3.102.74 \(\int \frac {e^{\frac {2 x^2}{1-8 x^2+16 x^4}} (-2 x+20 x^3-112 x^5+128 x^7)}{-1+12 x^2-48 x^4+64 x^6} \, dx\)

Optimal. Leaf size=28 \[ e^{\frac {2 x^2}{\left (1-4 x^2\right )^2}} x^2-\frac {\log (4)}{e} \]

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Rubi [F]  time = 0.91, 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^{\frac {2 x^2}{1-8 x^2+16 x^4}} \left (-2 x+20 x^3-112 x^5+128 x^7\right )}{-1+12 x^2-48 x^4+64 x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*x^2)/(1 - 8*x^2 + 16*x^4))*(-2*x + 20*x^3 - 112*x^5 + 128*x^7))/(-1 + 12*x^2 - 48*x^4 + 64*x^6),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{\frac {2 x^2}{1-8 x^2+16 x^4}} x \left (1-10 x^2+56 x^4-64 x^6\right )}{\left (1-4 x^2\right )^3} \, dx\\ &=2 \int \frac {e^{\frac {2 x^2}{1-8 x^2+16 x^4}} x \left (1-10 x^2+56 x^4-64 x^6\right )}{\left (1-4 x^2\right )^3} \, dx\\ &=\operatorname {Subst}\left (\int \frac {e^{\frac {2 x}{1-8 x+16 x^2}} \left (1-10 x+56 x^2-64 x^3\right )}{(1-4 x)^3} \, dx,x,x^2\right )\\ &=\operatorname {Subst}\left (\int \frac {e^{\frac {2 x}{(1-4 x)^2}} \left (1-10 x+56 x^2-64 x^3\right )}{(1-4 x)^3} \, dx,x,x^2\right )\\ &=\operatorname {Subst}\left (\int \left (e^{\frac {2 x}{(1-4 x)^2}}-\frac {e^{\frac {2 x}{(1-4 x)^2}}}{(-1+4 x)^3}-\frac {3 e^{\frac {2 x}{(1-4 x)^2}}}{2 (-1+4 x)^2}-\frac {e^{\frac {2 x}{(1-4 x)^2}}}{2 (-1+4 x)}\right ) \, dx,x,x^2\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {e^{\frac {2 x}{(1-4 x)^2}}}{-1+4 x} \, dx,x,x^2\right )\right )-\frac {3}{2} \operatorname {Subst}\left (\int \frac {e^{\frac {2 x}{(1-4 x)^2}}}{(-1+4 x)^2} \, dx,x,x^2\right )+\operatorname {Subst}\left (\int e^{\frac {2 x}{(1-4 x)^2}} \, dx,x,x^2\right )-\operatorname {Subst}\left (\int \frac {e^{\frac {2 x}{(1-4 x)^2}}}{(-1+4 x)^3} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 20, normalized size = 0.71 \begin {gather*} e^{\frac {2 x^2}{\left (1-4 x^2\right )^2}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x^2)/(1 - 8*x^2 + 16*x^4))*(-2*x + 20*x^3 - 112*x^5 + 128*x^7))/(-1 + 12*x^2 - 48*x^4 + 64*x^
6),x]

[Out]

E^((2*x^2)/(1 - 4*x^2)^2)*x^2

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fricas [A]  time = 0.99, size = 24, normalized size = 0.86 \begin {gather*} x^{2} e^{\left (\frac {2 \, x^{2}}{16 \, x^{4} - 8 \, x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x^7-112*x^5+20*x^3-2*x)*exp(x^2/(16*x^4-8*x^2+1))^2/(64*x^6-48*x^4+12*x^2-1),x, algorithm="fric
as")

[Out]

x^2*e^(2*x^2/(16*x^4 - 8*x^2 + 1))

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giac [A]  time = 0.15, size = 24, normalized size = 0.86 \begin {gather*} x^{2} e^{\left (\frac {2 \, x^{2}}{16 \, x^{4} - 8 \, x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x^7-112*x^5+20*x^3-2*x)*exp(x^2/(16*x^4-8*x^2+1))^2/(64*x^6-48*x^4+12*x^2-1),x, algorithm="giac
")

[Out]

x^2*e^(2*x^2/(16*x^4 - 8*x^2 + 1))

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maple [A]  time = 0.22, size = 25, normalized size = 0.89

method	result	size
risch	\(x^{2} {\mathrm e}^{\frac {2 x^{2}}{\left (2 x -1\right )^{2} \left (2 x +1\right )^{2}}}\)	\(25\)
gosper	\(x^{2} {\mathrm e}^{\frac {2 x^{2}}{16 x^{4}-8 x^{2}+1}}\)	\(26\)
norman	\(\frac {x^{2} {\mathrm e}^{\frac {2 x^{2}}{16 x^{4}-8 x^{2}+1}}-8 x^{4} {\mathrm e}^{\frac {2 x^{2}}{16 x^{4}-8 x^{2}+1}}+16 x^{6} {\mathrm e}^{\frac {2 x^{2}}{16 x^{4}-8 x^{2}+1}}}{\left (4 x^{2}-1\right )^{2}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((128*x^7-112*x^5+20*x^3-2*x)*exp(x^2/(16*x^4-8*x^2+1))^2/(64*x^6-48*x^4+12*x^2-1),x,method=_RETURNVERBOSE)

[Out]

x^2*exp(2*x^2/(2*x-1)^2/(2*x+1)^2)

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maxima [A]  time = 0.43, size = 52, normalized size = 1.86 \begin {gather*} x^{2} e^{\left (\frac {1}{8 \, {\left (4 \, x^{2} + 4 \, x + 1\right )}} + \frac {1}{8 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {1}{8 \, {\left (2 \, x + 1\right )}} + \frac {1}{8 \, {\left (2 \, x - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x^7-112*x^5+20*x^3-2*x)*exp(x^2/(16*x^4-8*x^2+1))^2/(64*x^6-48*x^4+12*x^2-1),x, algorithm="maxi
ma")

[Out]

x^2*e^(1/8/(4*x^2 + 4*x + 1) + 1/8/(4*x^2 - 4*x + 1) - 1/8/(2*x + 1) + 1/8/(2*x - 1))

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mupad [B]  time = 7.03, size = 24, normalized size = 0.86 \begin {gather*} x^2\,{\mathrm {e}}^{\frac {2\,x^2}{16\,x^4-8\,x^2+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*x^2)/(16*x^4 - 8*x^2 + 1))*(2*x - 20*x^3 + 112*x^5 - 128*x^7))/(12*x^2 - 48*x^4 + 64*x^6 - 1),x)

[Out]

x^2*exp((2*x^2)/(16*x^4 - 8*x^2 + 1))

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sympy [A]  time = 0.22, size = 20, normalized size = 0.71 \begin {gather*} x^{2} e^{\frac {2 x^{2}}{16 x^{4} - 8 x^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((128*x**7-112*x**5+20*x**3-2*x)*exp(x**2/(16*x**4-8*x**2+1))**2/(64*x**6-48*x**4+12*x**2-1),x)

[Out]

x**2*exp(2*x**2/(16*x**4 - 8*x**2 + 1))

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