3.68.57 \(\int \frac {-12 e^2-24 x \log (3)}{\log (3)} \, dx\)

Optimal. Leaf size=13 \[ -12 x \left (x+\frac {e^2}{\log (3)}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 17, normalized size of antiderivative = 1.31, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {9} \begin {gather*} -\frac {3 \left (2 x \log (3)+e^2\right )^2}{\log ^2(3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12*E^2 - 24*x*Log[3])/Log[3],x]

[Out]

(-3*(E^2 + 2*x*Log[3])^2)/Log[3]^2

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[(a*(b + c*x)^2)/(2*c), x] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {3 \left (e^2+2 x \log (3)\right )^2}{\log ^2(3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.62 \begin {gather*} -\frac {12 \left (e^2 x+\frac {1}{2} x^2 \log (9)\right )}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12*E^2 - 24*x*Log[3])/Log[3],x]

[Out]

(-12*(E^2*x + (x^2*Log[9])/2))/Log[3]

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fricas [A]  time = 0.46, size = 17, normalized size = 1.31 \begin {gather*} -\frac {12 \, {\left (x^{2} \log \relax (3) + x e^{2}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*x*log(3)-12*exp(2))/log(3),x, algorithm="fricas")

[Out]

-12*(x^2*log(3) + x*e^2)/log(3)

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giac [A]  time = 0.16, size = 17, normalized size = 1.31 \begin {gather*} -\frac {12 \, {\left (x^{2} \log \relax (3) + x e^{2}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*x*log(3)-12*exp(2))/log(3),x, algorithm="giac")

[Out]

-12*(x^2*log(3) + x*e^2)/log(3)

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maple [A]  time = 0.02, size = 15, normalized size = 1.15

method	result	size
gosper	\(-\frac {12 x \left (x \ln \relax (3)+{\mathrm e}^{2}\right )}{\ln \relax (3)}\)	\(15\)
norman	\(-12 x^{2}-\frac {12 \,{\mathrm e}^{2} x}{\ln \relax (3)}\)	\(16\)
risch	\(-12 x^{2}-\frac {12 \,{\mathrm e}^{2} x}{\ln \relax (3)}\)	\(16\)
default	\(\frac {-12 x^{2} \ln \relax (3)-12 \,{\mathrm e}^{2} x}{\ln \relax (3)}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-24*x*ln(3)-12*exp(2))/ln(3),x,method=_RETURNVERBOSE)

[Out]

-12*x*(x*ln(3)+exp(2))/ln(3)

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maxima [A]  time = 0.38, size = 17, normalized size = 1.31 \begin {gather*} -\frac {12 \, {\left (x^{2} \log \relax (3) + x e^{2}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*x*log(3)-12*exp(2))/log(3),x, algorithm="maxima")

[Out]

-12*(x^2*log(3) + x*e^2)/log(3)

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mupad [B]  time = 0.08, size = 18, normalized size = 1.38 \begin {gather*} -\frac {{\left (12\,{\mathrm {e}}^2+24\,x\,\ln \relax (3)\right )}^2}{48\,{\ln \relax (3)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*exp(2) + 24*x*log(3))/log(3),x)

[Out]

-(12*exp(2) + 24*x*log(3))^2/(48*log(3)^2)

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sympy [A]  time = 0.05, size = 15, normalized size = 1.15 \begin {gather*} - 12 x^{2} - \frac {12 x e^{2}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-24*x*ln(3)-12*exp(2))/ln(3),x)

[Out]

-12*x**2 - 12*x*exp(2)/log(3)

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