\(\int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx\) [3268]

Optimal result

Integrand size = 19, antiderivative size = 14 \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=5+\frac {3 e^{-x} \log (x)}{x^3} \]

[Out]

5+3*ln(x)/x^3/exp(x)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {6873, 12, 2326} \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\frac {3 e^{-x} \log (x)}{x^3} \]

[In]

Int[(3 + (-9 - 3*x)*Log[x])/(E^x*x^4),x]

[Out]

(3*Log[x])/(E^x*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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6873

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\frac {3 e^{-x} \log (x)}{x^3} \]

[In]

Integrate[(3 + (-9 - 3*x)*Log[x])/(E^x*x^4),x]

[Out]

(3*Log[x])/(E^x*x^3)

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86

method	result	size
norman	\(\frac {3 \ln \left (x \right ) {\mathrm e}^{-x}}{x^{3}}\)	\(12\)
risch	\(\frac {3 \ln \left (x \right ) {\mathrm e}^{-x}}{x^{3}}\)	\(12\)
parallelrisch	\(\frac {3 \ln \left (x \right ) {\mathrm e}^{-x}}{x^{3}}\)	\(12\)

[In]

int(((-3*x-9)*ln(x)+3)/exp(x)/x^4,x,method=_RETURNVERBOSE)

[Out]

3*ln(x)/x^3/exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\frac {3 \, e^{\left (-x\right )} \log \left (x\right )}{x^{3}} \]

[In]

integrate(((-3*x-9)*log(x)+3)/exp(x)/x^4,x, algorithm="fricas")

[Out]

3*e^(-x)*log(x)/x^3

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\frac {3 e^{- x} \log {\left (x \right )}}{x^{3}} \]

[In]

integrate(((-3*x-9)*ln(x)+3)/exp(x)/x**4,x)

[Out]

3*exp(-x)*log(x)/x**3

Maxima [F]

\[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\int { -\frac {3 \, {\left ({\left (x + 3\right )} \log \left (x\right ) - 1\right )} e^{\left (-x\right )}}{x^{4}} \,d x } \]

[In]

integrate(((-3*x-9)*log(x)+3)/exp(x)/x^4,x, algorithm="maxima")

[Out]

3*e^(-x)*log(x)/x^3 - 3*gamma(-3, x) - 3*integrate(e^(-x)/x^4, x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\frac {3 \, e^{\left (-x\right )} \log \left (x\right )}{x^{3}} \]

[In]

integrate(((-3*x-9)*log(x)+3)/exp(x)/x^4,x, algorithm="giac")

[Out]

3*e^(-x)*log(x)/x^3

Mupad [B] (verification not implemented)

Time = 9.37 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-x} (3+(-9-3 x) \log (x))}{x^4} \, dx=\frac {3\,{\mathrm {e}}^{-x}\,\ln \left (x\right )}{x^3} \]

[In]

int(-(exp(-x)*(log(x)*(3*x + 9) - 3))/x^4,x)

[Out]

(3*exp(-x)*log(x))/x^3