3.33.25 \(\int \frac {e^{-2-2 x^2} (-6-8 x^2+e^{6-x^2} (-30-20 x^2))}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx\)

Optimal. Leaf size=30 \[ \frac {2 e^{-2-2 x^2}}{x^2 \left (x+5 e^{6-x^2} x\right )} \]

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Rubi [F]  time = 1.36, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{-2-2 x^2} \left (-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )\right )}{x^4+25 e^{12-2 x^2} x^4+10 e^{6-x^2} x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-2 - 2*x^2)*(-6 - 8*x^2 + E^(6 - x^2)*(-30 - 20*x^2)))/(x^4 + 25*E^(12 - 2*x^2)*x^4 + 10*E^(6 - x^2)*x
^4),x]

[Out]

(2*E^(-8 - x^2))/(5*x^3) + (6*Defer[Int][1/((5*E^6 + E^x^2)*x^4), x])/(5*E^8) - (4*Defer[Int][1/((5*E^6 + E^x^
2)^2*x^2), x])/E^2 + (4*Defer[Int][1/((5*E^6 + E^x^2)*x^2), x])/(5*E^8)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )}{e^2 \left (5 e^6+e^{x^2}\right )^2 x^4} \, dx\\ &=\frac {\int \frac {-6-8 x^2+e^{6-x^2} \left (-30-20 x^2\right )}{\left (5 e^6+e^{x^2}\right )^2 x^4} \, dx}{e^2}\\ &=\frac {\int \left (-\frac {4}{\left (5 e^6+e^{x^2}\right )^2 x^2}-\frac {2 e^{-6-x^2} \left (3+2 x^2\right )}{5 x^4}+\frac {2 \left (3+2 x^2\right )}{5 e^6 \left (5 e^6+e^{x^2}\right ) x^4}\right ) \, dx}{e^2}\\ &=\frac {2 \int \frac {3+2 x^2}{\left (5 e^6+e^{x^2}\right ) x^4} \, dx}{5 e^8}-\frac {2 \int \frac {e^{-6-x^2} \left (3+2 x^2\right )}{x^4} \, dx}{5 e^2}-\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right )^2 x^2} \, dx}{e^2}\\ &=\frac {2 e^{-8-x^2}}{5 x^3}+\frac {2 \int \left (\frac {3}{\left (5 e^6+e^{x^2}\right ) x^4}+\frac {2}{\left (5 e^6+e^{x^2}\right ) x^2}\right ) \, dx}{5 e^8}-\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right )^2 x^2} \, dx}{e^2}\\ &=\frac {2 e^{-8-x^2}}{5 x^3}+\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right ) x^2} \, dx}{5 e^8}+\frac {6 \int \frac {1}{\left (5 e^6+e^{x^2}\right ) x^4} \, dx}{5 e^8}-\frac {4 \int \frac {1}{\left (5 e^6+e^{x^2}\right )^2 x^2} \, dx}{e^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 27, normalized size = 0.90 \begin {gather*} \frac {2 e^{-2-x^2}}{\left (5 e^6+e^{x^2}\right ) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 - 2*x^2)*(-6 - 8*x^2 + E^(6 - x^2)*(-30 - 20*x^2)))/(x^4 + 25*E^(12 - 2*x^2)*x^4 + 10*E^(6 -
x^2)*x^4),x]

[Out]

(2*E^(-2 - x^2))/((5*E^6 + E^x^2)*x^3)

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fricas [A]  time = 0.54, size = 32, normalized size = 1.07 \begin {gather*} \frac {2 \, e^{\left (-2 \, x^{2} + 12\right )}}{x^{3} e^{14} + 5 \, x^{3} e^{\left (-x^{2} + 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2-30)*exp(-x^2+6)-8*x^2-6)/(25*x^4*exp(-x^2+6)^2+10*x^4*exp(-x^2+6)+x^4)/exp(x^2+1)^2,x, alg
orithm="fricas")

[Out]

2*e^(-2*x^2 + 12)/(x^3*e^14 + 5*x^3*e^(-x^2 + 20))

