3.24.67 \(\int \frac {-80 x^2-40 x^4-5 x^6+e^{\frac {6 x^2}{4+x^2}} (-80-40 x^2-5 x^4)+e^{\frac {3 x^2}{4+x^2}} (64-160 x-64 x^2-80 x^3+4 x^4-10 x^5+(-16+16 x^2-x^4) \log (4))}{80 x^2+40 x^4+5 x^6+e^{\frac {6 x^2}{4+x^2}} (80+40 x^2+5 x^4)+e^{\frac {3 x^2}{4+x^2}} (160 x+80 x^3+10 x^5)} \, dx\)

Optimal. Leaf size=36 \[ 4-x+\frac {4 \left (x-\frac {1}{4} x \log (4)\right )}{5 \left (e^{\frac {3 x}{\frac {4}{x}+x}}+x\right )} \]

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Rubi [F]  time = 2.94, 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 {-80 x^2-40 x^4-5 x^6+e^{\frac {6 x^2}{4+x^2}} \left (-80-40 x^2-5 x^4\right )+e^{\frac {3 x^2}{4+x^2}} \left (64-160 x-64 x^2-80 x^3+4 x^4-10 x^5+\left (-16+16 x^2-x^4\right ) \log (4)\right )}{80 x^2+40 x^4+5 x^6+e^{\frac {6 x^2}{4+x^2}} \left (80+40 x^2+5 x^4\right )+e^{\frac {3 x^2}{4+x^2}} \left (160 x+80 x^3+10 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-80*x^2 - 40*x^4 - 5*x^6 + E^((6*x^2)/(4 + x^2))*(-80 - 40*x^2 - 5*x^4) + E^((3*x^2)/(4 + x^2))*(64 - 160
*x - 64*x^2 - 80*x^3 + 4*x^4 - 10*x^5 + (-16 + 16*x^2 - x^4)*Log[4]))/(80*x^2 + 40*x^4 + 5*x^6 + E^((6*x^2)/(4
 + x^2))*(80 + 40*x^2 + 5*x^4) + E^((3*x^2)/(4 + x^2))*(160*x + 80*x^3 + 10*x^5)),x]

[Out]

