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

Optimal. Leaf size=17 \[ e^{\frac {1}{5} \left (-\frac {150}{x^2}-x^2\right )} \]

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Rubi [A]  time = 0.15, antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6706} \begin {gather*} e^{-\frac {x^4+150}{5 x^2}} \end {gather*}

Antiderivative was successfully verified.

[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 6706

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

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Mathematica [A]  time = 0.07, size = 15, normalized size = 0.88 \begin {gather*} e^{-\frac {30}{x^2}-\frac {x^2}{5}} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.58, size = 11, normalized size = 0.65 \begin {gather*} e^{\left (-\frac {x^{4} + 150}{5 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.12, size = 12, normalized size = 0.71 \begin {gather*} e^{\left (-\frac {1}{5} \, x^{2} - \frac {30}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.04, size = 12, normalized size = 0.71

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.43, size = 12, normalized size = 0.71 \begin {gather*} e^{\left (-\frac {1}{5} \, x^{2} - \frac {30}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.99, size = 12, normalized size = 0.71 \begin {gather*} {\mathrm {e}}^{-\frac {30}{x^2}-\frac {x^2}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.12, size = 12, normalized size = 0.71 \begin {gather*} e^{\frac {- \frac {x^{4}}{5} - 30}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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