\(\int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx\) [2407]

Optimal result

Integrand size = 17, antiderivative size = 14 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=10+\frac {4 e^{-58-4 x}}{x^2} \]

[Out]

10+exp(ln(2)-2*x-29)^2/x^2

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} \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=\frac {4 e^{-4 x-58}}{x^2} \]

[In]

Int[(4*E^(-58 - 4*x)*(-2 - 4*x))/x^3,x]

[Out]

(4*E^(-58 - 4*x))/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 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}& = 4 \int \frac {e^{-58-4 x} (-2-4 x)}{x^3} \, dx \\ & = \frac {4 e^{-58-4 x}}{x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=\frac {4 e^{-58-4 x}}{x^2} \]

[In]

Integrate[(4*E^(-58 - 4*x)*(-2 - 4*x))/x^3,x]

[Out]

(4*E^(-58 - 4*x))/x^2

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(12\)
gosper	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)
norman	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)
parallelrisch	\(\frac {4 \,{\mathrm e}^{-4 x -58}}{x^{2}}\)	\(15\)
derivativedivides	\(-8 \left (-\ln \left (2\right )+29\right ) \left (-\frac {{\mathrm e}^{-4 x -58}}{2 x^{2}}+\frac {2 \,{\mathrm e}^{-4 x -58}}{x}-2 \,{\mathrm e}^{2 \ln \left (2\right )-58} \operatorname {Ei}_{1}\left (4 x \right )\right )+\frac {464 \,{\mathrm e}^{-4 x -58}}{x}-464 \,{\mathrm e}^{2 \ln \left (2\right )-58} \operatorname {Ei}_{1}\left (4 x \right )-\frac {112 \,{\mathrm e}^{-4 x -58}}{x^{2}}+8 \ln \left (2\right ) \left (\frac {{\mathrm e}^{-4 x -58}}{2 x^{2}}-\frac {2 \,{\mathrm e}^{-4 x -58}}{x}+2 \,{\mathrm e}^{2 \ln \left (2\right )-58} \operatorname {Ei}_{1}\left (4 x \right )\right )\)	\(148\)
default	\(-8 \left (-\ln \left (2\right )+29\right ) \left (-\frac {{\mathrm e}^{-4 x -58}}{2 x^{2}}+\frac {2 \,{\mathrm e}^{-4 x -58}}{x}-2 \,{\mathrm e}^{2 \ln \left (2\right )-58} \operatorname {Ei}_{1}\left (4 x \right )\right )+\frac {464 \,{\mathrm e}^{-4 x -58}}{x}-464 \,{\mathrm e}^{2 \ln \left (2\right )-58} \operatorname {Ei}_{1}\left (4 x \right )-\frac {112 \,{\mathrm e}^{-4 x -58}}{x^{2}}+8 \ln \left (2\right ) \left (\frac {{\mathrm e}^{-4 x -58}}{2 x^{2}}-\frac {2 \,{\mathrm e}^{-4 x -58}}{x}+2 \,{\mathrm e}^{2 \ln \left (2\right )-58} \operatorname {Ei}_{1}\left (4 x \right )\right )\)	\(148\)

[In]

int((-4*x-2)*exp(ln(2)-2*x-29)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

4*exp(-4*x-58)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=\frac {e^{\left (-4 \, x + 2 \, \log \left (2\right ) - 58\right )}}{x^{2}} \]

[In]

integrate((-4*x-2)*exp(log(2)-2*x-29)^2/x^3,x, algorithm="fricas")

[Out]

e^(-4*x + 2*log(2) - 58)/x^2

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=\frac {4 e^{- 4 x - 58}}{x^{2}} \]

[In]

integrate((-4*x-2)*exp(ln(2)-2*x-29)**2/x**3,x)

[Out]

4*exp(-4*x - 58)/x**2

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=64 \, e^{\left (-58\right )} \Gamma \left (-1, 4 \, x\right ) + 128 \, e^{\left (-58\right )} \Gamma \left (-2, 4 \, x\right ) \]

[In]

integrate((-4*x-2)*exp(log(2)-2*x-29)^2/x^3,x, algorithm="maxima")

[Out]

64*e^(-58)*gamma(-1, 4*x) + 128*e^(-58)*gamma(-2, 4*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=\frac {4 \, e^{\left (-4 \, x - 58\right )}}{x^{2}} \]

[In]

integrate((-4*x-2)*exp(log(2)-2*x-29)^2/x^3,x, algorithm="giac")

[Out]

4*e^(-4*x - 58)/x^2

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {4 e^{-58-4 x} (-2-4 x)}{x^3} \, dx=\frac {4\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-58}}{x^2} \]

[In]

int(-(exp(2*log(2) - 4*x - 58)*(4*x + 2))/x^3,x)

[Out]

(4*exp(-4*x)*exp(-58))/x^2