3.13.69 \(\int \frac {8 e^3 x}{64+192 x^2+144 x^4+e^2 (1+4 x^2+4 x^4)+e^{20} (1+4 x^2+4 x^4)+e (16+56 x^2+48 x^4)+e^{10} (-16-56 x^2-48 x^4+e (-2-8 x^2-8 x^4))} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^3}{6+e-e^{10}+\frac {1}{\frac {1}{2}+x^2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 38, normalized size of antiderivative = 1.65, number of steps used = 5, number of rules used = 5, integrand size = 96, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6, 12, 1989, 28, 261} \begin {gather*} -\frac {2 e^3}{\left (6+e-e^{10}\right ) \left (2 \left (6+e-e^{10}\right ) x^2-e^{10}+e+8\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8*E^3*x)/(64 + 192*x^2 + 144*x^4 + E^2*(1 + 4*x^2 + 4*x^4) + E^20*(1 + 4*x^2 + 4*x^4) + E*(16 + 56*x^2 +
48*x^4) + E^10*(-16 - 56*x^2 - 48*x^4 + E*(-2 - 8*x^2 - 8*x^4))),x]

[Out]

(-2*E^3)/((6 + E - E^10)*(8 + E - E^10 + 2*(6 + E - E^10)*x^2))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 e^3 x}{64+192 x^2+144 x^4+\left (e^2+e^{20}\right ) \left (1+4 x^2+4 x^4\right )+e \left (16+56 x^2+48 x^4\right )+e^{10} \left (-16-56 x^2-48 x^4+e \left (-2-8 x^2-8 x^4\right )\right )} \, dx\\ &=\left (8 e^3\right ) \int \frac {x}{64+192 x^2+144 x^4+\left (e^2+e^{20}\right ) \left (1+4 x^2+4 x^4\right )+e \left (16+56 x^2+48 x^4\right )+e^{10} \left (-16-56 x^2-48 x^4+e \left (-2-8 x^2-8 x^4\right )\right )} \, dx\\ &=\left (8 e^3\right ) \int \frac {x}{\left (8+e-e^{10}\right )^2+4 \left (6+e-e^{10}\right ) \left (8+e-e^{10}\right ) x^2+4 \left (6+e-e^{10}\right )^2 x^4} \, dx\\ &=\left (32 e^3 \left (6+e-e^{10}\right )^2\right ) \int \frac {x}{\left (2 \left (6+e-e^{10}\right ) \left (8+e-e^{10}\right )+4 \left (6+e-e^{10}\right )^2 x^2\right )^2} \, dx\\ &=-\frac {2 e^3}{\left (6+e-e^{10}\right ) \left (8+e-e^{10}+2 \left (6+e-e^{10}\right ) x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.91 \begin {gather*} -\frac {2 e^3}{\left (-6-e+e^{10}\right ) \left (-8-e+e^{10}-12 x^2-2 e x^2+2 e^{10} x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8*E^3*x)/(64 + 192*x^2 + 144*x^4 + E^2*(1 + 4*x^2 + 4*x^4) + E^20*(1 + 4*x^2 + 4*x^4) + E*(16 + 56*
x^2 + 48*x^4) + E^10*(-16 - 56*x^2 - 48*x^4 + E*(-2 - 8*x^2 - 8*x^4))),x]

[Out]

(-2*E^3)/((-6 - E + E^10)*(-8 - E + E^10 - 12*x^2 - 2*E*x^2 + 2*E^10*x^2))

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fricas [B]  time = 0.59, size = 66, normalized size = 2.87 \begin {gather*} -\frac {2 \, e^{3}}{72 \, x^{2} + {\left (2 \, x^{2} + 1\right )} e^{20} - 2 \, {\left (2 \, x^{2} + 1\right )} e^{11} - 2 \, {\left (12 \, x^{2} + 7\right )} e^{10} + {\left (2 \, x^{2} + 1\right )} e^{2} + 2 \, {\left (12 \, x^{2} + 7\right )} e + 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x*exp(3)/((4*x^4+4*x^2+1)*exp(5)^4+((-8*x^4-8*x^2-2)*exp(1)-48*x^4-56*x^2-16)*exp(5)^2+(4*x^4+4*x^
2+1)*exp(1)^2+(48*x^4+56*x^2+16)*exp(1)+144*x^4+192*x^2+64),x, algorithm="fricas")

[Out]

