3.19.3 \(\int \frac {e^{\frac {-5+16 x^2+16 x^4+e^{\frac {8}{e^3}} (x^2+x^4)+e^{\frac {4}{e^3}} (8 x^2+8 x^4)}{1+x^2}} (-42 x-64 x^3-32 x^5+e^{\frac {4}{e^3}} (-16 x-32 x^3-16 x^5)+e^{\frac {8}{e^3}} (-2 x-4 x^3-2 x^5))}{1+2 x^2+x^4} \, dx\)

Optimal. Leaf size=31 \[ 5-e^{\left (4+e^{\frac {4}{e^3}}\right )^2 x^2-\frac {5}{1+x^2}} \]

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Rubi [F]  time = 6.81, 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 {\exp \left (\frac {-5+16 x^2+16 x^4+e^{\frac {8}{e^3}} \left (x^2+x^4\right )+e^{\frac {4}{e^3}} \left (8 x^2+8 x^4\right )}{1+x^2}\right ) \left (-42 x-64 x^3-32 x^5+e^{\frac {4}{e^3}} \left (-16 x-32 x^3-16 x^5\right )+e^{\frac {8}{e^3}} \left (-2 x-4 x^3-2 x^5\right )\right )}{1+2 x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-5 + 16*x^2 + 16*x^4 + E^(8/E^3)*(x^2 + x^4) + E^(4/E^3)*(8*x^2 + 8*x^4))/(1 + x^2))*(-42*x - 64*x^3
- 32*x^5 + E^(4/E^3)*(-16*x - 32*x^3 - 16*x^5) + E^(8/E^3)*(-2*x - 4*x^3 - 2*x^5)))/(1 + 2*x^2 + x^4),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-5+16 x^2+16 x^4+e^{\frac {8}{e^3}} \left (x^2+x^4\right )+e^{\frac {4}{e^3}} \left (8 x^2+8 x^4\right )}{1+x^2}\right ) \left (-42 x-64 x^3-32 x^5+e^{\frac {4}{e^3}} \left (-16 x-32 x^3-16 x^5\right )+e^{\frac {8}{e^3}} \left (-2 x-4 x^3-2 x^5\right )\right )}{\left (1+x^2\right )^2} \, dx\\ &=\int \frac {\exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2+\left (4+e^{\frac {4}{e^3}}\right )^2 x^4}{1+x^2}\right ) \left (-42 x-64 x^3-32 x^5+e^{\frac {4}{e^3}} \left (1+\frac {e^{\frac {4}{e^3}}}{8}\right ) \left (-16 x-32 x^3-16 x^5\right )\right )}{\left (1+x^2\right )^2} \, dx\\ &=\int \left (-2 \exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2+\left (4+e^{\frac {4}{e^3}}\right )^2 x^4}{1+x^2}\right ) \left (4+e^{\frac {4}{e^3}}\right )^2 x-\frac {10 \exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2+\left (4+e^{\frac {4}{e^3}}\right )^2 x^4}{1+x^2}\right ) x}{\left (1+x^2\right )^2}\right ) \, dx\\ &=-\left (10 \int \frac {\exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2+\left (4+e^{\frac {4}{e^3}}\right )^2 x^4}{1+x^2}\right ) x}{\left (1+x^2\right )^2} \, dx\right )-\left (2 \left (4+e^{\frac {4}{e^3}}\right )^2\right ) \int \exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2+\left (4+e^{\frac {4}{e^3}}\right )^2 x^4}{1+x^2}\right ) x \, dx\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {\exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2}{1+x}\right )}{(1+x)^2} \, dx,x,x^2\right )\right )-\left (4+e^{\frac {4}{e^3}}\right )^2 \operatorname {Subst}\left (\int \exp \left (\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2}{1+x}\right ) \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 44, normalized size = 1.42 \begin {gather*} -e^{\frac {-5+\left (4+e^{\frac {4}{e^3}}\right )^2 x^2+\left (4+e^{\frac {4}{e^3}}\right )^2 x^4}{1+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-5 + 16*x^2 + 16*x^4 + E^(8/E^3)*(x^2 + x^4) + E^(4/E^3)*(8*x^2 + 8*x^4))/(1 + x^2))*(-42*x - 6
4*x^3 - 32*x^5 + E^(4/E^3)*(-16*x - 32*x^3 - 16*x^5) + E^(8/E^3)*(-2*x - 4*x^3 - 2*x^5)))/(1 + 2*x^2 + x^4),x]

[Out]

