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

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} $$\begin{array}{c}\frac{1}{4}{e}^{10-2e}{(1-2e^{e-5}x)}^2\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}{e}^{10-2e}{(1-2e^{-5+e}x)}^2\end{array}\end{array}$$

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

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 $$\begin{array}{c}(x^2{e}^{(e-5)}-x)e^{(-e+5)}\end{array}$$

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 $$\begin{array}{c}(x^2{e}^{(e-5)}-x)e^{(-e+5)}\end{array}$$

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	$x(x{\mathrm{e}}^{\mathrm{e}-5}-1){\mathrm{e}}^{5-\mathrm{e}}$	$19$
default	$({\mathrm{e}}^{\mathrm{e}-5}{x}^2-x){\mathrm{e}}^{5-\mathrm{e}}$	$22$

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 $$\begin{array}{c}(x^2{e}^{(e-5)}-x)e^{(-e+5)}\end{array}$$

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 $$\begin{array}{c}\frac{{(2x-{\mathrm{e}}^{5-\mathrm{e}})}^2}{4}\end{array}$$

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 $$\begin{array}{c}{x}^2-\frac{x{e}^5}{e^e}\end{array}$$

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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