\(\int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx\) [4027]

Optimal result

Integrand size = 17, antiderivative size = 14 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=3-\frac {5 e^{-4+x}}{2 x^3} \]

[Out]

-5/2/x^3/exp(-x+4)+3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2228} \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 e^{x-4}}{2 x^3} \]

[In]

Int[(E^(-4 + x)*(15 - 5*x))/(2*x^4),x]

[Out]

(-5*E^(-4 + x))/(2*x^3)

Rule 12

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

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-4+x} (15-5 x)}{x^4} \, dx \\ & = -\frac {5 e^{-4+x}}{2 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 e^{-4+x}}{2 x^3} \]

[In]

Integrate[(E^(-4 + x)*(15 - 5*x))/(2*x^4),x]

[Out]

(-5*E^(-4 + x))/(2*x^3)

Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(10\)
gosper	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(14\)
norman	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(14\)
parallelrisch	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(14\)
derivativedivides	\(-\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -8\right )}{12 x^{3}}+\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -14\right )}{12 x^{3}}\)	\(44\)
default	\(-\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -8\right )}{12 x^{3}}+\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -14\right )}{12 x^{3}}\)	\(44\)
meijerg	\(-\frac {15 \,{\mathrm e}^{x -16-{\mathrm e}^{-4} x} \left (\frac {{\mathrm e}^{12}}{3 x^{3}}+\frac {{\mathrm e}^{8}}{2 x^{2}}+\frac {{\mathrm e}^{4}}{2 x}+\frac {35}{36}-\frac {\ln \left (x \right )}{6}-\frac {i \pi }{6}-\frac {{\mathrm e}^{12} \left (22 x^{3} {\mathrm e}^{-12}+36 x^{2} {\mathrm e}^{-8}+36 \,{\mathrm e}^{-4} x +24\right )}{72 x^{3}}+\frac {{\mathrm e}^{12+{\mathrm e}^{-4} x} \left (4 x^{2} {\mathrm e}^{-8}+4 \,{\mathrm e}^{-4} x +8\right )}{24 x^{3}}+\frac {\ln \left (-{\mathrm e}^{-4} x \right )}{6}+\frac {\operatorname {Ei}_{1}\left (-{\mathrm e}^{-4} x \right )}{6}\right )}{2}-\frac {5 \,{\mathrm e}^{x -12-{\mathrm e}^{-4} x} \left (-\frac {{\mathrm e}^{8}}{2 x^{2}}-\frac {{\mathrm e}^{4}}{x}-\frac {11}{4}+\frac {\ln \left (x \right )}{2}+\frac {i \pi }{2}+\frac {{\mathrm e}^{8} \left (9 x^{2} {\mathrm e}^{-8}+12 \,{\mathrm e}^{-4} x +6\right )}{12 x^{2}}-\frac {{\mathrm e}^{8+{\mathrm e}^{-4} x} \left (3+3 \,{\mathrm e}^{-4} x \right )}{6 x^{2}}-\frac {\ln \left (-{\mathrm e}^{-4} x \right )}{2}-\frac {\operatorname {Ei}_{1}\left (-{\mathrm e}^{-4} x \right )}{2}\right )}{2}\)	\(207\)

[In]

int(1/2*(15-5*x)/x^4/exp(-x+4),x,method=_RETURNVERBOSE)

[Out]

-5/2/x^3*exp(x-4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 \, e^{\left (x - 4\right )}}{2 \, x^{3}} \]

[In]

integrate(1/2*(15-5*x)/x^4/exp(-x+4),x, algorithm="fricas")

[Out]

-5/2*e^(x - 4)/x^3

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=- \frac {5 e^{x - 4}}{2 x^{3}} \]

[In]

integrate(1/2*(15-5*x)/x**4/exp(-x+4),x)

[Out]

-5*exp(x - 4)/(2*x**3)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=\frac {5}{2} \, e^{\left (-4\right )} \Gamma \left (-2, -x\right ) + \frac {15}{2} \, e^{\left (-4\right )} \Gamma \left (-3, -x\right ) \]

[In]

integrate(1/2*(15-5*x)/x^4/exp(-x+4),x, algorithm="maxima")

[Out]

5/2*e^(-4)*gamma(-2, -x) + 15/2*e^(-4)*gamma(-3, -x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 \, e^{\left (x - 4\right )}}{2 \, x^{3}} \]

[In]

integrate(1/2*(15-5*x)/x^4/exp(-x+4),x, algorithm="giac")

[Out]

-5/2*e^(x - 4)/x^3

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}{2\,x^3} \]

[In]

int(-(exp(x - 4)*((5*x)/2 - 15/2))/x^4,x)

[Out]

-(5*exp(-4)*exp(x))/(2*x^3)