-E^((-5 + (4 + E^(4/E^3))^2*x^2 + (4 + E^(4/E^3))^2*x^4)/(1 + x^2))

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fricas [A]  time = 0.72, size = 50, normalized size = 1.61 \begin {gather*} -e^{\left (\frac {16 \, x^{4} + 16 \, x^{2} + {\left (x^{4} + x^{2}\right )} e^{\left (8 \, e^{\left (-3\right )}\right )} + 8 \, {\left (x^{4} + x^{2}\right )} e^{\left (4 \, e^{\left (-3\right )}\right )} - 5}{x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^5-4*x^3-2*x)*exp(4/exp(3))^2+(-16*x^5-32*x^3-16*x)*exp(4/exp(3))-32*x^5-64*x^3-42*x)*exp(((x^
4+x^2)*exp(4/exp(3))^2+(8*x^4+8*x^2)*exp(4/exp(3))+16*x^4+16*x^2-5)/(x^2+1))/(x^4+2*x^2+1),x, algorithm="frica
s")

[Out]

-e^((16*x^4 + 16*x^2 + (x^4 + x^2)*e^(8*e^(-3)) + 8*(x^4 + x^2)*e^(4*e^(-3)) - 5)/(x^2 + 1))

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giac [B]  time = 0.91, size = 103, normalized size = 3.32 \begin {gather*} -e^{\left (\frac {x^{4} e^{\left (8 \, e^{\left (-3\right )}\right )}}{x^{2} + 1} + \frac {8 \, x^{4} e^{\left (4 \, e^{\left (-3\right )}\right )}}{x^{2} + 1} + \frac {16 \, x^{4}}{x^{2} + 1} + \frac {x^{2} e^{\left (8 \, e^{\left (-3\right )}\right )}}{x^{2} + 1} + \frac {8 \, x^{2} e^{\left (4 \, e^{\left (-3\right )}\right )}}{x^{2} + 1} + \frac {16 \, x^{2}}{x^{2} + 1} - \frac {5}{x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^5-4*x^3-2*x)*exp(4/exp(3))^2+(-16*x^5-32*x^3-16*x)*exp(4/exp(3))-32*x^5-64*x^3-42*x)*exp(((x^
4+x^2)*exp(4/exp(3))^2+(8*x^4+8*x^2)*exp(4/exp(3))+16*x^4+16*x^2-5)/(x^2+1))/(x^4+2*x^2+1),x, algorithm="giac"
)

[Out]

-e^(x^4*e^(8*e^(-3))/(x^2 + 1) + 8*x^4*e^(4*e^(-3))/(x^2 + 1) + 16*x^4/(x^2 + 1) + x^2*e^(8*e^(-3))/(x^2 + 1)
+ 8*x^2*e^(4*e^(-3))/(x^2 + 1) + 16*x^2/(x^2 + 1) - 5/(x^2 + 1))

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maple [B]  time = 0.49, size = 62, normalized size = 2.00

method	result	size
risch	\(-{\mathrm e}^{\frac {{\mathrm e}^{8 \,{\mathrm e}^{-3}} x^{4}+8 \,{\mathrm e}^{4 \,{\mathrm e}^{-3}} x^{4}+16 x^{4}+{\mathrm e}^{8 \,{\mathrm e}^{-3}} x^{2}+8 \,{\mathrm e}^{4 \,{\mathrm e}^{-3}} x^{2}+16 x^{2}-5}{x^{2}+1}}\)	\(62\)
gosper	\(-{\mathrm e}^{\frac {{\mathrm e}^{8 \,{\mathrm e}^{-3}} x^{4}+8 \,{\mathrm e}^{4 \,{\mathrm e}^{-3}} x^{4}+16 x^{4}+{\mathrm e}^{8 \,{\mathrm e}^{-3}} x^{2}+8 \,{\mathrm e}^{4 \,{\mathrm e}^{-3}} x^{2}+16 x^{2}-5}{x^{2}+1}}\)	\(74\)
norman	\(\frac {-x^{2} {\mathrm e}^{\frac {\left (x^{4}+x^{2}\right ) {\mathrm e}^{8 \,{\mathrm e}^{-3}}+\left (8 x^{4}+8 x^{2}\right ) {\mathrm e}^{4 \,{\mathrm e}^{-3}}+16 x^{4}+16 x^{2}-5}{x^{2}+1}}-{\mathrm e}^{\frac {\left (x^{4}+x^{2}\right ) {\mathrm e}^{8 \,{\mathrm e}^{-3}}+\left (8 x^{4}+8 x^{2}\right ) {\mathrm e}^{4 \,{\mathrm e}^{-3}}+16 x^{4}+16 x^{2}-5}{x^{2}+1}}}{x^{2}+1}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^5-4*x^3-2*x)*exp(4/exp(3))^2+(-16*x^5-32*x^3-16*x)*exp(4/exp(3))-32*x^5-64*x^3-42*x)*exp(((x^4+x^2)
*exp(4/exp(3))^2+(8*x^4+8*x^2)*exp(4/exp(3))+16*x^4+16*x^2-5)/(x^2+1))/(x^4+2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-exp((exp(8*exp(-3))*x^4+8*exp(4*exp(-3))*x^4+16*x^4+exp(8*exp(-3))*x^2+8*exp(4*exp(-3))*x^2+16*x^2-5)/(x^2+1)
)

