3.53.40 \(\int \frac {1}{5} e^{-e^{24}} (e^5+6 x) \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{5} e^{-e^{24}} \left (3+\frac {e^5}{x}\right ) x^2 \]

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Rubi [A]  time = 0.00, antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {9} \begin {gather*} \frac {1}{60} e^{-e^{24}} \left (6 x+e^5\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5 + 6*x)/(5*E^E^24),x]

[Out]

(E^5 + 6*x)^2/(60*E^E^24)

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 {1}{60} e^{-e^{24}} \left (e^5+6 x\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{5} e^{-e^{24}} \left (e^5 x+3 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5 + 6*x)/(5*E^E^24),x]

[Out]

(E^5*x + 3*x^2)/(5*E^E^24)

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fricas [A]  time = 0.64, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{5} \, {\left (3 \, x^{2} + x e^{5}\right )} e^{\left (-e^{24}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(3*x^2 + x*e^5)*e^(-e^24)

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giac [A]  time = 0.22, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{5} \, {\left (3 \, x^{2} + x e^{5}\right )} e^{\left (-e^{24}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(3*x^2 + x*e^5)*e^(-e^24)

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maple [A]  time = 0.05, size = 20, normalized size = 0.87

method	result	size
default	\(\frac {{\mathrm e}^{-{\mathrm e}^{24}} \left (x \,{\mathrm e}^{5}+3 x^{2}\right )}{5}\)	\(20\)
risch	\(\frac {x \,{\mathrm e}^{-{\mathrm e}^{24}+5}}{5}+\frac {3 \,{\mathrm e}^{-{\mathrm e}^{24}} x^{2}}{5}\)	\(22\)
norman	\(\frac {3 \,{\mathrm e}^{-{\mathrm e}^{24}} x^{2}}{5}+\frac {{\mathrm e}^{-{\mathrm e}^{24}} {\mathrm e}^{5} x}{5}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/exp(exp(6)^4)*(x*exp(5)+3*x^2)

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maxima [A]  time = 0.34, size = 17, normalized size = 0.74 \begin {gather*} \frac {1}{5} \, {\left (3 \, x^{2} + x e^{5}\right )} e^{\left (-e^{24}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(3*x^2 + x*e^5)*e^(-e^24)

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mupad [B]  time = 0.06, size = 14, normalized size = 0.61 \begin {gather*} \frac {x\,{\mathrm {e}}^{-{\mathrm {e}}^{24}}\,\left (3\,x+{\mathrm {e}}^5\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-exp(24))*((6*x)/5 + (x*exp(5 - log(x)))/5),x)

[Out]

(x*exp(-exp(24))*(3*x + exp(5)))/5

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sympy [A]  time = 0.05, size = 22, normalized size = 0.96 \begin {gather*} \frac {3 x^{2}}{5 e^{e^{24}}} + \frac {x e^{5}}{5 e^{e^{24}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**2*exp(-exp(24))/5 + x*exp(5)*exp(-exp(24))/5

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