3.52.23 \(\int \frac {e^{2+\frac {e^2}{(16 x+8 x^3+x^5) \log ^2(81)}} (-4-5 x^2)}{(64 x^2+48 x^4+12 x^6+x^8) \log ^2(81)} \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {e^2}{x \left (4+x^2\right )^2 \log ^2(81)}} \]

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Rubi [A]  time = 0.73, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 6688, 6706} \begin {gather*} e^{\frac {e^2}{x \left (x^2+4\right )^2 \log ^2(81)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2 + E^2/((16*x + 8*x^3 + x^5)*Log[81]^2))*(-4 - 5*x^2))/((64*x^2 + 48*x^4 + 12*x^6 + x^8)*Log[81]^2),x
]

[Out]

E^(E^2/(x*(4 + x^2)^2*Log[81]^2))

Rule 12

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{2+\frac {e^2}{\left (16 x+8 x^3+x^5\right ) \log ^2(81)}} \left (-4-5 x^2\right )}{64 x^2+48 x^4+12 x^6+x^8} \, dx}{\log ^2(81)}\\ &=\frac {\int \frac {e^{2+\frac {e^2}{x \left (4+x^2\right )^2 \log ^2(81)}} \left (-4-5 x^2\right )}{x^2 \left (4+x^2\right )^3} \, dx}{\log ^2(81)}\\ &=e^{\frac {e^2}{x \left (4+x^2\right )^2 \log ^2(81)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.59, size = 20, normalized size = 1.00 \begin {gather*} e^{\frac {e^2}{x \left (4+x^2\right )^2 \log ^2(81)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + E^2/((16*x + 8*x^3 + x^5)*Log[81]^2))*(-4 - 5*x^2))/((64*x^2 + 48*x^4 + 12*x^6 + x^8)*Log[81
]^2),x]

[Out]

E^(E^2/(x*(4 + x^2)^2*Log[81]^2))

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fricas [B]  time = 0.50, size = 44, normalized size = 2.20 \begin {gather*} e^{\left (\frac {32 \, {\left (x^{5} + 8 \, x^{3} + 16 \, x\right )} \log \relax (3)^{2} + e^{2}}{16 \, {\left (x^{5} + 8 \, x^{3} + 16 \, x\right )} \log \relax (3)^{2}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-5*x^2-4)*exp(2)*exp(1/16*exp(2)/(x^5+8*x^3+16*x)/log(3)^2)/(x^8+12*x^6+48*x^4+64*x^2)/log(3)^
2,x, algorithm="fricas")

[Out]

e^(1/16*(32*(x^5 + 8*x^3 + 16*x)*log(3)^2 + e^2)/((x^5 + 8*x^3 + 16*x)*log(3)^2) - 2)

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giac [B]  time = 0.25, size = 140, normalized size = 7.00 \begin {gather*} e^{\left (\frac {2 \, x^{5} \log \relax (3)^{2}}{x^{5} \log \relax (3)^{2} + 8 \, x^{3} \log \relax (3)^{2} + 16 \, x \log \relax (3)^{2}} + \frac {16 \, x^{3} \log \relax (3)^{2}}{x^{5} \log \relax (3)^{2} + 8 \, x^{3} \log \relax (3)^{2} + 16 \, x \log \relax (3)^{2}} + \frac {32 \, x \log \relax (3)^{2}}{x^{5} \log \relax (3)^{2} + 8 \, x^{3} \log \relax (3)^{2} + 16 \, x \log \relax (3)^{2}} + \frac {e^{2}}{16 \, {\left (x^{5} \log \relax (3)^{2} + 8 \, x^{3} \log \relax (3)^{2} + 16 \, x \log \relax (3)^{2}\right )}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(2*x^5*log(3)^2/(x^5*log(3)^2 + 8*x^3*log(3)^2 + 16*x*log(3)^2) + 16*x^3*log(3)^2/(x^5*log(3)^2 + 8*x^3*log(
3)^2 + 16*x*log(3)^2) + 32*x*log(3)^2/(x^5*log(3)^2 + 8*x^3*log(3)^2 + 16*x*log(3)^2) + 1/16*e^2/(x^5*log(3)^2
+ 8*x^3*log(3)^2 + 16*x*log(3)^2) - 2)