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maxima [A]  time = 1.42, size = 37, normalized size = 1.19 \begin {gather*} -e^{\left (x^{2} e^{\left (8 \, e^{\left (-3\right )}\right )} + 8 \, x^{2} e^{\left (4 \, e^{\left (-3\right )}\right )} + 16 \, x^{2} - \frac {5}{x^{2} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^5-4*x^3-2*x)*exp(4/exp(3))^2+(-16*x^5-32*x^3-16*x)*exp(4/exp(3))-32*x^5-64*x^3-42*x)*exp(((x^
4+x^2)*exp(4/exp(3))^2+(8*x^4+8*x^2)*exp(4/exp(3))+16*x^4+16*x^2-5)/(x^2+1))/(x^4+2*x^2+1),x, algorithm="maxim
a")

[Out]

-e^(x^2*e^(8*e^(-3)) + 8*x^2*e^(4*e^(-3)) + 16*x^2 - 5/(x^2 + 1))

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mupad [B]  time = 1.59, size = 108, normalized size = 3.48 \begin {gather*} -{\mathrm {e}}^{\frac {16\,x^2}{x^2+1}}\,{\mathrm {e}}^{\frac {16\,x^4}{x^2+1}}\,{\mathrm {e}}^{-\frac {5}{x^2+1}}\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^{8\,{\mathrm {e}}^{-3}}}{x^2+1}}\,{\mathrm {e}}^{\frac {x^4\,{\mathrm {e}}^{8\,{\mathrm {e}}^{-3}}}{x^2+1}}\,{\mathrm {e}}^{\frac {8\,x^2\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-3}}}{x^2+1}}\,{\mathrm {e}}^{\frac {8\,x^4\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-3}}}{x^2+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(4*exp(-3))*(8*x^2 + 8*x^4) + exp(8*exp(-3))*(x^2 + x^4) + 16*x^2 + 16*x^4 - 5)/(x^2 + 1))*(42*x
+ exp(8*exp(-3))*(2*x + 4*x^3 + 2*x^5) + exp(4*exp(-3))*(16*x + 32*x^3 + 16*x^5) + 64*x^3 + 32*x^5))/(2*x^2 +
x^4 + 1),x)

[Out]

-exp((16*x^2)/(x^2 + 1))*exp((16*x^4)/(x^2 + 1))*exp(-5/(x^2 + 1))*exp((x^2*exp(8*exp(-3)))/(x^2 + 1))*exp((x^
4*exp(8*exp(-3)))/(x^2 + 1))*exp((8*x^2*exp(4*exp(-3)))/(x^2 + 1))*exp((8*x^4*exp(4*exp(-3)))/(x^2 + 1))

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sympy [B]  time = 0.40, size = 49, normalized size = 1.58 \begin {gather*} - e^{\frac {16 x^{4} + 16 x^{2} + \left (x^{4} + x^{2}\right ) e^{\frac {8}{e^{3}}} + \left (8 x^{4} + 8 x^{2}\right ) e^{\frac {4}{e^{3}}} - 5}{x^{2} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**5-4*x**3-2*x)*exp(4/exp(3))**2+(-16*x**5-32*x**3-16*x)*exp(4/exp(3))-32*x**5-64*x**3-42*x)*e
xp(((x**4+x**2)*exp(4/exp(3))**2+(8*x**4+8*x**2)*exp(4/exp(3))+16*x**4+16*x**2-5)/(x**2+1))/(x**4+2*x**2+1),x)

[Out]

-exp((16*x**4 + 16*x**2 + (x**4 + x**2)*exp(8*exp(-3)) + (8*x**4 + 8*x**2)*exp(4*exp(-3)) - 5)/(x**2 + 1))

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