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

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

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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 {gather*} -\frac {5 e^{x-4}}{2 x^3} \end {gather*}

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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

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 {gather*} -\frac {5 \, e^{\left (x - 4\right )}}{2 \, x^{3}} \end {gather*}

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 {gather*} -\frac {5 \, e^{\left (x - 4\right )}}{2 \, x^{3}} \end {gather*}

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	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(10\)
gosper	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(14\)
norman	\(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\)	\(14\)
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	\(-\frac {15 \,{\mathrm e}^{-x \,{\mathrm e}^{-4}-16+x} \left (\frac {{\mathrm e}^{12}}{3 x^{3}}+\frac {{\mathrm e}^{8}}{2 x^{2}}+\frac {{\mathrm e}^{4}}{2 x}+\frac {35}{36}-\frac {\ln \relax (x )}{6}-\frac {i \pi }{6}-\frac {{\mathrm e}^{12} \left (22 x^{3} {\mathrm e}^{-12}+36 x^{2} {\mathrm e}^{-8}+36 x \,{\mathrm e}^{-4}+24\right )}{72 x^{3}}+\frac {{\mathrm e}^{12+x \,{\mathrm e}^{-4}} \left (4 x^{2} {\mathrm e}^{-8}+4 x \,{\mathrm e}^{-4}+8\right )}{24 x^{3}}+\frac {\ln \left (-x \,{\mathrm e}^{-4}\right )}{6}+\frac {\expIntegralEi \left (1, -x \,{\mathrm e}^{-4}\right )}{6}\right )}{2}-\frac {5 \,{\mathrm e}^{x -12-x \,{\mathrm e}^{-4}} \left (-\frac {{\mathrm e}^{8}}{2 x^{2}}-\frac {{\mathrm e}^{4}}{x}-\frac {11}{4}+\frac {\ln \relax (x )}{2}+\frac {i \pi }{2}+\frac {{\mathrm e}^{8} \left (9 x^{2} {\mathrm e}^{-8}+12 x \,{\mathrm e}^{-4}+6\right )}{12 x^{2}}-\frac {{\mathrm e}^{8+x \,{\mathrm e}^{-4}} \left (3 x \,{\mathrm e}^{-4}+3\right )}{6 x^{2}}-\frac {\ln \left (-x \,{\mathrm e}^{-4}\right )}{2}-\frac {\expIntegralEi \left (1, -x \,{\mathrm e}^{-4}\right )}{2}\right )}{2}\)	\(207\)

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 {gather*} \frac {5}{2} \, e^{\left (-4\right )} \Gamma \left (-2, -x\right ) + \frac {15}{2} \, e^{\left (-4\right )} \Gamma \left (-3, -x\right ) \end {gather*}

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 {gather*} -\frac {5\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}{2\,x^3} \end {gather*}

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 {gather*} - \frac {5 e^{x - 4}}{2 x^{3}} \end {gather*}

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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