\(\int \frac {e^{\frac {-150-x^4}{5 x^2}} (300-2 x^4)}{5 x^3} \, dx\) [4147]

Optimal result

Integrand size = 30, antiderivative size = 17 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{\frac {1}{5} \left (-\frac {150}{x^2}-x^2\right )} \]

[Out]

exp(-1/5*x^2-30/x^2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6838} \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{-\frac {x^4+150}{5 x^2}} \]

[In]

Int[(E^((-150 - x^4)/(5*x^2))*(300 - 2*x^4))/(5*x^3),x]

[Out]

E^(-1/5*(150 + x^4)/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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{x^3} \, dx \\ & = e^{-\frac {150+x^4}{5 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{-\frac {30}{x^2}-\frac {x^2}{5}} \]

[In]

Integrate[(E^((-150 - x^4)/(5*x^2))*(300 - 2*x^4))/(5*x^3),x]

[Out]

E^(-30/x^2 - x^2/5)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71

method	result	size
gosper	\({\mathrm e}^{-\frac {x^{4}+150}{5 x^{2}}}\)	\(12\)
risch	\({\mathrm e}^{-\frac {x^{4}+150}{5 x^{2}}}\)	\(12\)
parallelrisch	\({\mathrm e}^{-\frac {x^{4}+150}{5 x^{2}}}\)	\(12\)
norman	\({\mathrm e}^{\frac {-x^{4}-150}{5 x^{2}}}\)	\(14\)

[In]

int(1/5*(-2*x^4+300)*exp(1/5*(-x^4-150)/x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

exp(-1/5*(x^4+150)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{\left (-\frac {x^{4} + 150}{5 \, x^{2}}\right )} \]

[In]

integrate(1/5*(-2*x^4+300)*exp(1/5*(-x^4-150)/x^2)/x^3,x, algorithm="fricas")

[Out]

e^(-1/5*(x^4 + 150)/x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{\frac {- \frac {x^{4}}{5} - 30}{x^{2}}} \]

[In]

integrate(1/5*(-2*x**4+300)*exp(1/5*(-x**4-150)/x**2)/x**3,x)

[Out]

exp((-x**4/5 - 30)/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{\left (-\frac {1}{5} \, x^{2} - \frac {30}{x^{2}}\right )} \]

[In]

integrate(1/5*(-2*x^4+300)*exp(1/5*(-x^4-150)/x^2)/x^3,x, algorithm="maxima")

[Out]

e^(-1/5*x^2 - 30/x^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx=e^{\left (-\frac {1}{5} \, x^{2} - \frac {30}{x^{2}}\right )} \]

[In]

integrate(1/5*(-2*x^4+300)*exp(1/5*(-x^4-150)/x^2)/x^3,x, algorithm="giac")

[Out]

e^(-1/5*x^2 - 30/x^2)

Mupad [B] (verification not implemented)

Time = 10.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {-150-x^4}{5 x^2}} \left (300-2 x^4\right )}{5 x^3} \, dx={\mathrm {e}}^{-\frac {30}{x^2}-\frac {x^2}{5}} \]

[In]

int(-(exp(-(x^4/5 + 30)/x^2)*(2*x^4 - 300))/(5*x^3),x)

[Out]

exp(- 30/x^2 - x^2/5)