3.41.27 $\int \frac{e^{-4+x}(15-5x)}{2x^4}dx$

Optimal. Leaf size=14 $$3-\frac{5e^{-4+x}}{2x^3}$$

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Rubi [A]  time = 0.03, antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 2, integrand size = 17, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.118, Rules used = {12, 2197} $$\begin{array}{c}-\frac{5e^{x-4}}{2x^3}\end{array}$$

Antiderivative was successfully verified.

[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 2197

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{2}\int \frac{e^{-4+x}(15-5x)}{x^4}dx\\ & =-\frac{5e^{-4+x}}{2x^3}\end{array}\end{array}$$

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Mathematica [A]  time = 0.00, size = 12, normalized size = 0.86 $$\begin{array}{c}-\frac{5e^{-4+x}}{2x^3}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*E^(-4 + x))/(2*x^3)

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fricas [A]  time = 0.57, size = 9, normalized size = 0.64 $$\begin{array}{c}-\frac{5e^{(x-4)}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*e^(x - 4)/x^3

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giac [A]  time = 0.14, size = 9, normalized size = 0.64 $$\begin{array}{c}-\frac{5e^{(x-4)}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*e^(x - 4)/x^3

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maple [A]  time = 0.09, size = 10, normalized size = 0.71

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.38, size = 19, normalized size = 1.36 $$\begin{array}{c}\frac{5}{2}{e}^{(-4)}\mathrm{\Gamma }(-2,-x)+\frac{15}{2}{e}^{(-4)}\mathrm{\Gamma }(-3,-x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.06, size = 9, normalized size = 0.64 $$\begin{array}{c}-\frac{5{\mathrm{e}}^{-4}{\mathrm{e}}^x}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.09, size = 12, normalized size = 0.86 $$\begin{array}{c}-\frac{5e^{x-4}}{2x^3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*exp(x - 4)/(2*x**3)

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