3.6.48 \(\int \frac {-25-11 e^2}{5 e^2} \, dx\)

Optimal. Leaf size=16 \[ 8-\frac {11 x}{5}-5 \left (-3+\frac {x}{e^2}\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {8} \begin {gather*} -\frac {\left (25+11 e^2\right ) x}{5 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-25 - 11*E^2)/(5*E^2),x]

[Out]

-1/5*((25 + 11*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 (25+11 e^2\right ) x}{5 e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.75 \begin {gather*} -\frac {11 x}{5}-\frac {5 x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 11*E^2)/(5*E^2),x]

[Out]

(-11*x)/5 - (5*x)/E^2

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fricas [A]  time = 0.57, size = 13, normalized size = 0.81 \begin {gather*} -\frac {1}{5} \, {\left (11 \, x e^{2} + 25 \, x\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-11*exp(2)-25)/exp(2),x, algorithm="fricas")

[Out]

-1/5*(11*x*e^2 + 25*x)*e^(-2)

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giac [A]  time = 0.30, size = 11, normalized size = 0.69 \begin {gather*} -\frac {1}{5} \, x {\left (11 \, e^{2} + 25\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-11*exp(2)-25)/exp(2),x, algorithm="giac")

[Out]

-1/5*x*(11*e^2 + 25)*e^(-2)

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maple [A]  time = 0.02, size = 14, normalized size = 0.88

method	result	size
default	\(\frac {\left (-11 \,{\mathrm e}^{2}-25\right ) {\mathrm e}^{-2} x}{5}\)	\(14\)
norman	\(-\frac {{\mathrm e}^{-2} \left (11 \,{\mathrm e}^{2}+25\right ) x}{5}\)	\(14\)
risch	\(-\frac {11 \,{\mathrm e}^{-2} x \,{\mathrm e}^{2}}{5}-5 \,{\mathrm e}^{-2} x\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-11*exp(2)-25)/exp(2),x,method=_RETURNVERBOSE)

[Out]

1/5*(-11*exp(2)-25)/exp(2)*x

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maxima [A]  time = 0.89, size = 11, normalized size = 0.69 \begin {gather*} -\frac {1}{5} \, x {\left (11 \, e^{2} + 25\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-11*exp(2)-25)/exp(2),x, algorithm="maxima")

[Out]

-1/5*x*(11*e^2 + 25)*e^(-2)

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mupad [B]  time = 0.00, size = 11, normalized size = 0.69 \begin {gather*} -x\,{\mathrm {e}}^{-2}\,\left (\frac {11\,{\mathrm {e}}^2}{5}+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-2)*((11*exp(2))/5 + 5),x)

[Out]

-x*exp(-2)*((11*exp(2))/5 + 5)

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sympy [A]  time = 0.04, size = 14, normalized size = 0.88 \begin {gather*} \frac {x \left (- \frac {11 e^{2}}{5} - 5\right )}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-11*exp(2)-25)/exp(2),x)

[Out]

x*(-11*exp(2)/5 - 5)*exp(-2)

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