3.102.46 \(\int \frac {1}{5} (e^2 (-18-4 x)+6 e^2 \log (3)) \, dx\)

Optimal. Leaf size=18 \[ \frac {2}{5} e^2 x (-x+3 (-3+\log (3))) \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12} \begin {gather*} \frac {6}{5} e^2 x \log (3)-\frac {1}{10} e^2 (2 x+9)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(-18 - 4*x) + 6*E^2*Log[3])/5,x]

[Out]

-1/10*(E^2*(9 + 2*x)^2) + (6*E^2*x*Log[3])/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 {1}{5} \int \left (e^2 (-18-4 x)+6 e^2 \log (3)\right ) \, dx\\ &=-\frac {1}{10} e^2 (9+2 x)^2+\frac {6}{5} e^2 x \log (3)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.06 \begin {gather*} -\frac {2}{5} e^2 \left (9 x+x^2-3 x \log (3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(-18 - 4*x) + 6*E^2*Log[3])/5,x]

[Out]

(-2*E^2*(9*x + x^2 - 3*x*Log[3]))/5

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fricas [A]  time = 0.43, size = 19, normalized size = 1.06 \begin {gather*} \frac {6}{5} \, x e^{2} \log \relax (3) - \frac {2}{5} \, {\left (x^{2} + 9 \, x\right )} e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6/5*exp(2)*log(3)+1/5*(-4*x-18)*exp(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.31, size = 19, normalized size = 1.06 \begin {gather*} \frac {6}{5} \, x e^{2} \log \relax (3) - \frac {2}{5} \, {\left (x^{2} + 9 \, x\right )} e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6/5*exp(2)*log(3)+1/5*(-4*x-18)*exp(2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 15, normalized size = 0.83

method	result	size
gosper	\(\frac {2 x \,{\mathrm e}^{2} \left (3 \ln \relax (3)-9-x \right )}{5}\)	\(15\)
risch	\(\frac {6 x \,{\mathrm e}^{2} \ln \relax (3)}{5}-\frac {2 x^{2} {\mathrm e}^{2}}{5}-\frac {18 \,{\mathrm e}^{2} x}{5}\)	\(21\)
default	\(\frac {6 x \,{\mathrm e}^{2} \ln \relax (3)}{5}+\frac {{\mathrm e}^{2} \left (-2 x^{2}-18 x \right )}{5}\)	\(22\)
norman	\(\left (\frac {6 \,{\mathrm e}^{2} \ln \relax (3)}{5}-\frac {18 \,{\mathrm e}^{2}}{5}\right ) x -\frac {2 x^{2} {\mathrm e}^{2}}{5}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6/5*exp(2)*ln(3)+1/5*(-4*x-18)*exp(2),x,method=_RETURNVERBOSE)

[Out]

2/5*x*exp(2)*(3*ln(3)-9-x)

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maxima [A]  time = 0.35, size = 19, normalized size = 1.06 \begin {gather*} \frac {6}{5} \, x e^{2} \log \relax (3) - \frac {2}{5} \, {\left (x^{2} + 9 \, x\right )} e^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6/5*exp(2)*log(3)+1/5*(-4*x-18)*exp(2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.10, size = 18, normalized size = 1.00 \begin {gather*} -\frac {{\mathrm {e}}^2\,\left (4\,x+18\right )\,\left (4\,x-12\,\ln \relax (3)+18\right )}{40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*exp(2)*log(3))/5 - (exp(2)*(4*x + 18))/5,x)

[Out]

-(exp(2)*(4*x + 18)*(4*x - 12*log(3) + 18))/40

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sympy [A]  time = 0.06, size = 27, normalized size = 1.50 \begin {gather*} - \frac {2 x^{2} e^{2}}{5} + x \left (- \frac {18 e^{2}}{5} + \frac {6 e^{2} \log {\relax (3 )}}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(6/5*exp(2)*ln(3)+1/5*(-4*x-18)*exp(2),x)

[Out]

-2*x**2*exp(2)/5 + x*(-18*exp(2)/5 + 6*exp(2)*log(3)/5)

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