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

Optimal. Leaf size=18 \[ 4+\frac {e^{-5+x}}{3}+x^2-\log (4) \]

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 0.72, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2194} \begin {gather*} x^2+\frac {e^{x-5}}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-5 + x)/3 + 2*x,x]

[Out]

E^(-5 + x)/3 + x^2

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} &=x^2+\frac {1}{3} \int e^{-5+x} \, dx\\ &=\frac {e^{-5+x}}{3}+x^2\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 13, normalized size = 0.72 \begin {gather*} \frac {e^{-5+x}}{3}+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-5 + x)/3 + 2*x,x]

[Out]

E^(-5 + x)/3 + x^2

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 12, normalized size = 0.67 \begin {gather*} x^{2} + e^{\left (x - \log \relax (3) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + e^(x - log(3) - 5)

________________________________________________________________________________________

giac [A]  time = 0.14, size = 14, normalized size = 0.78 \begin {gather*} x^{2} + e^{\left (x - \log \left (3 \, e^{5}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + e^(x - log(3*e^5))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 11, normalized size = 0.61

method	result	size
risch	\(\frac {{\mathrm e}^{x -5}}{3}+x^{2}\)	\(11\)
default	\(x^{2}+{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{5}\right )+x}\)	\(15\)
norman	\(x^{2}+{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{5}\right )+x}\)	\(15\)
derivativedivides	\(\left (-\ln \left (3 \,{\mathrm e}^{5}\right )+x \right )^{2}+2 \ln \left (3 \,{\mathrm e}^{5}\right ) \left (-\ln \left (3 \,{\mathrm e}^{5}\right )+x \right )+{\mathrm e}^{-\ln \left (3 \,{\mathrm e}^{5}\right )+x}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 10, normalized size = 0.56 \begin {gather*} x^{2} + \frac {1}{3} \, e^{\left (x - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + 1/3*e^(x - 5)

________________________________________________________________________________________

mupad [B]  time = 3.23, size = 10, normalized size = 0.56 \begin {gather*} \frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^x}{3}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + exp(x - log(3*exp(5))),x)

[Out]

(exp(-5)*exp(x))/3 + x^2

________________________________________________________________________________________

sympy [A]  time = 0.09, size = 10, normalized size = 0.56 \begin {gather*} x^{2} + \frac {e^{x}}{3 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2 + exp(-5)*exp(x)/3

________________________________________________________________________________________