\(\int \frac {-10+e^2}{e^2} \, dx\) [493]

Optimal result

Integrand size = 9, antiderivative size = 15 \[ \int \frac {-10+e^2}{e^2} \, dx=x-\frac {10 x-\log (3)}{e^2} \]

[Out]

x-(10*x-ln(3))/exp(1)^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {8} \[ \int \frac {-10+e^2}{e^2} \, dx=-\frac {\left (10-e^2\right ) x}{e^2} \]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.53 \[ \int \frac {-10+e^2}{e^2} \, dx=x-\frac {10 x}{e^2} \]

[In]

Integrate[(-10 + E^2)/E^2,x]

[Out]

x - (10*x)/E^2

Maple [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 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	\({\mathrm e}^{-2} x \,{\mathrm e}^{2}-10 \,{\mathrm e}^{-2} x\)	\(13\)
parallelrisch	\(\left ({\mathrm e}^{2}-10\right ) {\mathrm e}^{-2} x\)	\(13\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-10+e^2}{e^2} \, dx={\left (x e^{2} - 10 \, x\right )} e^{\left (-2\right )} \]

[In]

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

[Out]

(x*e^2 - 10*x)*e^(-2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.53 \[ \int \frac {-10+e^2}{e^2} \, dx=\frac {x \left (-10 + e^{2}\right )}{e^{2}} \]

[In]

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

[Out]

x*(-10 + exp(2))*exp(-2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.53 \[ \int \frac {-10+e^2}{e^2} \, dx=x {\left (e^{2} - 10\right )} e^{\left (-2\right )} \]

[In]

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

[Out]

x*(e^2 - 10)*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.53 \[ \int \frac {-10+e^2}{e^2} \, dx=x {\left (e^{2} - 10\right )} e^{\left (-2\right )} \]

[In]

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

[Out]

x*(e^2 - 10)*e^(-2)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.53 \[ \int \frac {-10+e^2}{e^2} \, dx=x\,{\mathrm {e}}^{-2}\,\left ({\mathrm {e}}^2-10\right ) \]

[In]

int(exp(-2)*(exp(2) - 10),x)

[Out]

x*exp(-2)*(exp(2) - 10)