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maple [A]  time = 0.35, size = 20, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{2}}{16 \left (x^{2}+4\right )^{2} \ln \relax (3)^{2} x}}\)	\(20\)
gosper	\({\mathrm e}^{\frac {{\mathrm e}^{2}}{16 x \left (x^{4}+8 x^{2}+16\right ) \ln \relax (3)^{2}}}\)	\(25\)
norman	\(\frac {x^{5} \ln \relax (3) {\mathrm e}^{\frac {{\mathrm e}^{2}}{16 \left (x^{5}+8 x^{3}+16 x \right ) \ln \relax (3)^{2}}}+16 x \ln \relax (3) {\mathrm e}^{\frac {{\mathrm e}^{2}}{16 \left (x^{5}+8 x^{3}+16 x \right ) \ln \relax (3)^{2}}}+8 x^{3} \ln \relax (3) {\mathrm e}^{\frac {{\mathrm e}^{2}}{16 \left (x^{5}+8 x^{3}+16 x \right ) \ln \relax (3)^{2}}}}{x \left (x^{2}+4\right )^{2} \ln \relax (3)}\)	\(104\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(-5*x^2-4)*exp(2)*exp(1/16*exp(2)/(x^5+8*x^3+16*x)/ln(3)^2)/(x^8+12*x^6+48*x^4+64*x^2)/ln(3)^2,x,meth
od=_RETURNVERBOSE)

[Out]

exp(1/16*exp(2)/(x^2+4)^2/ln(3)^2/x)

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maxima [B]  time = 0.62, size = 66, normalized size = 3.30 \begin {gather*} e^{\left (-\frac {x e^{2}}{64 \, {\left (x^{4} \log \relax (3)^{2} + 8 \, x^{2} \log \relax (3)^{2} + 16 \, \log \relax (3)^{2}\right )}} - \frac {x e^{2}}{256 \, {\left (x^{2} \log \relax (3)^{2} + 4 \, \log \relax (3)^{2}\right )}} + \frac {e^{2}}{256 \, x \log \relax (3)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(-5*x^2-4)*exp(2)*exp(1/16*exp(2)/(x^5+8*x^3+16*x)/log(3)^2)/(x^8+12*x^6+48*x^4+64*x^2)/log(3)^
2,x, algorithm="maxima")

[Out]

e^(-1/64*x*e^2/(x^4*log(3)^2 + 8*x^2*log(3)^2 + 16*log(3)^2) - 1/256*x*e^2/(x^2*log(3)^2 + 4*log(3)^2) + 1/256
*e^2/(x*log(3)^2))

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mupad [B]  time = 3.51, size = 32, normalized size = 1.60 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^2}{16\,\left ({\ln \relax (3)}^2\,x^5+8\,{\ln \relax (3)}^2\,x^3+16\,{\ln \relax (3)}^2\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*exp(exp(2)/(16*log(3)^2*(16*x + 8*x^3 + x^5)))*(5*x^2 + 4))/(16*log(3)^2*(64*x^2 + 48*x^4 + 12*x^
6 + x^8)),x)

[Out]

exp(exp(2)/(16*(8*x^3*log(3)^2 + x^5*log(3)^2 + 16*x*log(3)^2)))

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sympy [A]  time = 0.42, size = 22, normalized size = 1.10 \begin {gather*} e^{\frac {e^{2}}{16 \left (x^{5} + 8 x^{3} + 16 x\right ) \log {\relax (3 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(exp(2)/(16*(x**5 + 8*x**3 + 16*x)*log(3)**2))

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