-x - (12*(4 - Log[4])*Defer[Int][1/((2*I - x)*(E^((3*x^2)/(4 + x^2)) + x)^2), x])/5 - ((4 - Log[4])*Defer[Int]
[x/(E^((3*x^2)/(4 + x^2)) + x)^2, x])/5 + (12*(4 - Log[4])*Defer[Int][1/((2*I + x)*(E^((3*x^2)/(4 + x^2)) + x)
^2), x])/5 + ((4 - Log[4])*Defer[Int][(E^((3*x^2)/(4 + x^2)) + x)^(-1), x])/5 - ((6*I)/5)*(4 - Log[4])*Defer[I
nt][1/((2*I - x)*(E^((3*x^2)/(4 + x^2)) + x)), x] - ((6*I)/5)*(4 - Log[4])*Defer[Int][1/((2*I + x)*(E^((3*x^2)
/(4 + x^2)) + x)), x] - (96*(4 - Log[4])*Defer[Int][x/((E^((3*x^2)/(4 + x^2)) + x)^2*(4 + x^2)^2), x])/5 + (96
*(4 - Log[4])*Defer[Int][1/((E^((3*x^2)/(4 + x^2)) + x)*(4 + x^2)^2), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5 e^{\frac {6 x^2}{4+x^2}} \left (4+x^2\right )^2-5 x^2 \left (4+x^2\right )^2-e^{\frac {3 x^2}{4+x^2}} \left (160 x+80 x^3+10 x^5+16 (-4+\log (4))-16 x^2 (-4+\log (4))+x^4 (-4+\log (4))\right )}{5 \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {-5 e^{\frac {6 x^2}{4+x^2}} \left (4+x^2\right )^2-5 x^2 \left (4+x^2\right )^2-e^{\frac {3 x^2}{4+x^2}} \left (160 x+80 x^3+10 x^5+16 (-4+\log (4))-16 x^2 (-4+\log (4))+x^4 (-4+\log (4))\right )}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx\\ &=\frac {1}{5} \int \left (-5+\frac {x \left (16-16 x^2+x^4\right ) (-4+\log (4))}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2}-\frac {\left (16-16 x^2+x^4\right ) (-4+\log (4))}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2}\right ) \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \frac {16-16 x^2+x^4}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x \left (16-16 x^2+x^4\right )}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \left (\frac {1}{e^{\frac {3 x^2}{4+x^2}}+x}+\frac {96}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2}-\frac {24}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )}\right ) \, dx+\frac {1}{5} (-4+\log (4)) \int \left (\frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2}+\frac {96 x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2}-\frac {24 x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )}\right ) \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \frac {1}{e^{\frac {3 x^2}{4+x^2}}+x} \, dx+\frac {1}{5} (24 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )} \, dx-\frac {1}{5} (24 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )} \, dx-\frac {1}{5} (96 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx+\frac {1}{5} (96 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx\\ &=-x+\frac {1}{5} (4-\log (4)) \int \frac {1}{e^{\frac {3 x^2}{4+x^2}}+x} \, dx+\frac {1}{5} (24 (4-\log (4))) \int \left (-\frac {1}{2 (2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2}+\frac {1}{2 (2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2}\right ) \, dx-\frac {1}{5} (24 (4-\log (4))) \int \left (\frac {i}{4 (2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )}+\frac {i}{4 (2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )}\right ) \, dx-\frac {1}{5} (96 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx+\frac {1}{5} (96 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx\\ &=-x-\frac {1}{5} (6 i (4-\log (4))) \int \frac {1}{(2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )} \, dx-\frac {1}{5} (6 i (4-\log (4))) \int \frac {1}{(2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )} \, dx+\frac {1}{5} (4-\log (4)) \int \frac {1}{e^{\frac {3 x^2}{4+x^2}}+x} \, dx-\frac {1}{5} (12 (4-\log (4))) \int \frac {1}{(2 i-x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx+\frac {1}{5} (12 (4-\log (4))) \int \frac {1}{(2 i+x) \left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx-\frac {1}{5} (96 (4-\log (4))) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2 \left (4+x^2\right )^2} \, dx+\frac {1}{5} (96 (4-\log (4))) \int \frac {1}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right ) \left (4+x^2\right )^2} \, dx+\frac {1}{5} (-4+\log (4)) \int \frac {x}{\left (e^{\frac {3 x^2}{4+x^2}}+x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 32, normalized size = 0.89 \begin {gather*} \frac {1}{5} \left (-5 x-\frac {x (-4+\log (4))}{e^{3-\frac {12}{4+x^2}}+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-80*x^2 - 40*x^4 - 5*x^6 + E^((6*x^2)/(4 + x^2))*(-80 - 40*x^2 - 5*x^4) + E^((3*x^2)/(4 + x^2))*(64
 - 160*x - 64*x^2 - 80*x^3 + 4*x^4 - 10*x^5 + (-16 + 16*x^2 - x^4)*Log[4]))/(80*x^2 + 40*x^4 + 5*x^6 + E^((6*x
^2)/(4 + x^2))*(80 + 40*x^2 + 5*x^4) + E^((3*x^2)/(4 + x^2))*(160*x + 80*x^3 + 10*x^5)),x]

[Out]

(-5*x - (x*(-4 + Log[4]))/(E^(3 - 12/(4 + x^2)) + x))/5

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fricas [A]  time = 0.52, size = 49, normalized size = 1.36 \begin {gather*} -\frac {5 \, x^{2} + 5 \, x e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )} + 2 \, x \log \relax (2) - 4 \, x}{5 \, {\left (x + e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^4-40*x^2-80)*exp(3*x^2/(x^2+4))^2+(2*(-x^4+16*x^2-16)*log(2)-10*x^5+4*x^4-80*x^3-64*x^2-160*x
+64)*exp(3*x^2/(x^2+4))-5*x^6-40*x^4-80*x^2)/((5*x^4+40*x^2+80)*exp(3*x^2/(x^2+4))^2+(10*x^5+80*x^3+160*x)*exp
(3*x^2/(x^2+4))+5*x^6+40*x^4+80*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 49, normalized size = 1.36 \begin {gather*} -\frac {5 \, x^{2} + 5 \, x e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )} + 2 \, x \log \relax (2) - 4 \, x}{5 \, {\left (x + e^{\left (\frac {3 \, x^{2}}{x^{2} + 4}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^4-40*x^2-80)*exp(3*x^2/(x^2+4))^2+(2*(-x^4+16*x^2-16)*log(2)-10*x^5+4*x^4-80*x^3-64*x^2-160*x
+64)*exp(3*x^2/(x^2+4))-5*x^6-40*x^4-80*x^2)/((5*x^4+40*x^2+80)*exp(3*x^2/(x^2+4))^2+(10*x^5+80*x^3+160*x)*exp
(3*x^2/(x^2+4))+5*x^6+40*x^4+80*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.30, size = 29, normalized size = 0.81




