3.55.86 \(\int \frac {72+e^{2+x}+54 x}{3 e^2} \, dx\)

Optimal. Leaf size=24 \[ -2+\frac {e^x}{3}+\frac {(4+3 x)^2}{e^2}-\log (5) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2194} \begin {gather*} \frac {9 x^2}{e^2}+\frac {24 x}{e^2}+\frac {e^x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(72 + E^(2 + x) + 54*x)/(3*E^2),x]

[Out]

E^x/3 + (24*x)/E^2 + (9*x^2)/E^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (72+e^{2+x}+54 x\right ) \, dx}{3 e^2}\\ &=\frac {24 x}{e^2}+\frac {9 x^2}{e^2}+\frac {\int e^{2+x} \, dx}{3 e^2}\\ &=\frac {e^x}{3}+\frac {24 x}{e^2}+\frac {9 x^2}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 0.88 \begin {gather*} \frac {e^{2+x}+72 x+27 x^2}{3 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(72 + E^(2 + x) + 54*x)/(3*E^2),x]

[Out]

(E^(2 + x) + 72*x + 27*x^2)/(3*E^2)

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fricas [A]  time = 0.50, size = 17, normalized size = 0.71 \begin {gather*} \frac {1}{3} \, {\left (27 \, x^{2} + 72 \, x + e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(2)*exp(x)+54*x+72)/exp(2),x, algorithm="fricas")

[Out]

1/3*(27*x^2 + 72*x + e^(x + 2))*e^(-2)

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giac [A]  time = 1.99, size = 17, normalized size = 0.71 \begin {gather*} \frac {1}{3} \, {\left (27 \, x^{2} + 72 \, x + e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(2)*exp(x)+54*x+72)/exp(2),x, algorithm="giac")

[Out]

1/3*(27*x^2 + 72*x + e^(x + 2))*e^(-2)

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maple [A]  time = 0.04, size = 18, normalized size = 0.75

method	result	size
risch	\(24 \,{\mathrm e}^{-2} x +9 x^{2} {\mathrm e}^{-2}+\frac {{\mathrm e}^{x}}{3}\)	\(18\)
default	\(\frac {{\mathrm e}^{-2} \left (72 x +{\mathrm e}^{2} {\mathrm e}^{x}+27 x^{2}\right )}{3}\)	\(21\)
norman	\(24 \,{\mathrm e}^{-2} x +9 x^{2} {\mathrm e}^{-2}+\frac {{\mathrm e}^{x}}{3}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(exp(2)*exp(x)+54*x+72)/exp(2),x,method=_RETURNVERBOSE)

[Out]

24*exp(-2)*x+9*x^2*exp(-2)+1/3*exp(x)

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maxima [A]  time = 0.37, size = 17, normalized size = 0.71 \begin {gather*} \frac {1}{3} \, {\left (27 \, x^{2} + 72 \, x + e^{\left (x + 2\right )}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(2)*exp(x)+54*x+72)/exp(2),x, algorithm="maxima")

[Out]

1/3*(27*x^2 + 72*x + e^(x + 2))*e^(-2)

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mupad [B]  time = 0.07, size = 17, normalized size = 0.71 \begin {gather*} \frac {{\mathrm {e}}^x}{3}+24\,x\,{\mathrm {e}}^{-2}+9\,x^2\,{\mathrm {e}}^{-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2)*(18*x + (exp(2)*exp(x))/3 + 24),x)

[Out]

exp(x)/3 + 24*x*exp(-2) + 9*x^2*exp(-2)

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sympy [A]  time = 0.09, size = 19, normalized size = 0.79 \begin {gather*} \frac {9 x^{2}}{e^{2}} + \frac {24 x}{e^{2}} + \frac {e^{x}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(exp(2)*exp(x)+54*x+72)/exp(2),x)

[Out]

9*x**2*exp(-2) + 24*x*exp(-2) + exp(x)/3

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