-2*e^3/(72*x^2 + (2*x^2 + 1)*e^20 - 2*(2*x^2 + 1)*e^11 - 2*(12*x^2 + 7)*e^10 + (2*x^2 + 1)*e^2 + 2*(12*x^2 + 7
)*e + 48)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x*exp(3)/((4*x^4+4*x^2+1)*exp(5)^4+((-8*x^4-8*x^2-2)*exp(1)-48*x^4-56*x^2-16)*exp(5)^2+(4*x^4+4*x^
2+1)*exp(1)^2+(48*x^4+56*x^2+16)*exp(1)+144*x^4+192*x^2+64),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.19, size = 45, normalized size = 1.96

method	result	size
risch	\(-\frac {{\mathrm e}^{3}}{\left ({\mathrm e}^{10}-{\mathrm e}-6\right ) \left (x^{2} {\mathrm e}^{10}+\frac {{\mathrm e}^{10}}{2}-x^{2} {\mathrm e}-6 x^{2}-\frac {{\mathrm e}}{2}-4\right )}\)	\(45\)
gosper	\(-\frac {2 \,{\mathrm e}^{3}}{\left (2 x^{2} {\mathrm e}^{10}-2 x^{2} {\mathrm e}+{\mathrm e}^{10}-12 x^{2}-{\mathrm e}-8\right ) \left ({\mathrm e}^{10}-{\mathrm e}-6\right )}\)	\(50\)
norman	\(\frac {4 \,{\mathrm e}^{3} x^{2}}{\left ({\mathrm e}^{10}-{\mathrm e}-8\right ) \left (2 x^{2} {\mathrm e}^{10}-2 x^{2} {\mathrm e}+{\mathrm e}^{10}-12 x^{2}-{\mathrm e}-8\right )}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x*exp(3)/((4*x^4+4*x^2+1)*exp(5)^4+((-8*x^4-8*x^2-2)*exp(1)-48*x^4-56*x^2-16)*exp(5)^2+(4*x^4+4*x^2+1)*e
xp(1)^2+(48*x^4+56*x^2+16)*exp(1)+144*x^4+192*x^2+64),x,method=_RETURNVERBOSE)

[Out]

-exp(3)/(exp(10)-exp(1)-6)/(x^2*exp(10)+1/2*exp(10)-x^2*exp(1)-6*x^2-1/2*exp(1)-4)

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maxima [A]  time = 0.59, size = 47, normalized size = 2.04 \begin {gather*} -\frac {2 \, e^{3}}{2 \, x^{2} {\left (e^{20} - 2 \, e^{11} - 12 \, e^{10} + e^{2} + 12 \, e + 36\right )} + e^{20} - 2 \, e^{11} - 14 \, e^{10} + e^{2} + 14 \, e + 48} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x*exp(3)/((4*x^4+4*x^2+1)*exp(5)^4+((-8*x^4-8*x^2-2)*exp(1)-48*x^4-56*x^2-16)*exp(5)^2+(4*x^4+4*x^
2+1)*exp(1)^2+(48*x^4+56*x^2+16)*exp(1)+144*x^4+192*x^2+64),x, algorithm="maxima")

[Out]

-2*e^3/(2*x^2*(e^20 - 2*e^11 - 12*e^10 + e^2 + 12*e + 36) + e^20 - 2*e^11 - 14*e^10 + e^2 + 14*e + 48)

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mupad [B]  time = 0.18, size = 38, normalized size = 1.65 \begin {gather*} -\frac {2\,{\mathrm {e}}^3}{\left (\mathrm {e}-{\mathrm {e}}^{10}+6\right )\,\left (\left (2\,\mathrm {e}-2\,{\mathrm {e}}^{10}+12\right )\,x^2+\mathrm {e}-{\mathrm {e}}^{10}+8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x*exp(3))/(exp(2)*(4*x^2 + 4*x^4 + 1) - exp(10)*(exp(1)*(8*x^2 + 8*x^4 + 2) + 56*x^2 + 48*x^4 + 16) + e
xp(20)*(4*x^2 + 4*x^4 + 1) + exp(1)*(56*x^2 + 48*x^4 + 16) + 192*x^2 + 144*x^4 + 64),x)

[Out]

-(2*exp(3))/((exp(1) - exp(10) + 6)*(exp(1) - exp(10) + x^2*(2*exp(1) - 2*exp(10) + 12) + 8))

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sympy [B]  time = 0.75, size = 63, normalized size = 2.74 \begin {gather*} - \frac {8 e^{3}}{x^{2} \left (- 96 e^{10} - 16 e^{11} + 8 e^{2} + 96 e + 288 + 8 e^{20}\right ) - 56 e^{10} - 8 e^{11} + 4 e^{2} + 56 e + 192 + 4 e^{20}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x*exp(3)/((4*x**4+4*x**2+1)*exp(5)**4+((-8*x**4-8*x**2-2)*exp(1)-48*x**4-56*x**2-16)*exp(5)**2+(4*
x**4+4*x**2+1)*exp(1)**2+(48*x**4+56*x**2+16)*exp(1)+144*x**4+192*x**2+64),x)

[Out]

-8*exp(3)/(x**2*(-96*exp(10) - 16*exp(11) + 8*exp(2) + 96*E + 288 + 8*exp(20)) - 56*exp(10) - 8*exp(11) + 4*ex
p(2) + 56*E + 192 + 4*exp(20))

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