\(\int \frac {6 e^3-3 x}{e x^3} \, dx\) [5231]

Optimal result

Integrand size = 16, antiderivative size = 17 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=-2-\frac {3 \left (e^3-x\right )}{e x^2} \]

[Out]

-3*(-x+exp(3))/x^2/exp(1)-2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 37} \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=-\frac {3 \left (2 e^3-x\right )^2}{4 e^4 x^2} \]

[In]

Int[(6*E^3 - 3*x)/(E*x^3),x]

[Out]

(-3*(2*E^3 - x)^2)/(4*E^4*x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {6 e^3-3 x}{x^3} \, dx}{e} \\ & = -\frac {3 \left (2 e^3-x\right )^2}{4 e^4 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=\frac {-\frac {3 e^3}{x^2}+\frac {3}{x}}{e} \]

[In]

Integrate[(6*E^3 - 3*x)/(E*x^3),x]

[Out]

((-3*E^3)/x^2 + 3/x)/E

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {{\mathrm e}^{-1} \left (-3 \,{\mathrm e}^{3}+3 x \right )}{x^{2}}\)	\(15\)
gosper	\(-\frac {3 \left (-x +{\mathrm e}^{3}\right ) {\mathrm e}^{-1}}{x^{2}}\)	\(16\)
default	\(3 \,{\mathrm e}^{-1} \left (\frac {1}{x}-\frac {{\mathrm e}^{3}}{x^{2}}\right )\)	\(18\)
parallelrisch	\(-\frac {{\mathrm e}^{-1} \left (-3 x +3 \,{\mathrm e}^{3}\right )}{x^{2}}\)	\(18\)
norman	\(\frac {3 \,{\mathrm e}^{-1} x -3 \,{\mathrm e}^{-1} {\mathrm e}^{3}}{x^{2}}\)	\(21\)

[In]

int((6*exp(3)-3*x)/x^3/exp(1),x,method=_RETURNVERBOSE)

[Out]

exp(-1)*(-3*exp(3)+3*x)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=\frac {3 \, {\left (x - e^{3}\right )} e^{\left (-1\right )}}{x^{2}} \]

[In]

integrate((6*exp(3)-3*x)/x^3/exp(1),x, algorithm="fricas")

[Out]

3*(x - e^3)*e^(-1)/x^2

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=- \frac {- 3 x + 3 e^{3}}{e x^{2}} \]

[In]

integrate((6*exp(3)-3*x)/x**3/exp(1),x)

[Out]

-(-3*x + 3*exp(3))*exp(-1)/x**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=\frac {3 \, {\left (x - e^{3}\right )} e^{\left (-1\right )}}{x^{2}} \]

[In]

integrate((6*exp(3)-3*x)/x^3/exp(1),x, algorithm="maxima")

[Out]

3*(x - e^3)*e^(-1)/x^2

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=\frac {3 \, {\left (x - e^{3}\right )} e^{\left (-1\right )}}{x^{2}} \]

[In]

integrate((6*exp(3)-3*x)/x^3/exp(1),x, algorithm="giac")

[Out]

3*(x - e^3)*e^(-1)/x^2

Mupad [B] (verification not implemented)

Time = 10.67 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {6 e^3-3 x}{e x^3} \, dx=\frac {{\mathrm {e}}^{-1}\,\left (3\,x-3\,{\mathrm {e}}^3\right )}{x^2} \]

[In]

int(-(exp(-1)*(3*x - 6*exp(3)))/x^3,x)

[Out]

(exp(-1)*(3*x - 3*exp(3)))/x^2