\(\int \frac {e^{-x} (e^4 (-45-15 x)+e^x (24 x-3 x^2))}{x^4} \, dx\) [6575]

Optimal result

Integrand size = 32, antiderivative size = 25 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=\frac {3+\frac {3 \left (-4+\frac {5 e^{4-x}}{x}\right )}{x}}{x} \]

[Out]

(3*(5*exp(4)/exp(x)/x-4)/x+3)/x

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6874, 37, 2228} \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=\frac {15 e^{4-x}}{x^3}-\frac {3 (8-x)^2}{16 x^2} \]

[In]

Int[(E^4*(-45 - 15*x) + E^x*(24*x - 3*x^2))/(E^x*x^4),x]

[Out]

(15*E^(4 - x))/x^3 - (3*(8 - x)^2)/(16*x^2)

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]

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
]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3 (-8+x)}{x^3}-\frac {15 e^{4-x} (3+x)}{x^4}\right ) \, dx \\ & = -\left (3 \int \frac {-8+x}{x^3} \, dx\right )-15 \int \frac {e^{4-x} (3+x)}{x^4} \, dx \\ & = \frac {15 e^{4-x}}{x^3}-\frac {3 (8-x)^2}{16 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=\frac {3 \left (5 e^{4-x}+(-4+x) x\right )}{x^3} \]

[In]

Integrate[(E^4*(-45 - 15*x) + E^x*(24*x - 3*x^2))/(E^x*x^4),x]

[Out]

(3*(5*E^(4 - x) + (-4 + x)*x))/x^3

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {3 x -12}{x^{2}}+\frac {15 \,{\mathrm e}^{-x +4}}{x^{3}}\)	\(22\)
parts	\(-\frac {12}{x^{2}}+\frac {3}{x}+\frac {15 \,{\mathrm e}^{-x} {\mathrm e}^{4}}{x^{3}}\)	\(23\)
norman	\(\frac {\left (-12 \,{\mathrm e}^{x} x +3 \,{\mathrm e}^{x} x^{2}+15 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-x}}{x^{3}}\)	\(26\)
parallelrisch	\(\frac {\left (-12 \,{\mathrm e}^{x} x +3 \,{\mathrm e}^{x} x^{2}+15 \,{\mathrm e}^{4}\right ) {\mathrm e}^{-x}}{x^{3}}\)	\(26\)
default	\(-\frac {12}{x^{2}}+\frac {3}{x}-45 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{-x}}{3 x^{3}}+\frac {{\mathrm e}^{-x}}{6 x^{2}}-\frac {{\mathrm e}^{-x}}{6 x}+\frac {\operatorname {Ei}_{1}\left (x \right )}{6}\right )-15 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{-x}}{2 x^{2}}+\frac {{\mathrm e}^{-x}}{2 x}-\frac {\operatorname {Ei}_{1}\left (x \right )}{2}\right )\)	\(77\)

[In]

int(((-3*x^2+24*x)*exp(x)+(-15*x-45)*exp(4))/exp(x)/x^4,x,method=_RETURNVERBOSE)

[Out]

(3*x-12)/x^2+15/x^3*exp(-x+4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=\frac {3 \, {\left ({\left (x^{2} - 4 \, x\right )} e^{x} + 5 \, e^{4}\right )} e^{\left (-x\right )}}{x^{3}} \]

[In]

integrate(((-3*x^2+24*x)*exp(x)+(-15*x-45)*exp(4))/exp(x)/x^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=- \frac {12 - 3 x}{x^{2}} + \frac {15 e^{4} e^{- x}}{x^{3}} \]

[In]

integrate(((-3*x**2+24*x)*exp(x)+(-15*x-45)*exp(4))/exp(x)/x**4,x)

[Out]

-(12 - 3*x)/x**2 + 15*exp(4)*exp(-x)/x**3

Maxima [C] (verification not implemented)

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

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=15 \, e^{4} \Gamma \left (-2, x\right ) + 45 \, e^{4} \Gamma \left (-3, x\right ) + \frac {3}{x} - \frac {12}{x^{2}} \]

[In]

integrate(((-3*x^2+24*x)*exp(x)+(-15*x-45)*exp(4))/exp(x)/x^4,x, algorithm="maxima")

[Out]

15*e^4*gamma(-2, x) + 45*e^4*gamma(-3, x) + 3/x - 12/x^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=\frac {3 \, {\left (x^{2} - 4 \, x + 5 \, e^{\left (-x + 4\right )}\right )}}{x^{3}} \]

[In]

integrate(((-3*x^2+24*x)*exp(x)+(-15*x-45)*exp(4))/exp(x)/x^4,x, algorithm="giac")

[Out]

3*(x^2 - 4*x + 5*e^(-x + 4))/x^3

Mupad [B] (verification not implemented)

Time = 12.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-x} \left (e^4 (-45-15 x)+e^x \left (24 x-3 x^2\right )\right )}{x^4} \, dx=\frac {15\,{\mathrm {e}}^{4-x}-12\,x+3\,x^2}{x^3} \]

[In]

int((exp(-x)*(exp(x)*(24*x - 3*x^2) - exp(4)*(15*x + 45)))/x^4,x)

[Out]

(15*exp(4 - x) - 12*x + 3*x^2)/x^3