3.89.71 \(\int e^{5-e} (-1+2 e^{-5+e} x) \, dx\)

Optimal. Leaf size=15 \[ 25-e^{5-e} x+x^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.53, 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}{4} e^{10-2 e} \left (1-2 e^{e-5} x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(5 - E)*(-1 + 2*E^(-5 + E)*x),x]

[Out]

(E^(10 - 2*E)*(1 - 2*E^(-5 + E)*x)^2)/4

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}{4} e^{10-2 e} \left (1-2 e^{-5+e} x\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.93 \begin {gather*} -e^{5-e} x+x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(5 - E)*(-1 + 2*E^(-5 + E)*x),x]

[Out]

-(E^(5 - E)*x) + x^2

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fricas [A]  time = 0.60, size = 21, normalized size = 1.40 \begin {gather*} {\left (x^{2} e^{\left (e - 5\right )} - x\right )} e^{\left (-e + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(exp(1)-5)-1)/exp(exp(1)-5),x, algorithm="fricas")

[Out]

(x^2*e^(e - 5) - x)*e^(-e + 5)

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giac [A]  time = 0.14, size = 21, normalized size = 1.40 \begin {gather*} {\left (x^{2} e^{\left (e - 5\right )} - x\right )} e^{\left (-e + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(exp(1)-5)-1)/exp(exp(1)-5),x, algorithm="giac")

[Out]

(x^2*e^(e - 5) - x)*e^(-e + 5)

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maple [A]  time = 0.04, size = 15, normalized size = 1.00

method	result	size
risch	\(x^{2}-x \,{\mathrm e}^{5-{\mathrm e}}\)	\(15\)
norman	\(x^{2}-{\mathrm e}^{-{\mathrm e}} {\mathrm e}^{5} x\)	\(17\)
gosper	\(x \left (x \,{\mathrm e}^{{\mathrm e}-5}-1\right ) {\mathrm e}^{5-{\mathrm e}}\)	\(19\)
default	\(\left ({\mathrm e}^{{\mathrm e}-5} x^{2}-x \right ) {\mathrm e}^{5-{\mathrm e}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(exp(1)-5)-1)/exp(exp(1)-5),x,method=_RETURNVERBOSE)

[Out]

x^2-x*exp(5-exp(1))

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maxima [A]  time = 0.37, size = 21, normalized size = 1.40 \begin {gather*} {\left (x^{2} e^{\left (e - 5\right )} - x\right )} e^{\left (-e + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(exp(1)-5)-1)/exp(exp(1)-5),x, algorithm="maxima")

[Out]

(x^2*e^(e - 5) - x)*e^(-e + 5)

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mupad [B]  time = 5.24, size = 17, normalized size = 1.13 \begin {gather*} \frac {{\left (2\,x-{\mathrm {e}}^{5-\mathrm {e}}\right )}^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(5 - exp(1))*(2*x*exp(exp(1) - 5) - 1),x)

[Out]

(2*x - exp(5 - exp(1)))^2/4

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sympy [A]  time = 0.05, size = 12, normalized size = 0.80 \begin {gather*} x^{2} - \frac {x e^{5}}{e^{e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*exp(exp(1)-5)-1)/exp(exp(1)-5),x)

[Out]

x**2 - x*exp(5)*exp(-E)

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