (3)}{8 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6*x + 3*x^2 + (3 - 2*x)*Log[3])/(2*E^5),x]

[Out]

(-3*x^2)/(2*E^5) + x^3/(2*E^5) - ((3 - 2*x)^2*Log[3])/(8*E^5)

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 {\int \left (-6 x+3 x^2+(3-2 x) \log (3)\right ) \, dx}{2 e^5}\\ &=-\frac {3 x^2}{2 e^5}+\frac {x^3}{2 e^5}-\frac {(3-2 x)^2 \log (3)}{8 e^5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.24 \begin {gather*} \frac {x^3-\frac {1}{2} x^2 (6+\log (9))+x \log (27)}{2 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x + 3*x^2 + (3 - 2*x)*Log[3])/(2*E^5),x]

[Out]

(x^3 - (x^2*(6 + Log[9]))/2 + x*Log[27])/(2*E^5)

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fricas [A]  time = 0.91, size = 24, normalized size = 1.14 \begin {gather*} \frac {1}{2} \, {\left (x^{3} - 3 \, x^{2} - {\left (x^{2} - 3 \, x\right )} \log \relax (3)\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3-2*x)*log(3)+3*x^2-6*x)/exp(5),x, algorithm="fricas")

[Out]

1/2*(x^3 - 3*x^2 - (x^2 - 3*x)*log(3))*e^(-5)

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giac [A]  time = 0.16, size = 24, normalized size = 1.14 \begin {gather*} \frac {1}{2} \, {\left (x^{3} - 3 \, x^{2} - {\left (x^{2} - 3 \, x\right )} \log \relax (3)\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3-2*x)*log(3)+3*x^2-6*x)/exp(5),x, algorithm="giac")

[Out]

1/2*(x^3 - 3*x^2 - (x^2 - 3*x)*log(3))*e^(-5)

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maple [A]  time = 0.02, size = 25, normalized size = 1.19

method	result	size
gosper	\(-\frac {x \left (x \ln \relax (3)-x^{2}-3 \ln \relax (3)+3 x \right ) {\mathrm e}^{-5}}{2}\)	\(25\)
default	\(\frac {{\mathrm e}^{-5} \left (\ln \relax (3) \left (-x^{2}+3 x \right )+x^{3}-3 x^{2}\right )}{2}\)	\(28\)
risch	\(-\frac {{\mathrm e}^{-5} x^{2} \ln \relax (3)}{2}+\frac {x^{3} {\mathrm e}^{-5}}{2}+\frac {3 \,{\mathrm e}^{-5} \ln \relax (3) x}{2}-\frac {3 x^{2} {\mathrm e}^{-5}}{2}\)	\(32\)
norman	\(\frac {x^{3} {\mathrm e}^{-5}}{2}-\frac {\left (3+\ln \relax (3)\right ) {\mathrm e}^{-5} x^{2}}{2}+\frac {3 \,{\mathrm e}^{-5} \ln \relax (3) x}{2}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((3-2*x)*ln(3)+3*x^2-6*x)/exp(5),x,method=_RETURNVERBOSE)

[Out]

-1/2*x*(x*ln(3)-x^2-3*ln(3)+3*x)/exp(5)

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maxima [A]  time = 0.37, size = 24, normalized size = 1.14 \begin {gather*} \frac {1}{2} \, {\left (x^{3} - 3 \, x^{2} - {\left (x^{2} - 3 \, x\right )} \log \relax (3)\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3-2*x)*log(3)+3*x^2-6*x)/exp(5),x, algorithm="maxima")

[Out]

1/2*(x^3 - 3*x^2 - (x^2 - 3*x)*log(3))*e^(-5)

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mupad [B]  time = 0.06, size = 26, normalized size = 1.24 \begin {gather*} \frac {{\mathrm {e}}^{-5}\,x^3}{2}-\frac {{\mathrm {e}}^{-5}\,\left (\ln \relax (9)+6\right )\,x^2}{4}+\frac {3\,{\mathrm {e}}^{-5}\,\ln \relax (3)\,x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3*exp(-5))/2 - (x^2*exp(-5)*(log(9) + 6))/4 + (3*x*exp(-5)*log(3))/2

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sympy [B]  time = 0.08, size = 34, normalized size = 1.62 \begin {gather*} \frac {x^{3}}{2 e^{5}} + \frac {x^{2} \left (-3 - \log {\relax (3 )}\right )}{2 e^{5}} + \frac {3 x \log {\relax (3 )}}{2 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((3-2*x)*ln(3)+3*x**2-6*x)/exp(5),x)

[Out]

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

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