3.5.93 \(\int \frac {-10+e^2}{e^2} \, dx\)

Optimal. Leaf size=15 \[ x-\frac {10 x-\log (3)}{e^2} \]

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {8} \begin {gather*} -\frac {\left (10-e^2\right ) x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 + E^2)/E^2,x]

[Out]

-(((10 - E^2)*x)/E^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {\left (10-e^2\right ) x}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 8, normalized size = 0.53 \begin {gather*} x-\frac {10 x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^2)/E^2,x]

[Out]

x - (10*x)/E^2

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fricas [A]  time = 0.64, size = 11, normalized size = 0.73 \begin {gather*} {\left (x e^{2} - 10 \, x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)^2-10)/exp(1)^2,x, algorithm="fricas")

[Out]

(x*e^2 - 10*x)*e^(-2)

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giac [A]  time = 0.32, size = 8, normalized size = 0.53 \begin {gather*} x {\left (e^{2} - 10\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)^2-10)/exp(1)^2,x, algorithm="giac")

[Out]

x*(e^2 - 10)*e^(-2)

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maple [A]  time = 0.01, size = 13, normalized size = 0.87

method	result	size
default	\(\left ({\mathrm e}^{2}-10\right ) {\mathrm e}^{-2} x\)	\(13\)
norman	\(\left ({\mathrm e}^{2}-10\right ) {\mathrm e}^{-2} x\)	\(13\)
risch	\({\mathrm e}^{-2} x \,{\mathrm e}^{2}-10 \,{\mathrm e}^{-2} x\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)^2-10)/exp(1)^2,x,method=_RETURNVERBOSE)

[Out]

(exp(1)^2-10)/exp(1)^2*x

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maxima [A]  time = 0.70, size = 8, normalized size = 0.53 \begin {gather*} x {\left (e^{2} - 10\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)^2-10)/exp(1)^2,x, algorithm="maxima")

[Out]

x*(e^2 - 10)*e^(-2)

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mupad [B]  time = 0.00, size = 8, normalized size = 0.53 \begin {gather*} x\,{\mathrm {e}}^{-2}\,\left ({\mathrm {e}}^2-10\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2)*(exp(2) - 10),x)

[Out]

x*exp(-2)*(exp(2) - 10)

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sympy [A]  time = 0.05, size = 8, normalized size = 0.53 \begin {gather*} \frac {x \left (-10 + e^{2}\right )}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)**2-10)/exp(1)**2,x)

[Out]

x*(-10 + exp(2))*exp(-2)

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