3.101.16 \(\int \frac {2 e^4+6 x \log (9)}{e^4} \, dx\)

Optimal. Leaf size=15 \[ -5+2 x+\frac {3 x^2 \log (9)}{e^4} \]

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Rubi [A]  time = 0.00, antiderivative size = 22, normalized size of antiderivative = 1.47, 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 = {9} \begin {gather*} \frac {\left (3 x \log (9)+e^4\right )^2}{3 e^4 \log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^4 + 6*x*Log[9])/E^4,x]

[Out]

(E^4 + 3*x*Log[9])^2/(3*E^4*Log[9])

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 {\left (e^4+3 x \log (9)\right )^2}{3 e^4 \log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.93 \begin {gather*} 2 x+\frac {3 x^2 \log (9)}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^4 + 6*x*Log[9])/E^4,x]

[Out]

2*x + (3*x^2*Log[9])/E^4

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fricas [A]  time = 0.59, size = 16, normalized size = 1.07 \begin {gather*} 2 \, {\left (3 \, x^{2} \log \relax (3) + x e^{4}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3*x^2*log(3) + x*e^4)*e^(-4)

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giac [A]  time = 0.16, size = 16, normalized size = 1.07 \begin {gather*} 2 \, {\left (3 \, x^{2} \log \relax (3) + x e^{4}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3*x^2*log(3) + x*e^4)*e^(-4)

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maple [A]  time = 0.02, size = 14, normalized size = 0.93

method	result	size
risch	\(6 \ln \relax (3) x^{2} {\mathrm e}^{-4}+2 x\)	\(14\)
gosper	\(2 x \left ({\mathrm e}^{4}+3 x \ln \relax (3)\right ) {\mathrm e}^{-4}\)	\(18\)
default	\({\mathrm e}^{-4} \left (6 x^{2} \ln \relax (3)+2 x \,{\mathrm e}^{4}\right )\)	\(21\)
norman	\(\left (2 \,{\mathrm e}^{2} x +6 \,{\mathrm e}^{-2} \ln \relax (3) x^{2}\right ) {\mathrm e}^{-2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x*ln(3)+2*exp(2)^2)/exp(2)^2,x,method=_RETURNVERBOSE)

[Out]

6*ln(3)*x^2*exp(-4)+2*x

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maxima [A]  time = 0.35, size = 16, normalized size = 1.07 \begin {gather*} 2 \, {\left (3 \, x^{2} \log \relax (3) + x e^{4}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(3*x^2*log(3) + x*e^4)*e^(-4)

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mupad [B]  time = 0.13, size = 18, normalized size = 1.20 \begin {gather*} \frac {{\mathrm {e}}^{-4}\,{\left ({\mathrm {e}}^4+6\,x\,\ln \relax (3)\right )}^2}{6\,\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-4)*(2*exp(4) + 12*x*log(3)),x)

[Out]

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

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sympy [A]  time = 0.05, size = 14, normalized size = 0.93 \begin {gather*} \frac {6 x^{2} \log {\relax (3 )}}{e^{4}} + 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*x**2*exp(-4)*log(3) + 2*x

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