method result size



risch \(-x -\frac {2 x \left (\ln \relax (2)-2\right )}{5 \left ({\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}+x \right )}\) \(29\)
norman \(\frac {\left (-\frac {16}{5}+\frac {8 \ln \relax (2)}{5}\right ) {\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}+\left (-\frac {4}{5}+\frac {2 \ln \relax (2)}{5}\right ) x^{2} {\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}-4 x^{2}-x^{4}-4 x \,{\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}-{\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}} x^{3}}{x^{3}+x^{2} {\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}+4 x +4 \,{\mathrm e}^{\frac {3 x^{2}}{x^{2}+4}}}\) \(131\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^4-40*x^2-80)*exp(3*x^2/(x^2+4))^2+(2*(-x^4+16*x^2-16)*ln(2)-10*x^5+4*x^4-80*x^3-64*x^2-160*x+64)*ex
p(3*x^2/(x^2+4))-5*x^6-40*x^4-80*x^2)/((5*x^4+40*x^2+80)*exp(3*x^2/(x^2+4))^2+(10*x^5+80*x^3+160*x)*exp(3*x^2/
(x^2+4))+5*x^6+40*x^4+80*x^2),x,method=_RETURNVERBOSE)

[Out]

-x-2/5*x*(ln(2)-2)/(exp(3*x^2/(x^2+4))+x)

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maxima [A]  time = 0.66, size = 48, normalized size = 1.33 \begin {gather*} -\frac {5 \, x^{2} e^{\left (\frac {12}{x^{2} + 4}\right )} + 5 \, x e^{3} - 2 \, {\left (\log \relax (2) - 2\right )} e^{3}}{5 \, {\left (x e^{\left (\frac {12}{x^{2} + 4}\right )} + e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^4-40*x^2-80)*exp(3*x^2/(x^2+4))^2+(2*(-x^4+16*x^2-16)*log(2)-10*x^5+4*x^4-80*x^3-64*x^2-160*x
+64)*exp(3*x^2/(x^2+4))-5*x^6-40*x^4-80*x^2)/((5*x^4+40*x^2+80)*exp(3*x^2/(x^2+4))^2+(10*x^5+80*x^3+160*x)*exp
(3*x^2/(x^2+4))+5*x^6+40*x^4+80*x^2),x, algorithm="maxima")

[Out]

-1/5*(5*x^2*e^(12/(x^2 + 4)) + 5*x*e^3 - 2*(log(2) - 2)*e^3)/(x*e^(12/(x^2 + 4)) + e^3)

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mupad [B]  time = 1.51, size = 38, normalized size = 1.06 \begin {gather*} \frac {x^2-\frac {x\,\left (5\,x+2\,\ln \relax (2)-4\right )}{5}}{x+{\mathrm {e}}^{\frac {3\,x^2}{x^2+4}}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((6*x^2)/(x^2 + 4))*(40*x^2 + 5*x^4 + 80) + 80*x^2 + 40*x^4 + 5*x^6 + exp((3*x^2)/(x^2 + 4))*(160*x +
 2*log(2)*(x^4 - 16*x^2 + 16) + 64*x^2 + 80*x^3 - 4*x^4 + 10*x^5 - 64))/(exp((3*x^2)/(x^2 + 4))*(160*x + 80*x^
3 + 10*x^5) + exp((6*x^2)/(x^2 + 4))*(40*x^2 + 5*x^4 + 80) + 80*x^2 + 40*x^4 + 5*x^6),x)

[Out]

(x^2 - (x*(5*x + 2*log(2) - 4))/5)/(x + exp((3*x^2)/(x^2 + 4))) - x

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sympy [A]  time = 0.25, size = 27, normalized size = 0.75 \begin {gather*} - x + \frac {- 2 x \log {\relax (2 )} + 4 x}{5 x + 5 e^{\frac {3 x^{2}}{x^{2} + 4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**4-40*x**2-80)*exp(3*x**2/(x**2+4))**2+(2*(-x**4+16*x**2-16)*ln(2)-10*x**5+4*x**4-80*x**3-64*
x**2-160*x+64)*exp(3*x**2/(x**2+4))-5*x**6-40*x**4-80*x**2)/((5*x**4+40*x**2+80)*exp(3*x**2/(x**2+4))**2+(10*x
**5+80*x**3+160*x)*exp(3*x**2/(x**2+4))+5*x**6+40*x**4+80*x**2),x)

[Out]

-x + (-2*x*log(2) + 4*x)/(5*x + 5*exp(3*x**2/(x**2 + 4)))

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