3.86.72 \(\int \frac {e^2-10 x-9 x^2}{e^2} \, dx\)

Optimal. Leaf size=16 \[ 3+x-\frac {x^2 (5+3 x)}{e^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12} \begin {gather*} -\frac {3 x^3}{e^2}-\frac {5 x^2}{e^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2 - 10*x - 9*x^2)/E^2,x]

[Out]

x - (5*x^2)/E^2 - (3*x^3)/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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \left (e^2-10 x-9 x^2\right ) \, dx}{e^2}\\ &=x-\frac {5 x^2}{e^2}-\frac {3 x^3}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.12 \begin {gather*} x-\frac {5 x^2}{e^2}-\frac {3 x^3}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2 - 10*x - 9*x^2)/E^2,x]

[Out]

x - (5*x^2)/E^2 - (3*x^3)/E^2

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fricas [A]  time = 0.82, size = 20, normalized size = 1.25 \begin {gather*} -{\left (3 \, x^{3} + 5 \, x^{2} - x e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)-9*x^2-10*x)/exp(2),x, algorithm="fricas")

[Out]

-(3*x^3 + 5*x^2 - x*e^2)*e^(-2)

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giac [A]  time = 0.13, size = 20, normalized size = 1.25 \begin {gather*} -{\left (3 \, x^{3} + 5 \, x^{2} - x e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)-9*x^2-10*x)/exp(2),x, algorithm="giac")

[Out]

-(3*x^3 + 5*x^2 - x*e^2)*e^(-2)

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maple [A]  time = 0.03, size = 17, normalized size = 1.06

method	result	size
risch	\(x -5 x^{2} {\mathrm e}^{-2}-3 \,{\mathrm e}^{-2} x^{3}\)	\(17\)
gosper	\(x \left (-3 x^{2}+{\mathrm e}^{2}-5 x \right ) {\mathrm e}^{-2}\)	\(18\)
default	\(\left (-3 x^{3}-5 x^{2}+{\mathrm e}^{2} x \right ) {\mathrm e}^{-2}\)	\(21\)
norman	\(x -5 x^{2} {\mathrm e}^{-2}-3 \,{\mathrm e}^{-2} x^{3}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2)-9*x^2-10*x)/exp(2),x,method=_RETURNVERBOSE)

[Out]

x-5*x^2*exp(-2)-3*exp(-2)*x^3

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maxima [A]  time = 0.34, size = 20, normalized size = 1.25 \begin {gather*} -{\left (3 \, x^{3} + 5 \, x^{2} - x e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)-9*x^2-10*x)/exp(2),x, algorithm="maxima")

[Out]

-(3*x^3 + 5*x^2 - x*e^2)*e^(-2)

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mupad [B]  time = 0.03, size = 18, normalized size = 1.12 \begin {gather*} -x\,{\mathrm {e}}^{-2}\,\left (3\,x^2+5\,x-{\mathrm {e}}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-2)*(10*x - exp(2) + 9*x^2),x)

[Out]

-x*exp(-2)*(5*x - exp(2) + 3*x^2)

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sympy [A]  time = 0.05, size = 17, normalized size = 1.06 \begin {gather*} - \frac {3 x^{3}}{e^{2}} - \frac {5 x^{2}}{e^{2}} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(2)-9*x**2-10*x)/exp(2),x)

[Out]

-3*x**3*exp(-2) - 5*x**2*exp(-2) + x

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