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giac [B]  time = 0.20, size = 61, normalized size = 2.03 \begin {gather*} \frac {2 \, {\left (x^{2} e^{\left (-x^{2} + 6\right )} + 5 \, x^{2} e^{\left (-2 \, x^{2} + 12\right )}\right )}}{10 \, x^{5} e^{14} + x^{5} e^{\left (x^{2} + 8\right )} + 25 \, x^{5} e^{\left (-x^{2} + 20\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2-30)*exp(-x^2+6)-8*x^2-6)/(25*x^4*exp(-x^2+6)^2+10*x^4*exp(-x^2+6)+x^4)/exp(x^2+1)^2,x, alg
orithm="giac")

[Out]

2*(x^2*e^(-x^2 + 6) + 5*x^2*e^(-2*x^2 + 12))/(10*x^5*e^14 + x^5*e^(x^2 + 8) + 25*x^5*e^(-x^2 + 20))

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maple [A]  time = 0.18, size = 34, normalized size = 1.13

method	result	size
norman	\(\frac {2 \,{\mathrm e}^{-14} {\mathrm e}^{-2 x^{2}+12}}{x^{3} \left (1+5 \,{\mathrm e}^{-x^{2}+6}\right )}\)	\(34\)
risch	\(-\frac {2 \,{\mathrm e}^{-14}}{25 x^{3}}+\frac {2 \,{\mathrm e}^{-x^{2}-8}}{5 x^{3}}+\frac {2 \,{\mathrm e}^{-14}}{25 x^{3} \left (1+5 \,{\mathrm e}^{-x^{2}+6}\right )}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-20*x^2-30)*exp(-x^2+6)-8*x^2-6)/(25*x^4*exp(-x^2+6)^2+10*x^4*exp(-x^2+6)+x^4)/exp(x^2+1)^2,x,method=_RE
TURNVERBOSE)

[Out]

2/exp(7)^2*exp(-x^2+6)^2/x^3/(1+5*exp(-x^2+6))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left (4 \, x^{2} + 5 \, {\left (2 \, x^{2} + 3\right )} e^{\left (-x^{2} + 6\right )} + 3\right )} e^{\left (-2 \, x^{2} - 2\right )}}{10 \, x^{4} e^{\left (-x^{2} + 6\right )} + 25 \, x^{4} e^{\left (-2 \, x^{2} + 12\right )} + x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x^2-30)*exp(-x^2+6)-8*x^2-6)/(25*x^4*exp(-x^2+6)^2+10*x^4*exp(-x^2+6)+x^4)/exp(x^2+1)^2,x, alg
orithm="maxima")

[Out]

-2*integrate((4*x^2 + 5*(2*x^2 + 3)*e^(-x^2 + 6) + 3)*e^(-2*x^2 - 2)/(10*x^4*e^(-x^2 + 6) + 25*x^4*e^(-2*x^2 +
12) + x^4), x)

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mupad [B]  time = 2.02, size = 32, normalized size = 1.07 \begin {gather*} \frac {2\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{-2\,x^2}}{x^3\,{\mathrm {e}}^{14}+5\,x^3\,{\mathrm {e}}^{20}\,{\mathrm {e}}^{-x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- 2*x^2 - 2)*(8*x^2 + exp(6 - x^2)*(20*x^2 + 30) + 6))/(10*x^4*exp(6 - x^2) + 25*x^4*exp(12 - 2*x^2)
+ x^4),x)

[Out]

(2*exp(12)*exp(-2*x^2))/(x^3*exp(14) + 5*x^3*exp(20)*exp(-x^2))

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sympy [A]  time = 0.37, size = 51, normalized size = 1.70 \begin {gather*} \frac {2}{125 x^{3} e^{14} e^{6 - x^{2}} + 25 x^{3} e^{14}} + \frac {2 e^{6 - x^{2}}}{5 x^{3} e^{14}} - \frac {2}{25 x^{3} e^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-20*x**2-30)*exp(-x**2+6)-8*x**2-6)/(25*x**4*exp(-x**2+6)**2+10*x**4*exp(-x**2+6)+x**4)/exp(x**2+1
)**2,x)

[Out]

2/(125*x**3*exp(14)*exp(6 - x**2) + 25*x**3*exp(14)) + 2*exp(-14)*exp(6 - x**2)/(5*x**3) - 2*exp(-14)/(25*x**3
)

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